Challenging Probability Question

[tony1234]

Hi. I have a reasonably difficult probability question that I would really like some help with. If you don’t have some background in probability theory, you probably needn’t read farther.

If you do have some background in probability theory, I won’t waste your time explaining the basis for my interest. Rest assured, however, I am not just trying to get an answer for a homework problem set. I am really interested in the question, but don’t possess the training to answer it. Perhaps it will be easy for you.

The problem is below. Again, any help would be sincerely appreciated.

Many thanks,

Tony
tony12345678@yahoo.com

Question:

In Game 1, a fair coin is flipped repeatedly until it comes up heads. If it comes up heads after n tosses, you are awarded 3^n dollars.

In Game 2, a fair coin is flipped repeatedly until it comes up heads. If it comes up heads after n tosses, you are awarded 2*(3^n) dollars.

Game 1 and Game 2 are played repeatedly an equal number of times. As the number of plays increases, will the ratio of the average winnings of Game 1 and average winnings of Game 2 converge? If so, what will the ratio converge on? 1? ½? 0?

(The question appears nontrivial to me because, for both games, the expected utility of a single play is infinite.)

[Triskadecamus]

Pardon me,

Is there some arcane mathematical reason that doubling the payoff doesn't, oh, say, double the payoff?

Two times three to the nth is twice the magnitude of three to the nth, for all of the values of n which I have tried.

Am I missing something?

Tris

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Imagine my signature begins five spaces to the right of center.

[Zor]

Looks like 1/2 to me. I'll show you my reasoning if you tell me that's right :D

Chances are 1/3 that I'll be correct. Anyone else want to chip in and cover the other 2/3? :D

Note to tony1234: It's almost 3AM here, so let me know if it isn't 1/2 and I'll try something when I'm awake tomorrow. Good night...

[Andy]

This game's a con....but i'll still play :).

[AWB]

I assume that two separate fair coins are used. If so, the ratio of winnings will converge to 1:2.

If the two games are based on the same coin, the winnings ratio will be exactly 1:2.

Sound like great games. The player doesn't pay anything, and every once in a while he gets money. Count me in.

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Wrong thinking is punished, right thinking is just as swiftly rewarded. You'll find it an effective combination.

[AWB]

I just tried this game empirically. The ratio jumped all over the place. One game or the other would have a big streak of luck (several tails before a head came up), and its total winnings would jump so much that the ratio would be skewed for quite a while, even so much that the second game would be behind the first for quite a while.

So theoretically the ratio is 1:2. But in the realm of a few thousand experiments, the ratio jumped around from 1:5 to 6:1 (!).

I'll let it run longer as see where it goes.

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Wrong thinking is punished, right thinking is just as swiftly rewarded. You'll find it an effective combination.

[AWB]

Weird! I just ran it several times for 20,000,000 coin tosses, and the ratio seems to converge to around 0.309; that is, a little less than 1:3. Here's my program if anyone can read Pascal and see if my logic's faulty somewhere.
<BLOCKQUOTE><font size="1" face="Verdana, Arial">code:</font><HR><pre>
PROGRAM c3;

USES crt;

VAR coin1, coin2:integer;
pot1,pot2:real;
mp1,mp2:longint;
count:longint;

BEGIN
randomize;
pot1:=0; pot2:=0;
mp1:=1; mp2:=1;
count:=0;
REPEAT
count:=count+1;
coin1:=random(2); coin2:=random(2); { generates 0 or 1; 0 will be heads }
mp1:=mp1*3; mp2:=mp2*3;
{ mini-pot for streak of tails (Pascal doesn't do exponentation.) }
IF coin1=0 THEN BEGIN
pot1:=pot1+mp1;
mp1:=1;
END;
IF coin2=0 THEN BEGIN
pot2:=pot2+2*mp2;
mp2:=1;
END;
IF pot2&gt;0 THEN { If pot2 is non-zero (this prevents divide by 0 error) }
IF count MOD 10000 = 0 THEN { every 10000 coin flips show game status }
writeln(count:10,pot1:10,pot2:10,pot1/pot2:20:18);
UNTIL keypressed;
END.
[/code]

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Wrong thinking is punished, right thinking is just as swiftly rewarded. You'll find it an effective combination.

[AWB]

Geez, now I run it one more time, and the ratio is 0.80 (4:5).

I guess if either game gets a good jump ahead, even after a few million coin flips the ratio will still be skewed.



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Wrong thinking is punished, right thinking is just as swiftly rewarded. You'll find it an effective combination.

[ticker]

Originally posted by AWB:
Weird! I just ran it several times for 20,000,000 coin tosses, and the ratio seems to converge to around 0.309; that is, a little less than 1:3. Here's my program if anyone can read Pascal and see if my logic's faulty somewhere.
<BLOCKQUOTE><font size="1" face="Verdana, Arial">code:</font><HR><pre>
PROGRAM c3;

USES crt;

VAR coin1, coin2:integer;
pot1,pot2:real;
mp1,mp2:longint;
count:longint;

BEGIN
randomize;
pot1:=0; pot2:=0;
mp1:=1; mp2:=1;
count:=0;
REPEAT
count:=count+1;
coin1:=random(2); coin2:=random(2); { generates 0 or 1; 0 will be heads }
mp1:=mp1*3; mp2:=mp2*3;
{ mini-pot for streak of tails (Pascal doesn't do exponentation.) }
IF coin1=0 THEN BEGIN
pot1:=pot1+mp1;
mp1:=1;
END;
IF coin2=0 THEN BEGIN
pot2:=pot2+2*mp2;
mp2:=1;
END;
IF pot2&gt;0 THEN { If pot2 is non-zero (this prevents divide by 0 error) }
IF count MOD 10000 = 0 THEN { every 10000 coin flips show game status }
writeln(count:10,pot1:10,pot2:10,pot1/pot2:20:18);
UNTIL keypressed;
END.
[/code]


In the OP a game is defined as a series of tosses until heads and an equal number of games are played. In your code you have an equal number of tosses, not games.

[C K Dexter Haven]

IIRC, this is called the "St. Petersburg Paradox", and the first recorded mention of it was in the early 1700's in an article written by Nicolaus Bernoulli. The paradox is that the "expected value" of each game is infinite (unbounded.)

Expected value is the product of the likelihood (probability) of an outcome, times the amount paid upon that outcome.

SO, for Game 1:
Prob of Head on toss 1 = 1/2
Payout = $3, so 3/2 is expected outcome

Prob of tail on toss 1, head on toss 2 = 1/4, Payout = $9, so expected outcome = (3/2)^2

Prob of tail on toss 1,2; head on toss 3 = 1/8, Payout is $27, so expected outcome = (3/2)^3.

Etc.

Prob of tails on tosses 1 through n, head on toss n = (1/2)^n; payout is 3^n; so expected outcome is (3/2)^n.

The expected outcome for the game is the SUM of all those probabilities... which is an infinitely huge (unbounded) number. This is contrary to common sense, in some ways, but that's the way the math works.

That is to say, the chance of getting 20 tails in a row, then heads, is very slim ... (like, around 1 in a million) but the PAYOUT in that case is enormous, over $3 billion. If you have nothing else to do with your life, however, but play this game... sooner or later (like maybe after a million tosses) you will get that string of 20 tails with the big payout.

Since the expected values of both games is infinite, I am not sure what the "ratio" of two infinities would be. I would like to see what AWB gets when he fixes the program... but frankly, I would not be surprised if the ratio of winnings between the two games DOESN'T converge.

As I say, it's called a paradox. The simplistic view would be that, if heads comes up on the nth toss in Game 1 and Game 2, you get twice as much in Game 2, so the ratio must be 1/2. But I think the math runs deeper, in this case.

[Cabbage]

This seems to have some similarities to another game I've heard of. Two people, A and B, take a fair coin, and begin flipping the coin. At the end of each flip, A wins if the total heads count is greater than the total tails count, B wins if tails is greater than heads, tie if they're equal.

At first glance, it looks like the outcome would be that A and B remain fairly even. What tends to happen, however, is that there's a streak of heads (or tails) that is essentially "insurmountable". The ratio of the 2 counts will go to 1/2, of course, but one of the counts will tend to always be greater than the other.

Just a passing thought--those types of streaks, and the large numbers we're dealing with, seem to imply to me that this may actually not tend to converge.

[tony1234]

Thanks for your thoughts. In fact, I have tried simulating the problem on computer. I never ran over 160,000 plays, but, aat that level, the results were inconclusive. The problem obviously impicates very large numbers and it is difficult to say with confidence based on any empirical inquiry that the result *won't* converge.

I too suspect that the ration of winnings will not converge. Nevertheless, just because the average winnings from Game 1 and Game 2 will not coverge doesn't mean their ratio won't. So I still am interested in finding out the answer, preferably via a proof.

Finally, I throw out an additional question: Is the ratio does not converge, so that in the long run you cannot expect to acheive higher average winnings from either game, is there a rational reason to prefer one play of Game 2 to one play of Game 1?

Tony
tony12345678@yahoo.com

[Zor]

Did a little looking, and the best answer is that the ratio of winnings does converge to 1/2, but only so at infinity. Basically, that means before you reach infinity, you're going to keep getting conflicting results (I'll take Game #2 over Game #1 any time of the day though). Not very helpful? Well, that's why it's called a paradox I suppose :D The moral is that in practical situations, you can't assume that you can play an infinite number of games, and you must set a cut off point somewhere.

Perhaps this (http://www.mathpages.com/home/kmath200.htm) will be of some interest to you on a related matter...

[RM Mentock]

Originally posted by tony1234:
Finally, I throw out an additional question: Is the ratio does not converge, so that in the long run you cannot expect to acheive higher average winnings from either game, is there a rational reason to prefer one play of Game 2 to one play of Game 1?

Boy, that's a tough one. I would choose Game 2, because you always get twice the payout you would have gotten if you'd been playing Game 1.

------------------
rocks

[Cabbage]

Here's my idea for a rough proof; if anyone sees a flaw, let me know. (Sorry for the kind of bad notation, but I don't have much to work with).

Game 1: Change the game so that the game is over after n flips (for some fixed n). Then the expected payout is

S(n) = (sum from i=1 to n) (3/2)^i
.
.
.
.
Game 2: Same thing, for some fixed m. Expected payout is

D(m) = (sum from j=1 to m) 2 * (3/2)^j
.
.
.
Take the limit of S(n)/D(m) as n and m go to infinity--the limit doesn't exist, since m and n are not necessarily set to be equal at all times (i.e., they don't necessarily grow at the same rate).

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...ebius sig. This is a moebius sig. This is a mo...
(sig line courtesy of WallyM7)

[Zor]

...m and n are not necessarily set to be equal at all times...

I'm not so sure that will work, Cabbage. If you flip both coins at the same time, n=m is guranteed, and it's just the intervals between payoffs differing, right?

[Konrad]

There's no paradox. In math you deal with infinite series all the time. The ratio converges to 2:1.

Nice and intuitive, eh?

If you want proof it's just the ratio of the sums of all the payoffs times the probability of that payoff. (ie A sum of a sum). Since the probabilities for both games are the same you can just ignore that and take the ratio of 2*(3^n) to 3^n which is 2 for any n.

[Konrad]

Take the limit of S(n)/D(m) as n and m go to infinity--the limit doesn't exist, since m and n are not necessarily set to be equal at all times (i.e., they don't necessarily grow at the same rate).

They grow at the same rate if you treat them as a probability of payoff. Since this is a statistical problem that's all you're interested in anyway.

[Zor]

Sorry about the wording there. Was thinking about the Petersburg paradox... The answer for the OP is just 1/2 as stated...

[jcgmoi]

OK guys, I'm of 2 minds here. I agree the reasoning used to decide the winnings converge to a 1:2 ratio is persuasive. But consider the following argument:

As stated, Payoff(1) = Sum (3/2)^n as n ranges to infinity and Payoff(1) thus represents an infinitly large amount.

Now Payoff(2) = Sum (2 * (3/2)^n)
= 2 * Sum (3/2)^n
= 2 * Payoff(1)
= Payoff(1)!!

So the desired convergence is 1:1.

Any comments or clarifications will be appreciated.

[Cabbage]

They grow at the same rate if you treat them as a probability of payoff. Since this is a statistical problem that's all you're interested in anyway.

I'm not convinced of this, and since both the numerator and denominator are going to infinity, you have to be careful.

Say you play each game a million (or whatever) times. Take n to be the maximum number of flips that have occured in any of the Game 1 games. Take m to be the same thing for the Game 2 games. It's highly unlikely they will ever be equal, and even if they're close (within 1 of each other), that's enough to screw up the convergence.

I don't think m and n as I had them above will necessarily increase at the same rate, so I don't think there's a limit.

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...ebius sig. This is a moebius sig. This is a mo...
(sig line courtesy of WallyM7)

[tony1234]

[QUOTE]Originally posted by Cabbage:
[B] I'm not convinced of this, and since both the numerator and denominator are going to infinity, you have to be careful.

Cabbage, are you sure that the numerator (the average winings of Game 1) and the denominator (the average winnings of Game 2) are going to infinity? Certainly they will not be converging to any finite number, but does this imply that they will go to infinity? Perhaps they will, ah, just bounce around indefinitely, ie, not converge at all.

Also, the suggestion was made that we can consider variant games where the number of maximum flips allowable is limted to say, N, and then see what happens as we increase N. This approach, however, won't tell us what will happen when N is unlimited (as it is in the original question). Just because 2p/p=2 for every finite value of P does not mean that it equals 2 when P is infinite.

[Zor]

Cabbage, are you sure that the numerator (the average winings of Game 1) and the denominator (the average winnings of Game 2) are going to infinity?

I'm not Cabbage, but the numerator and denominator both approach infinity, as evident by the fact that they both have a larger than one value being raised to the nth power. This does not have any effect on their ratio however. See below...

Just because 2p/p=2 for every finite value of P does not mean that it equals 2 when P is infinite.

The basic rules of algebra still apply here. 2p/p will still converge to 2 even if p were infinite. Infinities of the same magnitude cancel out. I know that's not proper mathematical wording, but I'm just trying to get the point across. Since both the infinites are (3/2)^n, and not, say, n^2 and ln(n), you can safely cancel them out. The two values approach infinity at the same rate. There is no ambiguity here. The ratio of the two infinities converges to 1/2. Not all infinites are born equal, but these two are :)

And jcgmoi, you cannot assume that 2 * Payoff(1) = Payoff(1) since you're attempting to find the ratio of 2 * Payoff(1) / Payoff(1). That assumption you made is only valid after you've found the actual ratio to converge to infinity (e.g. you find the result to be 2 * infinity, and just write down infinity as your answer).

[Chrome Toaster]

hmmm.. bump? (see atmb)

[Konrad]

Cabbage sez:
since both the numerator and denominator are going to infinity, you have to be careful.

OK, I will be.

[Cabbage]

OK, look at this:

Fix n. Same type of modification to both the games as I used above--restrict game 1 to end after n flips (if necessary), restrict game 2 to end after n+1 flips. Take the ratio of the expected payouts. Now let n go to infinity. It won't converge to 1/2.

m and n as I had them were'nt equal to begin with. And they don't necessarily grow at the same rate, either. Certainly, there's an expected rate (value) here at which they'll grow, and that rate will be the same for both, but neither will actually follow that expected rate consistently; they'll kind of oscillate around that expected rate, averaging out to it, but not actually equalling it. Statistically, there's absolutely nothing wrong with that. That oscillation around the expected rate (which will be different for m and n) will be enough to prevent convergence.

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...ebius sig. This is a moebius sig. This is a mo...
(sig line courtesy of WallyM7)

[C K Dexter Haven]

We're not dealing with a straight-forward sequence (or ratio of sequences) here. We're dealing with the Expected Value, and the calculation of Expected Value is the divergent SERIES (sum of sequence).

That is, for Game 1, the Expected Value E(1)= SUM( 3/2)^n where n--&gt; infinity. That series (that is, the sum) diverges; the sum gets larger and larger without bound as n gets larger and larger. Don't talk to me about "equals" infinity; say instead, it diverges, it grows without bound.

For Game 2, the Expected Value E(2)= 2* SUM (3/2)^n, and this sum also diverges.

Is it the case that the ratio E(1)/E(2) = 1/2? I confess, I'm way too far away from infinite series to remember how that works. I can try to remember to look it up when I get some time with some reference books.

Or someone will cite the theorem before then, and I'll be embarrassed at not remembering.

Tony, you will NOT be able to have conclusive indications from a computer run that a series doesn't converge. Simple example, a series might seem to converge for the first billion terms, and then diverge... no computer run would show divergence. Similarly, you won't get proof of convergence from a computer run... You will only get an "indication", if you're lucky and if your run the simulation enough times (and don't ask me what "enough" means, a billion runs may not be enough.)

Practical Terms
Someone asked about "practical terms." In practical terms, of course, the whole point of the paradox is that you'd NEVER have a run of 10,000 tails. Well, OK, not NEVER, but you'd be running the computer a VERRRRRRY long time before you could expect such a thing to show up.

One of the paradox questions (Let's stick with Game 1) is: How much would you be willing to PAY to play this game? You pay $X, and then get a winning based on the coin toss. If you pay $7.60 ( = 3/2^5) and you get a head on before toss 5, you've lost. If you four tails and then a head on toss 5, you're even. If you get five tails (or more)and then a head, you've won.

If you pay $57.66 (= 3/2^10), then you win only if you get a string of at least 10 tails before you get the first head.

In theory, to make a "fair" game, you'd be willing to pay the Expected Value. In this case, that's Infinite! ... but this seems silly. You'd be hard pressed to find someone to bet $57.66 on getting a string of 10 tails before they get heads.

And of course, the paradox goes on... don't just play once! Play hundreds of times! You pay the $57.66 each time, and you lose and lose and lose, game after game, and then on the thousandth game or so (expectation of 1024 games), you win one! Well, you're way way far behind, of course -- you've lost every game but one so far. However, keep playing... sooner or later, after a billion years, maybe, you'll have a string of a thousand tails before you get heads... and THAT's when you'll get a huge payoff.

SO the paradox is that the math behind the game relies on an infinite number of plays; while the actual play of the game, you'd give up after a few billion games (even with the best computers wholly dedicated to this game, it could take thousands of years of computing time before you hit some of the huge payouts.)

[jcgmoi]

Zor says:

And jcgmoi, you cannot assume that 2 * Payoff(1) = Payoff(1) since you're
attempting to find the ratio of 2 * Payoff(1) / Payoff(1). That assumption
you made is only valid after you've found the actual ratio to converge to
infinity (e.g. you find the result to be 2 * infinity, and just write down
infinity as your answer).

But Zor, there is no assumption, only a paradox. Payoff(1) increases without
bound but it is a countable infinity. Twice a countable infinity is a countable infinity.

Look at it like this. Every 3rd positive integer is divisible by 3. Let's put those in S. Let's put all other positive integers in T. You would say T has twice as many members as S when in fact they contain the same number, as can be shown by pairing them off 1-1. And if you sum the elements of S and T you find they are equal, not to 'infinity', but to a number that increases
without bound. In fact, their sum is Payoff(1).

[CurtC]

CDtexthavn wrote:
Practical Terms
One of the paradox questions (Let's stick with Game 1) is: How much would you be willing to PAY to play this game? You pay $X, and then get a winning based on the coin toss. If you pay $7.60 ( = 3/2^5) and you get a head on before toss 5, you've lost. If you four tails and then a head on toss 5, you're even. If you get five tails (or more)and then a head, you've won.
If you pay $57.66 (= 3/2^10), then you win only if you get a string of at least 10 tails before you get the first head.I agree with most of your argument, but you messed up on the arithmetic on these two examples. If you pay $7.60 to play and you get a head on toss five, you're not even - you will have won 3^5 = 243 dollars. Pretty good deal.

And if you pay $57.66 and get a head on flip 10, then you'll collect over $59,000 dollars.

So in practical terms, I'd sure line up to play this game.

[tony1234]

Reading the discussion, I realize (with embarrasment) that I slightly mistated the question. Perhaps, that is the explanation for some of the disagreement. With respect to Game 2, I should have designated the number of coin flips as "m" not "n." As many grasped, I did not intend to assert that the pay-off of Game 2 was based on the same number of coin flips as Game 1. More explicitly: Game 1 and Game 2 are played simultaneously *but independently.* Though the total winning for each game will go up, they need not increase at the same rate for any given segment of sequential plays.

I also now realize that instead of asking, What will be the ratio of average winnings of the games?, I could simply have asked, What will be the ratio of the total winnings of the games?

Nevertheless, I am still interested in the more basic (?) question, Will the *average* winnings of either game converge to infinity or not converge? If the answer to this question is that it will converge, I think the answer to the ratio question is 1/2 because Game 2 will converge more quickly. But if the answer is that the average will not converge, that to my mind leaves the question open regarding the ratio of total winnings.

With respect to average winnings of either game, the answer is not clear to me. Intuitively, because extremely large payoffs will occasionally come up, the average winnings could converge to infinity. But perhaps they will come up so randomly that there will be no convergence at all.

CKDextHavn: I agree with much of what you have said. If you could find the relevant theorem (or even suggest to me the right reference work), I'd would asppreciate it.

Tony

[C K Dexter Haven]

I apologize, I did indeed screw up the math. The point is still a valid one, you pay $X and there's a certain point at which you win back WILDLY more than what you paid... after a sufficient number of tosses.

&lt;&lt; Will the *average* winnings of either game converge to infinity or not converge? &gt;&gt; NO, Tony, there's no such thing as "converge to infinity"... there's only converge (to some finite number) or diverge.

I have to think that, given a divergent series and another divergent series whose terms are all twice those of the first, that the ratio is undefined. "Infinity" divided by "infinity" is not defined.

Example: Take the series A = 1 + 1 + 1 + 1 + ...
which is divergent. Take the series B = 2 + 2 + 2 + ... which is divergent. The quotient A/B is undefined (not convergent).

Sure, you say, you could write B = 2 * A and then A/2*A = 1/2. But you could also write B as: (1 + 1) + (1 + 1) + (1 + 1) + ... and then, lo! B = A, so A/B = 1.

Or you could write A = (1 + 1 + 1 + 1) + ( 1 + 1 + 1 + 1) + ... = 4 + 4 + ... so then A/B = 4.

So, I am standing by my original guess, that the ratio of the two series does NOT converge.

Notice, that you can't pull this trick with a convergent series. Rearranging the terms of a convergent series (even one with pluses and minuses) does not affect the convergence. Rearranging the terms ofSo, I think this is another of those questions where "common sense" says that the ratio is 1/2, but "common sense" is mathematically not correct in this case. The ratio is not defined (that is, not convergent.)

Thus, when we have people running 100,000 trials, we get a DIFFERENT result each time.

[tony1234]

CKDextHavn,

I have to speak tentatively here because it's not my area, but I think you are confusing sums and series. The series .9, .99, .999, . . . converges to 1. The series 1, 2, 3, . . . converges to "infinity" (that is gets greater and greater without limit). The series 1, -2, 4, -8, 16 . . . does not converge at all. The average winnings for Game 1 will be a series, maybe sometime like 9, 6, 3, 18 . . . (I make these #s up). So it makes sense to ask whether the series will converge to infinity or not converge at all.

I agree with you that that infinity/infinity is undefined. Nevertheless I do not see how that answers the question. The total winnings of Game 1 and 2 will both increase without limits as the games are played repeatedly. But there still may be a well defined ratio of the winnings. Likewise, even if the average winnings of both games form a divergent series, it may be that there ratio is a convergent series--or at least you would have to refer me to a theorem to the contrary (I don't trust my intuitions).

In sum, I do not see how your discussion of expressions such as 1+1+1+1 . . . can answer the problem. But I may well be missing something, and I would be happy to be enlightenend.

Tony

[WhiteNight]

Ok, this was sort of interesting and I wanted to see what happened, so I wrote a program to test this.
http://www.members.home.com/whitenight/files/CoinToss.cpp - This is the C source http://www.members.home.com/whitenight/files/CoinToss.exe - This is the Win32 .exe

It's very fast, it plays about 70,000 games / second on my pentium 200.

I added in a bunch of features so you could analyze the results. The program features three main modes of operation, one where it shows the results of each series of tosses for each type of game, another where it shows only tosses that go past a certain threshhold (more than eight tosses in a series, etc) or silent mode where it only displays results when you hit a key to exit. Needless to say, silent is fastest, but the least informative.

I'll detail the options, though it comes with some help.
Options: -t # -g # -z # -s -h

-g: Display every Nth game, speeds simulation. Default 1
This lets you check the stats every so many games, without exiting, without having to see each turn.

-t: Display only games that go past N tosses, speeds simulation. Default 8
This shows you just the exceptional turns, so you don't watch the massively more common 2 and 3 toss turns.

-z: Game Types. The number of types of games played. Default 2

The game defaults to two game types, it will work with up to ten. The payback is x(3^n) where x is the type of game. This lets you see if a trend continues.

-s: Silent. Display no output except the final results.
Will show only the final results. Use this if you want to let it run while you go eat and check the results later.

Run the program, hit a key to exit. Read the output. It's pretty simple.


And, because I'm bored, and want to show off my mad ASM Sk1llz :) here's a little demo I wrote. 1238 bytes, with 768B pallette. My first (and hopefully only) x86 all-ASM project.
http://www.members.home.com/whitenight/files/fire.com


Enjoy. BTW, please let me know if you see any problems.

[tony1234]

CKD, you write:

I still content that the ratio of the two games is undefined. The "average" winning of Game (1) (that is, the expected value) is unbounded, converges to infinity, is undefined. Ditto the "average" winning of Game (2). The ratio of the two is thus not defined -- there is no single numeric value (finite or infinite) that can represent that quantity. It's like 0/0, it is not a defined quantity.

I don't mean to be tedious, but if the average winnings of games 1 and 2 will both converge to infinity (given enough plays), why do you think their ratio is undefined (will not converge)? It seems more plausible to me that the average winnings of game 2 would converge to infinity faster than those of game 1, or at least the rates of convergence could be meaningfully compared. Clearly the series of average winnings cannot converge on a finite number, or that number would equal an amount you should be willing to pay to play the game, and all amounts are two small (the expected value of a play is infinite). But mightn't the average winnings, as the number of plays increases, not converge at all, i.e., simply bounce around over an unlimited range in a purely random way?

Finally, for what its worth, playing a game with a payoff of 2^n for a million plays produced suprizingly consistent average winnings: 20, 22, 24, 19.

WhiteKnight, I tried running your program, but was unable to find any controls. When I hit a key, the program window shut down. But thanks anyway!

Tony

[Zor]

When you're evaluating a limit such as this...

lim { f(n) } as n --&gt; x

You cannot assume that n=x right off the bat. The same goes for evaluating any limit, including infinity. Take the following limit for example...

lim { sin(n) / n } as n --&gt; 0

If you just set n=0 right off the bat, you'd get 0/0 as an answer. The correct solution, however, is 1, which you can get by applying l'Hopital's rule (i.e. find the ratio of the rate at which the numerator and denominator approach 0). Now if we return to the original question...

lim { (3/2)^n / 2*(3/2)^n } as n --&gt; infinity

Again, you cannot set n to infinity right away; they cancel out first. Furthermore, you cannot set the 2 n's to different values. They are exactly the same by definition; n represents the maximum number of flips that are allowed to be played within a single game. We evaluate the ratio of the average winnings by continuously raising the value of n, simultaneously for both games. You cannot raise the maximum number of flips allowed for Game #1 first, compare its average winnings to that of Game #2, then raise the maximum number of flips allowed for Game #2 later. The maximum number of flips allowed must be the same for both games. That is why you can cancel the two (3/2)^n out and get 1/2.

Take another limit as an example of this...

lim { n / n^2 } as n --&gt; infinity

Now what is your answer? Undefined? No, it's 0, because the denominator approaches infinity faster than the numerator. Here's one last example...

lim { n / n } as n --&gt; infinity

So what is it this time? Undefined? Nope, it's 1, because here the denominator approaches infinity at the same rate the numerator does.


p.s. Just in case any of you have forgotten or are unfamiliar with l'Hopital's rule, it's mathematical form is as follows:

lim { f(n)/g(n) } as n --&gt; infinity = lim { f'(n)/g'(n) } as n --&gt; infinity

Under the conditions that the ratios are in of the indeterminate forms 0/0 or Infinity/Infinity.

[tony1234]

Zor,

I agree with your math, I just don't see how the equations you advance are relevant to solving the problem. Perhaps you could explain this more clearly.

It seems to me that you are imagining versions of Game 1 and 2 where the maximum number of flips is limited and then examining the ratio of winnings as that limit is increased. Is that right? If so, I don't not think it is a correct approach to the problem. To put it simply, in the actual problem, n, the maximum permitted number of flips, is infinite. Examining bounded variants of the game is misleading. It's like saying that because n/n, n to infinity, equals 1, then when n=infinity, n/n equals 1. But it's undefined.

At least so it seems to me.

Tony

[Cabbage]

lim { (3/2)^n / 2*(3/2)^n } as n --&gt; infinity

Again, you cannot set n to infinity right away; they cancel out first. Furthermore, you
cannot set the 2 n's to different values. They are exactly the same by definition; n
represents the maximum number of flips that are allowed to be played within a single
game. We evaluate the ratio of the average winnings by continuously raising the value
of n, simultaneously for both games. You cannot raise the maximum number of flips
allowed for Game #1 first, compare its average winnings to that of Game #2, then raise the maximum number of flips allowed for Game #2 later. The maximum number of flips
allowed must be the same for both games. That is why you can cancel the two (3/2)^n out and get 1/2.

First, I just want to mention that the limit mentioned there at the top should actually have sums in both the numerator and denominator.

Suppose that the ratios actually do converge. I claim that I should be able to model it the way I did, i.e, by allowing m and n to be different, letting both go off to infinity, and taking the limit. This will still take the limit of the ratios of the partial sums as both partial sums go to infinity.

As I mentioned before, however, say I take the top sum to be from 1 through n, the bottom from 1 through n+1 (sure, I'm setting the max number of allowed flips to one more in the second game, but in a moment I'm going to let n go to infinity, so that won't matter after all). Take the ratio. Now take the limit of that ratio as n goes to infinity. If the ratio of the two infinite series is going to make any sense at all, this limit should be 1/2 (as it would be if I had the sum from 1 through n in both top and bottom). It's not.

This describes a way we can model the ratios of the infinite series where we do not get 1/2 as their limit. In fact, we can try to model their ratio many different ways, and get many different answers.

Conclusion: This limit we're trying to find simply does not exist; the ratios do not converge.

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[tony1234]

Cabbage,

Interesting answer. perhaps it is right. It's the most persuasive thing so far, I think. I'll have to think about it more (even though I don't think I'll be able to conclusively evaluate it). But let me ask you this:

You argue that the ratio doesn't exist because different ways of "modelling" the ratio result in different answers. But let's say that this was not the case. Let's say that for a different pair of games 9with different payoff functions), the various models you consider (n, n; n, n+1; n, n+2, etc.) all converged on 1/2, as n when to infinity. Would you then conclude that the ratio was 1/2? Based on my previous post, as I explained to Zor (please see that post), this would NOT justify the conclusion that the true ratio was 1/2. In other words, I have doubts whether the modelling technique you use is correct in the first place, so that, when it produces different answers, you are warranted in concluding that there is no single correct answer. The failure of an invalid technique to produce an aswer does not mean there is no answer.

Tony

[Cabbage]

Let's say that for a different pair of games with different payoff functions), the various models you consider (n, n; n, n+1; n, n+2, etc.) all converged on 1/2, as n when to infinity. Would you then conclude that the ratio was 1/2?

Yeah, I would, if ALL the "models" converged to 1/2. That means (n,n), (n,n+1), (n,n+2), (n,n^2), (n^3,the nth prime); ANY of the uncountably many possible pairs of increasing sequences of integers.

For example, if we changed it like this: Game 1, where the payoff on the winning flip was just $1 each time, and Game 2 where the payoff was $2 each time, then the ratio of payoffs as you played the games over and over would converge to 1/2.

As for justifying the model I used, first off, it doesn't make sense to compare winnings after the games have been played an infinite number of times--both will have won an infinite amount of money, you can't say this infinite is a certain percentage of that, it doesn't make sense. What we can do is play each game an ever increasing number of times (so the payoff will be finite at all times) and compare those ratios and see how they behave in the long run.

We can do this by considering the expected payout. The expected payout is the sum of all the possible payouts, weighted with their corresponding probabilities. In general, the more a game is played, the closer the average winnings per game will come to the expected payout. An important point: Statistics says nothing (or if it does, then not enough, in this case) about this rate of convergence. Sometimes, by pure luck, your average winnings may exactly match the expected payout early on. Other times, it may take a while to settle down and get close. The rate of convergence is not consistent, all we know is that it eventually will converge.

That's not the case in this game, however--the expected payout is infinite, which is why it's called a paradox. It does imply, however, that the more the game is played, the average winnings will increase without bound.

So what to do about this? We're playing a finite number of games, and seeing what happens to the ratio of winnings as the number of games increase. If we play a finite number of games, there will also be an expected maximum number of flips having occured in that game at some point. If I play the game only once, probability says I can expect to flip the coin once or twice. If I play the game 1000 times, probability says that the expected maximum number of flips that occured in a single game is 10; 1000000 games, an expected max number of 20 flips has occured (I'm not claiming these numbers are accurate, just illustrating the point). The more games, the larger the expected max number of flips.

So if we play a finite number of games, we can safely go ahead and limit the number of flips allowed in a game, because if you play a set of n games, then play another set of n games, over and over, on average the maximum number of flips occuring in a single game will converge to the expected maximum number of flips.

Allowing only a finite number of flips like this has the benefit of giving a finite expected payoff for each game (the sum from i=1 to n I mentioned in a previous post). Now we can take the ratio of these payoffs.

If we play the "finite flips" version of the game over and over, the ratio of the payoffs the games will converge to the ratio of the expected values.

Now, we take the step we've been working towards all along--what happens to this ratio we just obtained as we play more and more games?

First of all, we have to start increasing the max number of flips allowed--the more games played, the larger the expected max number of flips becomes. Correspondingly, this will increase the expected payoff, which will start going to infinity for both games.

Here's where the important point I mentioned above comes in. There's nothing guaranteeing the expected payoff in Game 1 will be concurrent with the expected payoff in Game 2, they may converge at different rates (not to mention the "max number of flips in a single game" may increase at different rates as well, which only makes things worse). Sometimes 1 may increase faster, sometimes 2, there will be nothing consistent about it.

That's why we have to consider all the possible different "models" of this convergence I mentioned at the beginning. Using the different models sometimes the ratio converges to 1/2, sometimes to 1/3, sometimes it just bounces around and never settles down at all.

Since we have no consistency at this point, we're done, we've found out that the ratios will not converge.

Sorry I got kind of long winded, but there are some subtle points that need to be watched out for. Interesting question, though.

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[tony1234]

Cabbage,

You say that as the number of plays increases, the ratio of winnings of Game 1 and Game 2 will not converge. You also say that we can say nothing about the comparative rates of convergence to infinity of the average winnings of each game. Does this mean that after, say, a large finite number of plays of the games, we can say nothing about which game's total winnings will likely be greater (because the average winnings of each will be bouncing around unpredictably toward infinity)?

***

If so (and here I move from probability to decision theory), is there a reason to prefer one play of Game 2 to one play of Game 1? [If not, is there a reason to prefer one play of Game 2 to the Game 3, which is the ***actual*** outcome of Game 1 plus a dollar. (Game 3 obviously must be prefered to Game 1)?]

[Cabbage]

Yeah, I would say there is a reason to play Game 2 over Game 1. Over a course of many games, the probability that Game 2 will give a larger overall payout is greater than the probability that Game 1 will (simply because it does have double the payout, after all). Over this course, the actual results won't follow that probability exactly, and there still certainly exists the possibility, however unlikely it gets to be in the long run, that game 1's payout will even be larger than game 2.

Overall, I guess what it amounts to is this: The ratios won't converge. However, the ratios will probably have a tendency to stay within a certain range (for example, the ratio may tend to stay between 1/4 and 3/4, but not actually converge to anything). It would be difficult to actually give a possible range for the successive values of the ratio, and the corresponding probability that the ratio will remain in that interval for a given length of time, but I think it could probably be done. I'm not going to try to do it though. :)

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[Splat]

At last I have something modestly useful to say....

The programs posted above for empirical testing of this problem will work well enough for small numbers of flips, usually about 30,000 depending on your compiler and the initial seed you use. The trouble is that the random number generator built in to most programming tools is pretty sad. We can't get really random numbers from a computer, so the best we can hope for is a pseudo-random sequence. The easiest way to get such a sequence is a linear congruential generator (see Knuth, Seminumerical Algorithms for the gory details), and all such generators produce a sequence which eventually repeats. Most compilers use a signed 16 bit int for results, hence the 30,000 limit. We can do a little better without undue exertion by using an unsigned long (32 bits in Visual C++, your compiler may vary) and constants cribbed from Numerical Recipes in C. The period is still only 1.7 million, but if we're doing runs of a few hundred thousand flips that's plenty.

The sample code below has a better_rand() function and a test of the period length for better_rand() and rand(). If you compile this with Visual C++ you'll notice that of you seed rand() with 55 you'll get a period just over 3,000....

Of course a better simulation still misses the point-- if you're free to play either game forever your winnings per game will keep increasing as you play. Still, I can't pass up the opportunity to spread the word about bad random number generator.

#include &lt;stdlib.h&gt;
#include &lt;iostream.h&gt;


const unsigned long RAND_M = 1771875;
const unsigned long RAND_A = 2416;
const unsigned long RAND_C = 374441;

unsigned long better_rand_current;

void better_seed(unsigned long seed) {
better_rand_current = seed;
}


unsigned long better_rand() {
better_rand_current = (RAND_A * better_rand_current + RAND_C) % RAND_M;
return better_rand_current;
}


int main(int argc, char* argv[])
{

unsigned int ct;
int base;

srand(55);

ct = 1;

base = rand();
while (base != rand())
ct++;
cout &lt;&lt; ct &lt;&lt; endl;
cout &lt;&lt; RAND_MAX &lt;&lt; endl;


unsigned long better_base;
better_seed(1);
better_base = better_rand();
ct = 1;
while (better_base != better_rand())
ct ++;
cout &lt;&lt; ct &lt;&lt; endl;
return 0;

}

[Splat]

You can trust your computer to act unpredictably. It does it whenever you don't want it to!

[C K Dexter Haven]

My arithmetic error was confusing the original statement of payout (3^n) with the expected value terms (3/2)^n. Sorry about that.

The point is still the same. How much do you pay to play?

Tony, if you understand the difference between "converge to infinity" and "equals infinity" (as opposed to other divergence), taht's fine. My concern was that most posters didn't, and are cheerfully dividing one infinity by another infinity.

My point is that you can't do that. When you are dealing with divergent sequences (or the sum of a divergent sequence) -- and I am using the term divergent to mean not convergent,whether it increases without bound or not -- it is NOT TRUE that A/2*A = 1/2. I used the example of 1's and 2's because it's easier to manipulate than your example of 3^n and 2*3^n -- or, more correctly, than your expected values of (3/2)^n and 2*(3/2)^n.

The example of the 1's and 2's was meant to show that one cannot "cancel out" the infinite sum in the numerator and denominator.

I still content that the ratio of the two games is undefined. The "average" winning of Game (1) (that is, the expected value) is unbounded, converges to infinity, is undefined. Ditto the "average" winning of Game (2). The ratio of the two is thus not defined -- there is no single numeric value (finite or infinite) that can represent that quantity. It's like 0/0, it is not a defined quantity.

[tony1234]

Splat,

Very interesting. That explains, I take it, why it is that when I run these coin-flipping experiments a million times on my computer, the outcomes tend to be more consistent than probability theory would predict.

If you can't trust your computer to act unpredictably, who can you trust?

Tony

[C K Dexter Haven]

Yes, I agree (and apologize if my wording was off) that, given two divergent infinite series, one cannot necessarily conclude that the ratio is defined or not. I wanted to make the point that merely "cancelling" out the factor (3^n) is NOT the way to determine the ratio.

Cabbage, thanks for the insights. You've explained very well why, in simulating a small finite number of tries (like a million), you would get results that are fairly consistent ... you probably haven't run enough tries to get a string of 100 heads, for instance.

On random number generation, that's also a good point, that computer simulation may give limited insight.

Computer simulation can give a good indication, but it doesn't PROVE anything.

[Konrad]

Yes, I agree (and apologize if my wording was off) that, given two divergent infinite series, one cannot necessarily conclude that the ratio is defined or not. I wanted to make the point that merely "cancelling" out the factor (3^n) is NOT the way to determine the ratio.

Yes it is the way. It's called the limit test, you take the limit as n goes to infinity of infinite series divided by another and you can get a consistent limit even if neither of the series converge on their own. If the limit exists then the ratio of the two series converges. This is elementary calculus.

Although it is true that the limit test will not work for ANY 2 series, it will work for these two.

Or you could write A = (1 + 1 + 1 + 1) + ( 1 + 1 + 1 + 1) + ... = 4 + 4 + ... so then A/B = 4.

No mater how you rewrite it you won't get anything but 2:1. If you rewrite 2 as 1 + 1 you have just doubled the number of terms. You're making the false assumptions that by taking the first n terms of each series and dividing by each other will give you the same result as using the sums of the series.

Splat: The reason the computer simulations are getting different answers is because the error increases as you increase n. If you want to get the same answers with a finite number of terms then you'd have to calculate the sum of the probability times the payoff. That's just (1/2^n) times the series.

[jcgmoi]

Konrad: Would you tell me the limit of SUM[(3/2)^n] as n tends to infinity from 1?

[jcgmoi]

Konrad said:
quote:

It's called the limit test, you take the limit as n goes to infinity of infinite series divided by another and you can get a consistent limit even if neither of the series converge on their own.


jcg sez:

quote:

Would you tell me the limit of SUM[(3/2)^n] as n tends to infinity from 1?

Need I say more?

Since you're urging me to take the limit of an infinite series, yeah, you need to say quite a bit more about what that limit is and what this 'limit test' of yours comprises.

I'm assuming you're thinking of L'Hospital's rule, where

lim f(x)/g(x) = lim f'(x)/g'(x).

My source, Mathematics from the Birth of Numbers by Jan Gullberg (which I recommend highly, by the way, even though it makes no mention of the 'limit test') says:
'After differentiation the quotient sometimes still appears in indeterminate form. The rule is then applied--in a repetitive manner-- to the quotients of the
derivatives'.

So lim (3^n)/2(3^n) =
lim(n)(3^(n-1))/(2n)(3^(n-1)) =
lim (3^(n-1))/2(3^(n-1)).

This is the same form as the original we wanted to reduce, so it's still indeterminate, so on the assumption the
limit of the ratio exists it can't be found by this method. I say the ratio of the payoffs for the 2 games is indeterminate. If you say otherwise, I'm sure you'll post again. Next time I'd appreciate content over
attitude.

[Cabbage]

Konrad:
Yes it is the way. It's called the limit test, you take the limit as n goes to infinity of infinite series divided by another and you can get a consistent limit even if neither of the series converge on their own. If the limit exists then the ratio of the two series converges. This is elementary calculus.

I'm not sure what it is you're referring to as the "limit test", but I do not believe it means what you think it means.

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[Konrad]

It's called the limit test, you take the limit as n goes to infinity of infinite series divided by another and you can get a consistent limit even if neither of the series converge on their own.

jcg sez:
Would you tell me the limit of SUM[(3/2)^n] as n tends to infinity from 1?

Need I say more?

[Konrad]

jcg sez:
Since you're urging me to take the limit of an infinite series, yeah, you need to say quite a bit more about what that limit is and what this 'limit test' of yours comprises.

First of all, the limit test applies to the quotients of two series not to the limit of one series, which is what you first asked and that is why I had to quote my own post.

Second of all, the limit test applies to the quotients of two series and not to the quotient of two functions like L'Hôpital's rule does. You can't apply L'Hôpital's rule to series in the way you are.

Cabbage sez:
I'm not sure what it is you're referring to as the "limit test", but I do not believe it means what you think it means.

Let's a random elementary calculus book here.... Calculus of a Single Variable by Larson.

Page 577 under the heading of "Limit Comparison Test" in chapter 8 "Infinite Series".

"Suppose that An &gt; 0, Bn &gt; 0 and
lim(An/Bn) = L as n -&gt; infinity
where L is finite and positive.
Then the two series SUM(An) and SUM(Bn) either both converge or both diverge."

Conversely, it is obvious that it must be possible for the quotient of two divergent series to have a finite limit.

Technical note: To be perfectly accurate this proof works under the assumption that the limit of SUM(An)/SUM(Bn) from 0 to n as n -&gt; infinity = An/Bn. In this case it is obvious that it is so but if you don't believe me here is the proof:

CKDextHavn's objection was that 2 could be rewritten as 1 + 1 thus changing the rate of divergence of the series. This is not so because when dealing with series we always use limits. The limit of
SUM(2) from 0 to n
= lim SUM(1 + 1) from 0 to n
= lim SUM(1) from 0 to 2n (&lt;= this 2n is the critical part)

As n goes to infinity SUM(1) from 0 to 2n will diverge just as quickly as SUM(2) from 0 to n.

You can't just plug in infinity into n and assume that SUM(3^n) = SUM (2*(3^n)) as CK did.

Therefore once the factor of 2 is removed each series diverges at an equal rate no matter how you rewrite them. Therefore the ratio of their sums is just the ratio of their terms.

[jcgmoi]

Konrad: Thanks for elaborating.

You say in describing the limit comparison test:
"Suppose that An &gt; 0, Bn &gt; 0 and
lim(An/Bn) = L as n -&gt; infinity where L is
finite and positive. Then the two series
SUM(An) and SUM(Bn) either both converge or
both diverge."

I don't get your point. Let's say your application of the test is correct. So the 'series SUM(An) and SUM(Bn) either both converge or both diverge'.
But that's not in question: we agree they're divergent.

You say...the limit test applies to the quotients of two series and not to the quotient of two functions like L'Hôpital's rule does. You can't apply L'Hôpital's
rule to series in the way you are.

I'm not using series. The function for the value of the nth partial sum of game 1 payoff is f(n) = 3[(3/2)^n - 1]. For game 2, g(n) is twice that. The limit of their ratio as n approcaches infinity is indeterminate and repeated application of L'Hospital's rule yields repeated indeterminate results. So the ratio of the payoffs is indeterminate.

[Konrad]

The point about rewriting 1 + 1 + 1 ... is NOT that it diverges any faster or slower by being re-written, but that it DIVERGES. And that the "ratio" of (1 + 1 + 1 + ...)/(2 + 2 + 2 + ...) is undefined.

No it's not undefined. The ratio of SUM(1)/SUM(2) from 0 to n is 1/2 no matter how you rewrite it. You can rewrite SUM(1)/SUM(2) from 0 to n as SUM(1/2) from 0 to n and then take the limit as n goes to infinity. This is perfectly "legal" arithmetic.

You said:
It is certainly possible for SUM(An) and SUM(Bn) both to diverge and yet the quotient converges... but that does not mean that ALL such do.

I said:Conversely, it is obvious that it must be possible for the quotient of two divergent series to have a finite limit.

Notice the word possible in there. I was just saying that since it's possible that contradicts your statement that the ratio of the two doesn't have a limit just because they are divergent. What you said was:

I have to think that, given a divergent series and another divergent series whose terms are all twice those of the first, that the ratio is undefined. "Infinity" divided by "infinity" is not defined.

According to my book infinity over infinity can be defined. In fact there's plenty of ways that it can be defined. For example the ratio convergence test for infinite series relies on the fact that infinity over infinity can be greater or less than 1.

That ratio is NOT 1/2, that ratio is defined, just as 0/0 is undefined; and you cannot simply "cancel" the infinite divergent sums.

Again, the limit of 0/0 can be defined. L'Hôpital's rule is just one way of doing this.

Also please note, and this is the key: there is a difference between SUM(An/Bn) and SUM(An)/SUM(Bn).

Yes, that's why I added my little technical note. Like I said it's perfectly legal to rewrite [SUM(A) from 0 to n]/[SUM(B) from 0 to n] as
SUM(A/B) from 0 to n
Your trick of rewriting 2 as 1 + 1 would just change the limits of the sum.

Again, if you don't believe me look up the ratio test. You take An/(An+1) as n goes to infinity. In 90% of cases you'll end up with something like 3n/5n. By your logic this test would never work because both 3n and 5n are infinity therefore the ratio would always be undefined.

[Cabbage]

Konrad:
"Suppose that An &gt; 0, Bn &gt; 0 and lim(An/Bn) = L as n -&gt; infinity where L is finite and positive. Then the two series SUM(An) and SUM(Bn) either both converge or both diverge."

Conversely, it is obvious that it must be possible for the quotient of two divergent
series to have a finite limit.

Technical note: To be perfectly accurate this proof works under the assumption that
the limit of SUM(An)/SUM(Bn) from 0 to n as n -&gt; infinity = An/Bn.

You're using the limit comparison test wrong; it says nothing about the ratio of the sum of two infinite series.

The technical note is wrong. Take these series:

1. SUM (1/n^2) = (pi^2)/6
2. SUM (1/n^4) = (pi^4)/90

Take the ratio of the second over the first. The technical note claims the ratio of these infinite series is (1/n^4)/(1/n^2) = 0 as n goes to infinity (the ratio of corresponding terms). I think it's obvious that the ratio of the infinite series is actually [(pi^4)/90]/[(pi^2)/6)].

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[C K Dexter Haven]

&lt;&lt; You can't just plug in infinity into n and assume that SUM(3^n) = SUM (2*(3^n)) as CK did. &gt;&gt;

Sorry, Konrad, but "just plugging infinity into n" is EXACTLY what I was saying NOT (repeat, NOT) to do.

The point about rewriting 1 + 1 + 1 ... is NOT that it diverges any faster or slower by being re-written, but that it DIVERGES. And that the "ratio" of (1 + 1 + 1 + ...)/(2 + 2 + 2 + ...) is undefined. That ratio is NOT 1/2, that ratio is defined, just as 0/0 is undefined; and you cannot simply "cancel" the infinite divergent sums.

Now, the Limit Theorem says that if LIM (An/Bn) = L as n --&gt; infinty, then SUM(An) and SUM(Bn) either both converge or both digerge.

It is certainly possible for SUM(An) and SUM(Bn) both to diverge and yet the quotient converges... but that does not mean that ALL such do. ((And that's not the converse of the statement; the converse is that if one of SUM(An) or SUM(Bn) diverge, then the lim(An/Bn) must diverge.))

Examples of where the quotient of two infinite divergent sequences produces a convergent sequence would be An = a^n (for a &gt; 0) and Bn = n! The series SUM(a^n/n!) is convergent, but (say a&gt;1) both a^n and n! are divergent sequences.

Also please note, and this is the key: there is a difference between SUM(An/Bn) and SUM(An)/SUM(Bn). We are dealing in this example with SUM(An)/SUM(Bn) which is the quotient of the sums. We are NOT here dealihng with SUM(An/Bn) which is the sum of the quotients.
The two are not equivalent, except in the trivial cases.

[Konrad]

Ignore that part. SUM(A)/SUM(B) doesn't equal SUM(A/B). Don't know why I wrote that.

Here's an even easier proof that it converges to 1/2:

SUM(2*(3^n)) = 2*SUM(3^n)

SUM(3^n)/[2*SUM(3^n)] = [SUM(3^n)/SUM(3^n)]/2
= 1/2

There. I don't think it's possible to get it any simpler than that.

[Konrad]

And the technical note is right. I didn't say it applies to any series just the series we're working on.

[Cabbage]

SUM(3^n)/[2*SUM(3^n)] = [SUM(3^n)/SUM(3^n)]/2 = 1/2

But the sum is infinity; infinity is not a real number, and you can't just say that inf/inf = 1. And, specifically, the two sums are "independent" of each other, one corresponds to game 1, the other to game 2.

In other words, the fact that the expected payout is infinite means that, if you play either game over and over, the average winnings will increase without bound. But we don't know how fast it will increase (I suppose you could also find an expected rate, but, still, that wouldn't be concrete, just a tendency over many repetitions. The point being that one game's average winnings may increase faster than another's, and not at a constant rate, either. As I said before, if you look at the ratio of

(SUM i=1 to n) (3/2)^i, and
(SUM j=1 to n+1) 2*(3/2)^j,

take the limit as n goes to infinity, and try to calculate the ratio we did previously we don't get 1/2. But both are still the exact same sums we had before, it all comes down to how they're represented (sure, for each finite n, there's an extra term in the second sum; that simply corresponds to the fact that, on a given series of trials, the average winnings of game 2 may increase quicker than that of game 1 (and I don't mean just double). Different representations give different limits, therefore there's no limit for the ratio of these infinite series.

A more intuitive idea: I feel pretty confident that if you modelled this on any computer, regardless of memory or speed, eventually it would "seem" that the ratios are converging (I don't think it would necessarily be to 1/2, though).

I think there will probably be arbitrarily long periods when the ratio becomes "relatively constant", so to speak. After each of those long periods, though, there will be another game which screws up the "convergence" and kicks the ratio out to some other value. It may seem to settle down at .5 for a while; then along comes a huge payoff, changing the ratio to .6, where it may seem to settle down for another long while.

I think this will happen because after several plays, the winnings of each game will be so huge, that the normal 2 toss, 3 toss, 10 toss, whatever, will do virtually nothing to change the ratio. The winnings for an individual game are boundless, though. Every time the ratio seems to settle down, the total winnings are still finite. So one game with a really huge payoff is all it takes to change the ratio and eliminate convergence. And we're playing the games over and over without end, so this will happen again and again.

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[Konrad]

I already said that it doesn't matter if the games are independent of each other. We're not calculating the results of a specific game but the probability of payout for each game.

We have SUM(probability of a result)(payoff for the result)

Since the probability of getting to n flips for each game is the same the ratio just reduces to SUM(payoff for game 1 of n flips)/SUM(payoff for game 2 of n flips) from 0 to infinity

And will you stop saying infinity over infinity is undefined? I already gave examples (like the ratio convergence test) where it isn't so unless you can prove that calculus is wrong I don't see how you can say that.

[Cabbage]

Since the probability of getting to n flips for each game is the same the ratio just reduces to SUM(payoff for game 1 of n flips)/SUM(payoff for game 2 of n flips) from 0 to infinity.

We're talking about expected values and how, on average, the payoffs will be "close" to the expected values. Here, however, "close" is not good enough--we need exactness for this limit to exist, not just a likelihood of closeness, because of the huge payoffs.

And will you stop saying infinity over infinity is undefined? I already gave examples (like the ratio convergence test) where it isn't so unless you can prove that calculus is wrong I don't see how you can say that.

No, I will continue to say infinity over infinity is undefined, as is 0/0, or any other "indeterminate form". Calculus does not define infinity over infinity, calculus develops the notion that the ratio of two functions can have a limiting value as x goes to infinity. At no time is infinity divided by infinity.

For example, lim (x^2 + 1)/x^2 = 1 as x goes to infinity. Does that mean (inf^2 + 1)/inf^2 = 1? Of course not. It merely means that as x gets arbitrarily large, (x^2 + 1)/x^2 gets arbitrarily close to 1. No arithmetic using infinity is involved, it is still left undefined.

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[tony1234]

Cabbage and Chronos,

I've been following the discussion and finding it very edifying. I don't have a lot to add particularly, although I am glad that it seems to generally be returning to the probability issues initially raised by my question, rather than reamining stuck on calculus issues in the abstract. i hope your positions on these matters eventually "converges."

Cabbagge, a few posts ago offered a narrative argument for why there would be no convergence. I realize that you are not offering it as a proof, but to my mind, it is no more than suggestive. Yes, eventually really huge sums will come up, but that will only result in the respective total pots being increased enormounsly so that next time you need a *really* huge payoff to knock the convergence off. I don't know whether you can expect these really huge sums to come up *frequently enough* to do this forever. Intuitively, it's not obvious to me.

The expected payoff of each game for a single play is infinite. Can you prove from *this fact alone* that the long-term ratios are undefined? It seems a tempting inference because the ratio of the expected payoffs is undefined and the long-run is just the manifestation of the expectations of the short-run (if you know what I mean). Is it really so straightforward? Stated conversely, can you prove that there are no games with infinite expected utilities for a play which will a have average winning ratio that will converge?

tony1234

[Chronos]

Wow, tony, I didn't even realize I'd posted in this thread yet! :) I do have something to add, though, that will hopefully throw a monkey wrench into a lot of arguments (including, perhaps, my own)-- Hey, we've got to keep this interesting, after all. First, some hypotheticals: First off, suppose there's two tables in the casino, and at both of them, game 1 is being played (payoff is 3^n dollars) Which table is preferable? Clearly, since it's the exact same game at both tables, it doesn't matter.
Second hypothetical: Suppose we have one table, but with two players, player 1 and player 2. The same coin is flipped for both; payoff is 3^n for player 1, and 2*3^n for player 2. In other words, no matter what the payoff for a given play, player 2 ALWAYS wins twice as much as player 1. In this case, if you had the choice of being player 1 or player 2, which would you choose? I think that it's pretty obvious that player 2 would be the better choice, but I also rather suspect that this is the weak point in my argument.
OK, hypothetical number three: two tables, for each table there's one coin and two players, player 1 and 2. Same setup at each table as in previous example, player 2 wins 2*3^n, player 1 only wins 3^n. Now, suppose you walk into the casino, and there's already a person registered as player 1 at the first table, and someone registered as player 2 at the second table. Again, you have a choice of playing as player 1 or player 2. If my reasoning is correct (no guarantees there), then you should choose to play at the table where you get to be player 2. But this is just like the original question, choosing from two independent possible games. Can anyone show me if/where the error in my reasoning is?

On a completely different note, as to how much a person should be willing to pay to play this game: I would not be willing to pay any more than $3, the minimum possible payout, for the simple reason that if someone's offering me a game with a theoretically infinite payout, it's got to be rigged :)

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"There are only two things that are infinite: The Universe, and human stupidity-- and I'm not sure about the Universe"
--A. Einstein

[tony1234]

Whoops, Chronos, I'm getting our threads tangled! I meant to write "Cabbage and Konrad" in my earlier post. Sorry.

Now let me think about your hypos.

Tony

p.s. In the finite-universe thread I meant "isotropic," not "isomorphic" obviously. Also, given your handle, you might want to take a look at my post on the metric-time thread.

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[Cabbage]

OK, I guess I'll try one more time. First, I claim that, over repeated playings of either game, there will be no limit to the max payoff on a given game. There will be a game that pays off 3^n (or 2*3^n, for Game 2) regardless if n is 100, 1000, googolplex, whatever. The probability of this happening is 100%. It can be proven, but I don't want to make this overly long, but, basically, remember: We are playing an endless streak of games here, we ain't gonna stop playing. Ever.

We would expect that, out of 2^n flips, there will be exactly one payoff of 3^n dollars in Game 1. (2*3^n in Game 2, of course). It's likely that this will happen in one game before another--I don't think this even requires proof. I'm claiming that, whichever game gets that payoff first, it's going to screw up the ratio so that it will never be able to settle down to 1/2. That's going to happen for n=1,2,3,4,...; in other words, it's always going happen, it will never stop screwing up the ratio.

Here's why it will screw up the ratio:

Consider the first 2^n flips. As I said, we would expect exactly one of those to have a 3^n dollar winner. The later that win happens, the less likely it is to screw up the ratio. (If it happens early, the ratio will really be lopsided in favor of one or the other. Later on, the total winnings of each game will be large, making a change in one of the total winnings not as significant to the ratio).

Now that that's established, let's assume that the 3^n win happens later--say exactly on the (2^n) + 1 flip for one of the games (but not both, since, as I've said, it's unlikely the two games would match up that well). If it screws the ratio up here, it certainly would have screwed it up earlier.

Through the 2^n flips (with no 3^n winner or greater winner yet), what is the total expected winning for each game? Half the games are expected to win on the first flip, a quarter on the second, and so on. The expected winnings are:

(SUM i=1 to n-1) [2^(n-i)]*(3^i)

= 2*(3^n) - (2^n)*3

Of course, the corresponding expected winnings for game 2 will be exactly twice that.

OK, now for the big question: Somewhere along those 2^n flips, we will expect a 3^n (or greater) winner. How much will that knock this ratio off of 1/2?

If game 1 wins the big one first, the ratio will be:

(1/4) * [3^n - 2^n]/[3^(n-1) - 2^(n-1)]

which has limit 3/4 as n goes to infinity. (NOT 1/2).

If game 2 wins it first, the ratio will be:

[2*3^(n-1) - 2^n]/[6*3^(n-1) - 2^(n+1)]

which has limit limit 1/3 as n goes to infinity. (NOT 1/2)

So if Game 1 is the first to win 3^n (win the game on the nth flip), it screws up the ratio. If Game 2 is the first to win 2 * 3^n (win the game on the nth flip), it screws up the ratio. This will happen over and over again, for each value of n. It won't converge. Each time a new big win comes (and it will come, 'cause we never stop playing the game), it screws up the ratio--screws up the convergence.

I don't think I'm capable of making it any clearer than that.

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[Cabbage]

Chronos:

I don't see any error in what you said. Yeah, I think you should play game 2--in a finite series of games, you would "expect" to win more in Game 2 than in Game 1. (I'm using "expect" somewhat loosely here--not the strict probability definition).

All I'm claiming is that the ratio of winnings of Game 1 to Game 2 will not converge to 1/2. The ratio will jump around.

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[tony1234]

Ok, Chronos, I've thought about your hypos. Your presentation of the hypos seems admirably clear. Nevertheless, I don't see quite what you are driving at--that is, why there's something problematic. Regarding Hypo 1, I agree that you should be indifference between the tables. Regarding Hypo 2, I feel pretty firmly you should prefer the game with the 2*3^n payoff (where there a dependence between the games). Rearding Hypo 3, you should prefer to be at the table where you are Player 2. One way of arriving at this conclusion is that you are indifferent between being Player 1 at either table and you prefer to be Player 2 at a given table to Player 1 at that table. By transitivy of preferes, . . . . So I think that all of these results are intuitively consistent. perhaps you can clarifiy, where's the rub?

But this is the interesting point. If all of this is correct, it seems you have a very easy proof that Game 2 in my original question must be prefered to Game 1, even if the ratio of there average winnings does not converge. The point of this thread for me, in a sense, has been to trying to get a proof via probabilty theory (in quantitative terms) of a result--Game 2 is preferable-- that seems obvious via that basic ulitity theory argument you have advanced.

Finally, if anybody is interested, I am currently trying to write a paper (hopefully for publication) on a series of hypos not terribly differ from the one Chronos mentions. My hypos are slightly more complicated and present *very* paradoxical results. I'd be delighted to discuss my specific hypos and theories with anyone, but think that it would be only appropriate to do so off-list. At a minimum you would have to read a 10-page draft paper of mine.

But its nice to know others are thinking about things similarly.

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[Cabbage]

A couple of corrections I just noticed I need to make on my long post above. In many places I accidentally used the word "flips" when I should have used the word "games".

Also, when I was figuring up the total expected winnings through the first 2^n games (not flips), assuming there had been no 3^n winner or greater, I under counted the total winnings by a little (basically, only by a couple of games). That doesn't make enough of a difference to change my original argument, though.

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[C K Dexter Haven]

Konrad, you are simply WRONG on many counts.

You say: &lt;&lt; The ratio of SUM(1)/SUM(2) from 0 to n is 1/2 no matter how you rewrite it. &gt;&gt;

This is just plain false (assuming the notation SUM is used the way I've been using it, to mean the infinite sum.) Would you say that SUM(1)/SUM(4) = 1/4? But note that SUM(4) = SUM(2), by simple re-parenthesis SUM(2) = 2 + 2 + 2 ... = (2+2) + (2+2) + (2+2) + ...

IF the series under consideration were convergent, then you can rearrange, factor, etc and you won't change the values. But when an infinite series is DIVERGENT, simple arithmetic (like SUM(2) = SUM (2*1) = 2*SUM(1)) just doesn't work.

You need to step back and rethink. You've got some misconceptions about how limits and infinite series work.

You are also wrong to say that 0/0 is defined. 0/0 is NOT defined. Period.

NOTE: f(x)/g(x) can be defined as f(x) and g(x) both go to zero; but that's a VERY different animal from 0/0. A function approaching a limit is NOT the same as a function being equal to the limit point. Therein lies the secret of much of calculus, which I taught lo these many years ago.

[Cabbage]

And actually, even if an infinite series does converge, in some cases you still have to be careful with rearrangements. For example, the series

1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 +...

can be made to converge to any real number by an appropriate rearrangement of the terms.

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[tony1234]

Cabbage,

I'm working through your reasoning. I certainly appreciate you efforts to explain this, but I need a bit a clarification.

You write:

"Through the 2^n flips (with no 3^n winner or greater winner yet), what is the total expected winning for each game? Half the games are expected to win on the first flip, a quarter on the second, and so on. The expected winnings are:

(SUM i=1 to n-1) [2^(n-i)]*(3^i)

= 2*(3^n) - (2^n)*3"

This cannot be literally correct. The expected winnings for every series of throws is not a number, but a probability distribution. (Well, if it were a number the number would be "infinity" in the sense that we say "the expected winnings from a single throw should be valued at infinitey, i.e., an amount higher than any finite number.") That is, there would be a probabilty associated with every possible total of your winnings, which are infinite in number. Do you mean to say that 2*(3^n) - (2^n)*3 represents a sum that you would have .5 confidence your total winnings would not exceed (so those winnings or less are "expected") or that 2*(3^n) - (2^n)*3 is the most likely total winnings? Neither seems to work. If n=3, then we are talking about the expected winnings over 8 games. When n=3, the expression equals 54-24, or 30. But that is much too low on either interpretation.

I certainly follow the general approach of your argument, I just am having some trouble on the details.

Tony



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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[Cabbage]

It's the same kind of thing as before when I was talking about the expected maximum number of flips. I was doing the expected value of total winnings after 2^n games, assuming that no game had won 3^n up to that point. The assumption allows us to go work with finite values instead of infinite, just like I did before. And the argument that we can actually make this assumption was that, if a game had won 3^n before that point, it would have REALLY screwed up the ratios--the ratios get harder to change the more total winnings you have. If you can show that a win of 3^n will significantly change the ratio after 2^n games have occured and the winnings have built up, certainly it would have changed the ratios if it happened earlier. (In other words, if you compare (1+x)/2 to (10000+x)/20000, if x is large enough to change the latter "significantly" from 1/2, it will definitely change the former ratio a lot from 1/2.

So, for example, the case when n=3 (played 8 games), and none of the games have won 3^3 = 27 dollars or more. We would expect, then, that half of the games have won on the first flip, 4*3 = 12 dollars. Half of what's left would win on the second flip, 2*9 = 18. Here's where I mentioned I forgot to add in a couple of games, let's go ahead and assume that they won 9 each, as well, for a total winning of 12+18+18=48 (making the winnings as large as possible, within a certain range of probability, so that the ratio will be hard to change). The the corresponding expected total winnings of the other game would be twice that, or 96, keeping the ratio at 1/2 (The actual ratio won't necessarily be 1/2, but since one of the main questions seems to be whether or not the ratio converges to 1/2, let's try and keep it as close as we can and see what happens.

But it won't stay at 1/2. For any given game, there's a 1/4 chance you'll win $27 dollars or more. Through playing 8 games, that would be expected to happen at least once (twice, actually). But if the ratio is around 48/96, whichever game wins $27 (or more) is going to significantly screw up the ratio. Suppose on the 9th game that one game wins 27, the other 9 (or 18, depending on which game). (48+27)/(96+18) = .66, (48+9)/(96+27) = .46, both far from .5. This isn't unique to n=3, each time one game or another has a big payout of 3^n or 2*3^n for the first time, for either game, that win will generally be big enough to change the ratio significantly. Enough to keep it from converging.

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[Konrad]

CK sez:
Konrad, you are simply WRONG on many counts.

En Garde! CKDextHavn, you are WRONG on many counts! There, I countered your wild assertion with one of my own.

This is just plain false (assuming the notation SUM is used the way I've been using it, to mean the infinite sum.) Would you say that SUM(1)/SUM(4) = 1/4? But note that SUM(4) = SUM(2), by simple re-parenthesis SUM(2) = 2 + 2 + 2 ... = (2+2) + (2+2) + (2+2) + ...

Haven't I already disproved this twice in this thread? Wasn't disproving this silly idea a question on the Descartes math competition? (A question which I got right) And, no, sum from 0 to n does not mean infinite sum. It means sum from 0 to n. (How can you say an infinite sum from 0 to n???) If you want an infinite sum you have to take the limit as n goes to infinity. Either way, as I said many times already, rewriting 4 as 2 + 2 would simply give you the limit as n goes to infinity of SUM(2) from 0 to to 2n. That diverges just as quickly as SUM(4) from 0 to n.

IF the series under consideration were convergent, then you can rearrange, factor, etc and you won't change the values. But when an infinite series is DIVERGENT, simple arithmetic (like SUM(2) = SUM (2*1) = 2*SUM(1)) just doesn't work

And you found this information where exactly? You're telling me that 2(SUM(1)) from 0 to n doesn't equal SUM(2) from 0 to n for any n?

You need to step back and rethink. You've got some misconceptions about how limits and infinite series work.

Let's try not to be patronizing here. Especially when one of us obviously doesn't know what they're talking about.

You are also wrong to say that 0/0 is defined. 0/0 is NOT defined. Period.

Here is what I said (with some added asterixes):Again, the ***limit*** of 0/0 ***can*** be defined. L'Hôpital's rule is just one way of doing this.

Would you at least do me the honour, wise and sumpremely correct being that you are, of reading my posts before you respond to them? Is it really so much to ask that you at least make incorrect assertions about what I have written rather than doing so about what I have not?

Can you just tell me one thing? Do you agree or disagree with the statement that SUM(A)/SUM(A) both from 0 to n = 1. If you agree with this statement then tell me if you agree or disagree with the statement that the limit of that equals 1 as n goes to infinity.

Cabbage: I think the function you are talking about (1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 +...) can be made to sum to any number by rearanging the terms FOR A FINITE n but it will always diverge in limit as n goes to infinity. That's a big difference.

[Konrad]

Woops, I missed a clever yet easy jab.

CK sez:
Therein lies the secret of much of calculus, which I taught lo these many years ago.

Yes, you obviously taught it before these 2 chaps named Newton and Leibniz made many changes to it. You should brush up on the latest publications by Euler and Riemann, there's been a lot of changes since you first learned it.

Gotta go tutor my friend for his multivariable calculus exam now...

[Cabbage]

Cabbage: I think the function you are talking about (1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 +...) can be made to sum to any number by rearanging the terms FOR A FINITE n but it
will always diverge in limit as n goes to infinity. That's a big difference.

Wrong. Review "alternating series" and "conditional convergence" (as opposed to "absolute convergence").

Gotta run right now though, I'll try and post more later.

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[jcgmoi]

konrad says:
SUM(3^n)/[2*SUM(3^n)] = [SUM(3^n)/SUM(3^n)]/2 = 1/2

cabbage responds:
But the sum is infinity; infinity is not a real number, and you can't just say that inf/inf = 1.

cabbage says later:
(1/4) * [3^n - 2^n]/[3^(n-1) - 2^(n-1)] which has limit 3/4 as n goes to infinity.

Aren't you committing the same error you nail
Konrad for? Isn't the 2nd ratio the same
indeterminate form as the 1st? Then how
do you evaluate the limit?

Re the infinite series
S = (1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 +...).

Using methods of integration, you can show
S = ln2 (natural log of 2).

[C K Dexter Haven]

&lt;&lt; Do you agree or disagree with the statement that SUM(A)/SUM(A) both from 0 to n = 1. &gt;&gt;

I disagree with the statement if SUM(A) is divergent, but I agree with the statement if SUM(A) is convergent.

&lt;&lt; If you agree with this statement then tell me if you agree or disagree with the statement that the limit of that equals 1 as n goes to infinity.&gt;&gt;

Again, if the SUM(A) is convergent, then I agree that the limit SUM(A)/SUM(A) is 1. If SUM(A) is divergent, then I disagree that the limit is 1. That's the whole point of this argument, that SUM(1)/SUM(2) is not 1/2.

My point about rearranging the terms is that SUM(1) and SUM(2) and SUM(3) and SUM(4) all diverge. Yes, they all diverge to a countable infinity (and the same countable infinity, if you like), but that's not the point. When dealing with the quotient of such sums, there is no reason to take SUM(1)/SUM(2) to be any different from SUM(1)/SUM(4) or SUM(1)/SUM(1). The ratio of countable infinity to countable infinity is not defined.

[Konrad]

I disagree with the statement if SUM(A) is divergent, but I agree with the statement if SUM(A) is convergent.

But SUM(A) from 0 to n has to be convergent since we're not talking about limits here. n must be finite. So basically you're saying you agree.

Now if SUM(A)/SUM(B) = 1/2 for all finite n do you agree with the statement that if we take the limit as n goes to infinity it should also go to 1/2.

If not, why not? If it is true for any n it must be true when you take a limit. If you plot the ratio versus n it's a straight line.

[Cabbage]

jcgmoi:
Aren't you committing the same error you nail Konrad for? Isn't the 2nd ratio the same indeterminate form as the 1st? Then how
do you evaluate the limit?

No, you can find the limit with a little algebra. Try dividing both top and bottom by 3^n, it makes it a little easier to see the limit.

Re the infinite series
S = (1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 +...).

Using methods of integration, you can show
S = ln2 (natural log of 2).

And, if you rearrange the terms of the series, you can make it converge to any other number you like: pi, e, -4365.45, whatever. Order is important.

Konrad:

I would agree that

lim [(SUM from i=1 to n) (3/2)^i]/[(SUM from i=1 to n) 2*(3/2)^i]

as n goes to infinity is 1/2. We're taking the ratio of the nth partial sum each time (which is 1/2 always), so the limit is 1/2, of course.

This is not the same, however, as (and I'm hoping this comes out readable):

(lim as m goes to infinity) [(SUM from i=1 to m) (3/2)^i]
---------------------------------------------------------------
(lim as n goes to infinity) [(SUM from i=1 to n) 2*(3/2)^i]

which doesn't exist. For example, as I've said before, take the ratio of the jth partial sum in the top, and the (j+1)st partial sum in the bottom. This isn't 1/2, and will not converge to 1/2. m and n are independent of each other, the "limit", if you want to call it such, will be different things, depending on what rates m and n increase, so there's actually no limit at all.

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[tony1234]

If I follow this, I think the point is that while:

Lim a f(a)/Lim a f'(a) = Lim a [f(a)/f'(a)]

is true,

Lim a f(a)/Lim b f(b) = Lim a [f(a)/f(b)]

is false. Indeed, the left side of the equality is a number, the right a function.

Hope me notation is clear. I haven't done any calculus for years . . . .

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[manhattan]

:: Test Post. Please ignore this post. If there are multiposts above, please ignore them too.::

[Cabbage]

Lim a f(a)/Lim a f'(a) = Lim a [f(a)/f'(a)]

Actually, what I'm trying to say is that this is not true (and the derivatives have nothing to do with what I'm trying to say, I haven't used any derivatives. You can replace the f's with fs and it still won't be true).

On the left hand side, you may as well replace the a's on the bottom with b's, actually; it will mean the same thing--a is just a "dummy" variable in each limit, anyway; the a on top has nothing to do with the a on the bottom. Writing it this way makes it clearer that there's no kind of "dependence" going on between the two.

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[december]

CKD appears a bit lonesome in his staunch defence of truth, justice and the American way, so let me try to provide support.

Consider two infinite sequences A(n) and B(n). If they both converge to finite, non-zero limits, then we all agree that

limit A(n)/limit B(n) = limit{A(n)/B(n)}

(By the way, "Limit A(n)" is defined as a number that A(n) gets closer to and stays closer to than any preset small distance for all n larger than some value.)

If both sequences diverge to infinity, or both converge to zero, then there is simply no meaning to the expression

limit A(n)/limit B(n)

We can write the symbol, but it does not represent anything. It's undefined.

Take the example

A(n) = 1,2,4,8,16,...
B(n) = 2,4,8,16,...


B(n)/A(n) = 2 for all n. So, you might think that

limit B(n)/limit A(n) = 2

But, B(n) and A(n) are the same sequence, except for the first term. So you could also argue that

limit B(n) = limit A(n), or

limit B(n)/limit A(n) = 1

This appears to be a paradox, which helps illustrate why there is no natural way to define Infinity/Infinity.

The actual problem is even more interesting in that involves a series of random numbers. I appreciate being introduced to it.

[Cabbage]

CKD appears a bit lonesome in his staunch defence of truth, justice and the American way, so let me try to provide support.

Perhaps I should change my UserName to "Chopped Liver". ;)


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[C K Dexter Haven]

I didn't think I was alone, when I had Truth on my side, but thanks, December, for the support. Actuaries know these things.

I have reread some of the posts, and I think December has hit the nail on the head: the problem is indeed that the quotient of the limits is not necessarily the limit of the quotients.

In this case, we are dealing with the Expected Values of the games, which are determined by an infinite sum SUM(An) and SUM(Bn) where n goes to infinity.

Konrad, you have apparently been dealing with finite sums, and I have no problem with what you say regarding finite sums... or even absolutely convergent sums.

But in this case, the expected winnings from each Game an INFINITE (and divergent) sum. OK, "converges to infinity" if you want, but the point is that the sum is unbounded... that is, the partial sums can be made as large as one would like, by choosing enough terms.

Now, it is true that the infinite sum is the limit of the partial sums from 0 to n as n --&gt; infinity. OK, let's call SAn the partial sum of the A1 + A2 +...+An. Then the infinite sum SUM(An) as n --&gt; infinity is lim(SAn) as n--&gt;infinity.

Similarly for SBn as the partial sum from 0 to n of B1 + B2 + ... + Bn.

In the situation we are dealing with, the quotient of all the partial sums SAn/SBn are all 1/2 (the question of whether we are dealing with SAn/SBm, I will let slide.)

But, as December rightly notes, it is NOT the case that lim(SAn)/lim(SBn) = lim(SAn/SBn) as n--&gt; infinity. It can be true, but it is not necessarily true, if the limit is unbounded (or if the limit of the denominator is zero, for instance.) (Damn, I wish I could do math notation here.) Hence, stuff like the limit test, ratio test, L'Hopital's Rule, etc.

I stand where I've stood: the expected outcomes are infinite (unbounded) and the ratio of the expected outcomes is also (in this case) indeterminate/infinite/undefined.

One last comment: Konrad, I see no reason for the gratuitous insults. I said you were wrong, and wrong you have been, in several matters (including thinking that SUM(An)/SUM(Bn) = SUM(An/Bn).) I did not insult you personally, and I see no reason for you to insult me. You want to do that, I'll meet you in the Pit.

[Konrad]

Cabbage sez:
This is not the same, however, as (and I'm hoping this comes out readable):

(lim as m goes to infinity) [(SUM from i=1 to m) (3/2)^i] ---------------------------------------------------------------
(lim as n goes to infinity) [(SUM from i=1 to n) 2*(3/2)^i]

OK, now you have to think what you're doing here for a moment. You can't just blindly divide 2 sums for no reason. You gotta ask yourself what is this ratio that I'm finding? I can say 5/7 = 10/14 but that has nothing to do with the problem.

Now what we are trying to find out is the ratio of expected payoffs for the games. Yes, the games ARE independent of each other when you play them. BUT, what is the ratio of their payoffs? The expected payoff for a game is SUM(probability of getting to n flips)(payoff for n flips). If we were playing 2 games and trying to find out the ratio that way you would be right to say that we couldn't do it because one series could go faster than the other. What you are saying by writing that limit the way you have is that you have two independent games. That's not what we're trying to find here. We're trying to find the ratios of average payoff per play. When you try to find that you have to compare each payoff for a game of n flips.

This way the two series are not independent because it's not the ratios for two games but the ratio of the payout for each game if it goes to n flips.

If I'm saying I'm summing the ratios of the expected payoffs then right away I can write the sum as SUM(2^n/2^n)(2*(3^n)/(3^n)). I don't even have to start out with two series because it's just a simple formula for finding the expected payoff for one play of any game.

So basically what I'm trying to say is you have to ask yourself why we're trying to find the ratio SUM(A)/SUM(B). Since we are just finding the ratios of expected payoff for n flips we KNOW they are not independent. No matter how you rewrite it will be the same.

You want some more evidence? What if the game was played with a 4 sided die (where you have to keep landing on the same side to keep playing) instead of a coin? It shouldn't change the ratios of the payoff should it? But there I could prove easily (SUM(x^n) = 1/1-x) that you'd get a 2:1 ratio. The ratio of 2(3^n)/3^n is still divergent on top and bottom yet we still get an answer that isn't. Or if you want to look at it another you could say the player has to flip the coin twice each turn and have it land heads both times. (That is n is now half the number of flips) How would something like this affect the game? Logically, it shouldn't affect the ratios. Yet, (1/4^n)(3^n) is convergent so the ratio will give 2:1. If you assume that the original game has an undefined expected payoff then you're saying that by making the player flip the coin twice instead of once per turn suddenly the game changes and there is a defined ratio. But really haven't changed the game so that doesn't make sense. It gets harder to get to a big payoff but the ratio of the payoffs for the two games hasn't changed. This is a logical inconsistency that proves the original problem must have a defined ratio. It's like saying that in the original problem it wouldn't matter which table you'd go to in a casino but if the suddenly the made both games harder then it would matter which table you went to.

CK sez:
But, as December rightly notes, it is NOT the case that im(SAn)/lim(SBn) = lim(SAn/SBn) as n--&gt; infinity. It can be true, but it is not necessarily true, if the limit is unbounded (or if the limit of the denominator is zero, for instance.)

But it is true in certain cases. For example the limit of y=x over y=2x as x goes to infinity. These are two straight lines that go to infinity just as our sums do yet their limit is 1/2. This is the exact same thing as the two series except it's continous.

I've shown that in the limit test two divergent series can have a ratio. Logically, if it's possible and these 2 series don't have a ratio then there must be some reason why. You still haven't shown me why they don't. ***It's not enough to say they can't because they're divergent!*** According to the limit test you CAN have divergent series with a ratio. If nothing else I would like you to answer this question for me.

I said you were wrong, and wrong you have been, in several matters (including thinking that SUM(An)/SUM(Bn) = SUM(An/Bn).) I did not insult you personally, and I see no reason for you to insult me.

You were being patronizing, and that's insulting.

Apart for the SUM(A)/SUM(B) = SUM (A/B), which I corrected myself and which wasn't even necessery to my proof, what have I said that was wrong? Where are these "several matters"? Why do you see it so important to latch on to something so trivial as that little mistake which had nothing to do with my proof? Is it because you need to prove to yourself that you are correct? Because you can't find any holes in my reasoning apart from saying "No, you can't do that because it's divergent." Do you have any proof that you can't use normal alegbra in those cases? You're just saying "No, you can't". You can't write 2SUM(A) = SUM(2A)??? Says who?

Konrad, I see no reason for the gratuitous insults. I said you were wrong, and wrong you have been, in several matters (including thinking that SUM(An)/SUM(Bn) = SUM(An/Bn).) I did not insult you personally, and I see no reason for you to insult me. You want to do that, I'll meet you in the Pit.

Ha! You thin-skinned weenie! You call those gratuitous insults? Come to the pit and I'll give you such a mocking that the very earth will tremble! When I'm through with you statisticians will scare their children by telling them that if they don't behave CKDextHavn will teach them mathematics.

[C K Dexter Haven]

Aha! Enlightment comes.

Konrad persists: &lt;&lt; The expected payoff for a game is SUM(probability of getting to n flips)(payoff for n flips). &gt;&gt;

Well... sort of. The expected payoff of Game (1) is the INFINITE SUM (probability of getting to n flips) x (payoff for n flips), as n goes from 1 to infinity. And that value is unbounded.

Ditto for Game (2).

&lt;&lt;Since we are just finding the ratios of expected payoff for n flips we KNOW they are not independent. &gt;&gt;

Here's where the enlightenment falls. Konrad and I are talking about different games. So, I say: NO. We are NOT just finding the ratio of expected payoff for n flips. That's a different question: If you flip the coin n times, what's the ratio of the likely payoff from Game(1) to the likely payoff from Game(2)?

That's the easy question, because if there are only n flips (n being finite), there is a finite sum in the numerator, and a finite sum in the denominator. That's a different game.

The game being asked about is, you flip a coin repeatedly UNTIL IT COMES UP HEADS, however many flips that may take. The Game Konrad is talking about is, you flip a coin n times where n is fixed.

Before you go throwing your insults around or taking offense because of the way you read my tone, Konrad, you might ask whether there is some base different understanding of what's going on.

I stand where I stood: in the game as stated, the expected value of numerator and denominator are unbounded and the ratio is not convergent.

Can one have infinite sums where the ratio is convergent? Sure. But this example ain't one of them.

[Konrad]

I am talking about what you originally thought I was. SUM(probability of getting to n flips)(payoff for n flips) from 0 to n as n goes to infinity is the expected payoff *per game*.

[Cabbage]

I'm starting to wonder if we're talking about the same thing, too:

If we were playing 2 games and trying to find out the ratio that way you would be right to say that we couldn't do it because one series could go faster than the other. What you are saying by writing that limit the way you have is that you have two independent games. That's not what we're trying to find here. We're trying to find the ratios of average payoff per play. When you try to find that you have to compare each payoff for a game of n flips.

Is this the source of the confusion? I can think of two scenarios:

1. Two people are at the same table, flipping the same coin, for game after game. One player gets the Game 1 payoffs, the other gets the Game 2 payoffs; however, winnings for both are determined by the same flips of the coin.

2. Two people are at different tables, flipping different coins, for game after game. One player is playing Game 1, the other, Game 2. Each plays an equal number of games, but with diferent coins, resulting in a different number of flips for corresponding games.

Konrad, I know we both agree on 1. At the end of each game, player 2 gets exactly twice as much as player 1. The ratio of winnings will always be 1/2, because their games have the same outcome (same number of flips for both players for each game) each time. The ratio is constant, so it will converge to 1/2, of course.

Do we agree on 2? Corresponding games will be different for each player. The first game, one might win $3, the other might win $54, and so on.

I agree that in 1. the ratio converges to 1/2. I've been arguing that in 2., the ratios will not be expected to converge, they will jump around, as the different players get succesively larger payoffs from time to time.

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[tony1234]

Just for the record, at about post 25 in this thread, I stated:

Originally posted by tony1234:
Reading the discussion, I realize (with embarrasment) that I slightly mistated the question. Perhaps, that is the explanation for some of the disagreement. With respect to Game 2, I should have designated the number of coin flips as "m" not "n." As many grasped, I did not intend to assert that the pay-off of Game 2 was based on the same number of coin flips as Game 1. More explicitly: Game 1 and Game 2 are played simultaneously *but independently.* Though the total winning for each game will go up, they need not increase at the same rate for any given segment of sequential plays.
Tony

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[C K Dexter Haven]

Thanks, Cabbage, I think you were more succinct than I in describing the two different games. And yes, Tony, I've been responding to what you were asking.

There is no question about the first interpretation. On any play of the game, player of Game 1 gets half the amount of player in Game 2, since the same toss governs both games.

So, now we await Konrad's response.

[zut]

The ratio of the payout of these two games is, as was correctly pointed out before, undefined, specifically because the payout is an *infinite*, *unbounded* sum. I think everyone is in agreement that the expected payoffs for the two games are:

Payoff[1] = SUM[n=1 to infinity][(3/2)^n]
Payoff[2] = SUM[n=1 to infinity][2*(3/2)^n]

which, of course, assumes that there are NO time or money limits; i.e., it is technically possible that a game could last for one flip, or ten flips, or a googol flips.

Now, consider the following:
P[2]/P[1] = S[1-inf][2*(3/2)^n]/S[1-inf][(3/2)^n]
where I've used some notational abbreviation that is hopefully obvious. I think everyone agrees with this statement. The problem comes in when you do this:

P[2]/P[1] = S[1-inf][2*(3/2)^n]/S[1-inf][(3/2)^n] = 2*S[1-inf][(3/2)^n]/S[1-inf][(3/2)^n] = 2

However, I can also say (pay attention here!) that

P[2] = S[1-inf][2*(3/2)^n] = S[1-inf][(3^n/(2^(n-1))] = S[1-inf][3*(3/2)^(n-1)] = S[0-inf][3*(3/2)^n] = 3 + S[1-inf][3*(3/2)^n]

so now
P[2]/P[1] = (3+S[1-inf][3*(3/2)^n])/S[1-inf][(3/2)^n] = 3/S[1-inf][(3/2)^n] + 3*S[1-inf][(3/2)^n]/S[1-inf][(3/2)^n] = 0+3

So, since 2=P[2]/P[1]=3, then 2=3. But 2 does not equal 3, so there must be something wrong here. The fallacy is the assumption that ratios of infinite, diverging sums are defined. QED.

If you disagree with this, I can only think of two reasons why:
1) You don't believe that the payoff for the game is really an *infinite* sum. ANS: If there is a real limitation on time or money, of any sort (like, you have to complete the game before the universe collapses at the end of time), then we're talking about a finite sum, and the ratio *is* defined, and the argument above is false. But that is *not* the game I'm talking about, and I don't think it's the game the OP was talking about, either.
2) You think there is some trickery in the mathematical manipulation, or a "loss" of terms. ANS: Nope, everything is legit for an infinite sum. If you don't believe me (and I suppose there is no reason why you should, other than the stellar clarity of my arguments), look up the following: Bryan Bunch, 1982 "Mathematical Fallacies and Paradoxes", Dover, ISBN 0-486-29664-4. This contains a pretty complete discussion of the fallacy of comparing infinite sums, and also a short discussion of the Petersburg paradox (which is what this thread is about).

[tony1234]

Zut,

Welcome to the conversation.

Quick question from a person who took only two courses in college math 22 years.

We all agree that *if* Game 1 and Game 2 were dependent, i.e., their payoffs depended upon the results of the same coin-flipping event, the ratio of their payoffs would be 1/2. Is your proof inconsistent with that fact? In other words, where in your proof is the assumption that the games are not dependent?

In any case, I look forward to studying your proof more carefully when I have some more time, as well as review the article you recommend.

Finally, do you know how to quanitify (mathematically express) the long-run advantage in aggregate expected payoffs of Game 2 over Game 1. If there will be no fixed ratio approached, what can be said with precision about their expected comparative payoffs? (I realize that Cabbage early opined on this question in general terms.) For example, is there an upper bound of their expected ratios as the number of plays increases?

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

Tony: Zut's contradiction starts with:
P[2]/P[1] = S[1-inf][2*(3/2)^n]/S[1-inf][(3/2)^n]

In the notation I was using, that's = SUM(An)/SUM(Bn)

This assumes that P[2] and P[1] are independent. If they are not independent (both are based on the same flips), then:
P[2]/P[1] = SUM[1-inf](2*(3/2)^n)/((3/2)^n)

That's SUM(An/Bn)... a different animal.

* * * * * * * * * *

The whole paradox in this game is that the mathematical "expected value" is counter to common sense. Common sense says that you will never get a string of a hundred tails in a row before you get the first heads.

Mathematics says, that's not quite right. If you flip the coin once a second for a few billion years, even waiting a second between tries after you got the first heads, you'll eventually get that string of 100 tails.

Common sense says, a few billion years is pretty much "never."

There are techniques to map the likelihood of outcomes, and thus to come up with a probability range for distributions. In computer simulations, if you're going to play the game 250,000 times for instance, your longest string of tails will likely be around 16 or 17.



[Note: This message has been edited by CKDextHavn]

[zut]

tony,

The proof I posed is strictly for independent games; for *dependent* games, it breaks down because the initial suppositions are no longer valid. For a dependent game, you are saying "Flip the coin, pay player one according to a certain set of rules, and then pay player two twice as much," which equates to:
Payoff[1] = SUM[n=1 to infinity][(3/2)^n]
Payoff[2] = 2*Payoff[1]
which, of course, is a different problem than the one that (I think) you originally posed.

A rather poor analogy (sorry, the best I can do) is this: You and I both roll one die each (independently). If the die comes up a 6, the rolling player receives one dollar. Now, if we play n times, what is the probability we each walk away with the *same* amount of money? Intuitively, this probability decreases with increasing n, and is always less than 100%. OK, what about if money is paid to *both* of us based on the *same* die roll? Then the probability we each walk away with the same amount of money is exactly 100%, always, irrespective of n. Let me emphasize that this example is not meant to reflect on the actual probabilities in the Petersburg paradox (or even ratios of probabilities), but rather is intended to illustrate how problems change according to initial assumptions about the dependence or independence of events.

Also, the long-run advantage in aggregate expected payoffs of Game 2 over Game 1 is...undefined. I realize that this is intuitively unsatisfying, but there it is. The problem is that the size of the payoff keeps increasing *faster* than the probability of achieving them decreases; more to the point, this increase continues to *infinity*. If there were any finite limit, you would expect a payoff ratio of 1:2. However, since there is no limit, there is the tendency for single games in a large series to upset the applecart by introducing extremely large payoffs.

Put another way, the more often you play, the more likely high payoffs (with low individual event probability) are to occur. Thus, no matter how often you play, your total payoff is highly influenced by the presence or absence of individual, low-probability events.

Put a third way, the method I used in my previous post can be (fallaciously) used to show that the ratio of payouts from games one and two can be *anything.* 1:2 or 1:3 or 3:1 or 1:100000000. So you can't even say that game 2 has an advantage over game 1. The only thing you *can* say is that you can't say anything.

Like I said, this is intuitively unsatisfying, but unfortunately it's the nature of infinity to not live up to our finite expectations.

[tony1234]

Zut,

"So you can't even say that game 2 has an advantage over game 1." ??? What is your reaction to Cabbage's post 594?

Tony

[zut]

tony,

Ummm...sounds to me like Cabbage and I are saying the same thing. Unfortunately, this is confused because I'm not sure which post you are referring to (for some reason, all of his/her [sorry, Cabbage, I'm new] posts are #594). However, it seems to me that Cabbage is saying that, if you were to actually play these two games independently, you would find that the payoff ratio is inconsistent. Moreover, it will jump around in response to individual large payoffs in one or the other of the games, and you *can't* predict where it will end up after x number of games, no matter how large x is. That's pretty much what I'm saying.

[zut]

CKDextHavn, I might disagree with you, depending on what you meant. You said,

Originally posted by CKDextHavn:
On a finite game level, however, where we can only play the game a dozen times or so, we know that the distribution of likely outcomes favours the game that pays double.

1) If you meant, "We restrict the number of tosses in a game to be less than some finite number" then I agree with your conclusion.

2) If you meant, "If we play both game(1) and game(2) a dozen times each, it is probable that the payoff from game(2) will be more than the payoff from game(1) more often than not" then I agree.

3) If you meant, "If we play both game(1) and game(2) a dozen times each, the total expected payoff from game(2) is more than the expected payoff from game(1)" then I disagree.

Upon rereading what you said, I suspect you meant option 2 above. However, there's a difference between "How much more will you win with game(2)?" (which has no answer) and "How much more *often* will you win with game(2)?" (which, I think, does have an answer). Which, upon rereading yet again, looks like your point. However, the difference between these two, although not subtle at all in mathematics, *is* rather subtle in an intuitive, verbal exchange.

[C K Dexter Haven]

&lt;&lt; However, there's a difference between "How much more will you win with game(2)?" (which has no answer) and "How much more *often* will you win with game(2)?" (which, I think, does have an answer). Which, upon rereading yet again, looks like your point. However, the difference between these two, although not subtle at all in mathematics, *is* rather subtle in an intuitive, verbal exchange. &gt;&gt;

Yes, fair enough. Common sense says that a game with a higher payout will probably pay more IF YOU PLAY OFTEN ENOUGH... but still a small (finite) number of games. Sorry for being sloppy in the wording.

Also, zut, I have removed your multiple postings. Once you hit SUBMIT REPLY, the reply has been sent... even though you may have to wait a bit to see it. Hitting SUBMIT REPLY multiple times, results in multiple posts. It is certainly a legitimate debating technique or political ploy, to repeat yourself over and over and over until your opponents give up in exasperation.... but it's not recommended here

[tony1234]

All of Cabbage's post are 594, you say? Well, now that you mention it . . . (I'm new to these parts myself).

What I was refering to was this:

Originally posted by Cabbage:

Overall, I guess what it amounts to is this: The ratios won't converge. However, the ratios will probably have a tendency to stay within a certain range (for example, the ratio may tend to stay between 1/4 and 3/4, but not actually converge to anything). It would be difficult to actually give a possible range for the successive values of the ratio, and the corresponding probability that the ratio will remain in that interval for a given length of time, but I think it could probably be done. I'm not going to try to do it though. :)

[/B]

Now, I entirely grasp the point that the ratio will not not converge (even if I don't entirely grasp the math--just starting to read up on it). But even it the ratios don't converge, there is no reason that something could not be said about the limit of the *average ratio* or the *likely* upper bound of the ratio as the number of plays goes to infinity. I don't think that you would deny this, Zut, but your earlier statement about there being no advantage to playing Game 2 seemed to imply you did.

Now, CDK has just stated,

"Tony, in terms of your latest question, whether we can't even tell whether Game(1) or Game(2) is better... the answer is in my last post. In a strict mathematical sense, no, we can't. The expected values are unbounded/infinite for both games."

Again, this seems to imply that probability theory (that branch of math intended to model our intuitions about expected outcomes in the presence of uncertainty) can't say anything in favor of Game 2. And again, I don't think CDK intends to say this much.

The intended interest of my original question was that even where the expected utilities two lotteries were equivalent (same order of infinity) we could meaningfully enquire about the relationship of their actual payoffs (in the long run), and perhaps, say something about the ratio of their payoffs. (At the beginning, I of course was not sure that there would be no convergence.) Even if these ratios don't converge, I assume that the preference we intuitively have for Game 2 will predictably be manifested over repeated plays of the game.

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[manhattan]

I apologize unreservedly. The above comment was totally uncalled for. I'd edit it out, but I believe that moderators should not edit their words, even the very stupid ones.

I is not your fault that our software is inadequate, and it is not your fault that the NASDAQ is cratering.

I will get to each and every multipost this evening, and return you to your thread.

[manhattan]

Heh. The comment didn't go through. Well, suffice it to say that it was wholly inappropriate and fully deserved the above apology.

[C K Dexter Haven]

I will try to post this only once, relying on Manhatt to eliminate the duplicate posts. I tried to do some of that, but the response time was too slow (at five minutes per deletion.... well...)

You can create a probability distribution, as follows:
GAME(1):
Probability of payoff of $3 = 50%
2 tails;ity of payoff of $9 = 25%
Probability of payoff of $27 = 12.5%
Probability of payoff of $81 = 6.25%
Probability of payoff of $243 = 3.125%
Probability of payoff of $729 = 1.56325%
Probability of payoff of $2187 = 0.78125%
Etc etc

Thus, you can see that there is a 98% likelihood that the payoff is $2187 or less.

Similarly for Game 2, with a 98% likelihood that the payoff is $4374 or less.

So, what would be a "fair" charge for you to play? Well, if you played 100 times, (assuming that your flips were exactly according to the theoretic distribution) you would probably have won around $4826. So, a
"fair" price would be around $48 per play. A good deal. However (and this is the point of the paradox) ... you could have had that wonderful string of 10 tails that would have won you $59,000 or so.

[zut]

Originally posted by tony1234:
Now, I entirely grasp the point that the ratio will not not converge (even if I don't entirely grasp the math--just starting to read up on it). But even it the ratios don't converge, there is no reason that something could not be said about the limit of the *average ratio* or the *likely* upper bound of the ratio as the number of plays goes to infinity. I don't think that you would deny this, Zut, but your earlier statement about there being no advantage to playing Game 2 seemed to imply you did.

Well... maybe I would deny it. Depends on what you mean. I don't think you can say anything about the "average ratio" because, just like the game itself, the average ratio is skewed by individual, low probability, high payoff events. (Now, I'll admit that this is about the edge of my knowledge/intuition/experience, so I'm willing to be wrong here.) However, I think you could say something about the *median* ratio, which I'd bet was about 1:2.

Originally posted by tony1234:
The intended interest of my original question was that even where the expected utilities two lotteries were equivalent (same order of infinity) we could meaningfully enquire about the relationship of their actual payoffs (in the long run), and perhaps, say something about the ratio of their payoffs. (At the beginning, I of course was not sure that there would be no convergence.) Even if these ratios don't converge, I assume that the preference we intuitively have for Game 2 will predictably be manifested over repeated plays of the game.

I would say that if you ran a large number of trials, Game 2 will outperform Game 1 *most of the time*. You can (I think) even calculate the probabilities of *how often* this would happen. However, you cannot say *by how much*. You cannot even say that the total amount won in Game 2 is more than in Game 1.

[Konrad]

OK let me explicitly state what I'm calculating. We have two players they each go out and play one game to completion. One player has the first game and the other has the second.

What is the ratio of the expected winnings of player 1 to player 2, for one complete game.

That, I believe, is what we were all talking about from the beginning.

I'm still waiting for a response to my post pointing out the inconsistencies when you flip a coin twice per turn.

Also I think I've established that you can have two things that tend to infinity that have a defined ratio. (Just making sure we all agree here, right? Limit test, remember?) So if it is possible you have to prove why it isn't so in this case. The burden of proof is on you to show why I can't use regular algebra. It's one thing to say 3SUM(x) doesn't equal SUM(3x) but if you say that then you have to give a reason why.

Again, if you respond to nothing else in this post tell me why regular algebra doesn't work in this particular case. And don't just say because they're divergent, I've already shown that something which goes to infinity/infinity can work with regular algebra. You can say 2(1 + 1) = 2*2 and I can object that you can't do that. But that's not a proof. I can't just say "You can't do that" and have a proof.

[tony1234]

[QUOTE]Originally posted by zut:
[B]The ratio of the payout of these two games is, as was correctly pointed out before, undefined, specifically because the payout is an *infinite*, *unbounded* sum. I think everyone is in agreement that the expected payoffs for the two games are:

Payoff[1] = SUM[n=1 to infinity][(3/2)^n]
Payoff[2] = SUM[n=1 to infinity][2*(3/2)^n]

which, of course, assumes that there are NO time or money limits; i.e., it is technically possible that a game could last for one flip, or ten flips, or a googol flips.

Now, consider the following:
P[2]/P[1] = S[1-inf][2*(3/2)^n]/S[1-inf][(3/2)^n]
where I've used some notational abbreviation that is hopefully obvious. I think everyone agrees with this statement. The problem comes in when you do this:

P[2]/P[1] = S[1-inf][2*(3/2)^n]/S[1-inf][(3/2)^n] = 2*S[1-inf][(3/2)^n]/S[1-inf][(3/2)^n] = 2

However, I can also say (pay attention here!) that

P[2] = S[1-inf][2*(3/2)^n] = S[1-inf][(3^n/(2^(n-1))] = S[1-inf][3*(3/2)^(n-1)] = S[0-inf][3*(3/2)^n] = 3 + S[1-inf][3*(3/2)^n]

so now
P[2]/P[1] = (3+S[1-inf][3*(3/2)^n])/S[1-inf][(3/2)^n] = 3/S[1-inf][(3/2)^n] + 3*S[1-inf][(3/2)^n]/S[1-inf][(3/2)^n] = 0+3

So, since 2=P[2]/P[1]=3, then 2=3. But 2 does not equal 3, so there must be something wrong here. The fallacy is the assumption that ratios of infinite, diverging sums are defined. QED.

Zut,

I'm not convinced by this reasoning (even though I am strongly inclined to the *conclusion* that the ratio of average winnings will not converge). The problem with your reasoning is that P[2]/P[1] is the ratio of the expected winnings on a single play. I'm willing to agree that's undefined. The issue, however, is not what is the ratio of the expected winnings; it's what is the *expected ratio of the actual winnings*. Perhaps these are the same, but I don't think you have proved it.

By the way CKD, the real point of the StP paradox is not that the value of the lottery is greater than any finite number (because you can get an unlimited string of tails), but that the value of the lottery is greater than any finite value *and* the actual return will always be less than expected, i.e., a finite amount. Computing expected utility in the usual way (sum of discounted possible returns) will always result, paradoxically, in defeated expectations.

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[Cabbage]

Konrad:

In the variation you mentioned of flipping the coin twice per turn, the expected payout is still infinite, (SUM) (3^n)*[3^(n-1)/4^n]. We'll still have the same problem.

I realize you can easily rig up a game where the expected payout will be finite, if that is what you were trying to do, so, for the sake of argument, I'll assume that you did.

That would be a BIG difference though. One way to interpret it is this way:

1. If the expected payout is finite, the average payoffs will tend to converge as more games are played.

2. If the expected payout is infinite (what we've been looking at), the average payoffs will tend to increase without bound as more games are played.

Earlier, December noted:
limit A(n)/limit B(n) = limit{A(n)/B(n)}

If both sequences diverge to infinity, or both converge to zero, then there is simply no meaning to the expression.

Konrad countered:
But it is true in certain cases. For example the limit of y=x over y=2x as x goes to infinity. These are two straight lines that go to infinity just as our sums do yet their limit is 1/2. This is the exact same thing as the two series except it's continous.

...but that's precisely an example of where it's not true!

lim (x/2x) = 1/2, but

(lim x)/(lim 2x) is undefined, (inf/inf).

An important thing to note here is that the former (the limit of a quotient) is easy to deal with, while the latter (quotient of limits) is not. The mistake you seem to be making is that you take the latter, think (mistakenly) that it's an indeterminate form, and proceed to rewrite it as the former, use algebra, and find a limit which never existed to begin with.

An indeterminate form, for example, would be lim [f(x)/g(x)], where both f and g go to infinity. This is not the same as [lim f(x)]/[lim g(x)]. The top is infinite; the bottom is infinite. In this case, we're not comparing "rates of increase" or anything, just attempting to divide infinity by infinity, period. It's undefined.

In the problem we've been dealing with, we're trying to make sense of the ratio of the two expected values, both of which diverge to infinity. In other words, we're trying to take the quotient of the limits, which, just as in the above example, is undefined.

Read a calculus text. You can break up a quotient of limits into the limit of the quotients only if all limits exist. ("Exist" implying they must be finite, as well).

To top it all off, even if we try to make sense of it by rewriting it as a limit of quotients (which, really, we can't do to begin with, but, for the sake of argument, I'm willing to go along with here), we can get different values (2, 3, ...) as pointed out before, by algebraic manipulation of the sums.

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[C K Dexter Haven]

The "post number" changes each time, to reflect the total number of posts by that poster. I know, I know, but complain about it in a different forum. That's the way it works. So you can't identify someone's post by the number.

I think at this point everyone is agreed, except for Konrad, from whom we are waiting to hear now that we think we figured out the difference.

Tony, in terms of your latest question, whether we can't even tell whether Game(1) or Game(2) is better... the answer is in my last post. In a strict mathematical sense, no, we can't. The expected values are unbounded/infinite for both games.

On a finite game level, however, where we can only play the game a dozen times or so, we know that the distribution of likely outcomes favours the game that pays double. Especially if you're using a computer simulation random-number generator that always follows the same sequence (which puts us in the case of really using the same coin toss for both games, rather than independent tosses!)

[Cabbage]

Konrad:

I've reread some of your posts and think that I finally see the approach you're taking to the problem:

OK let me explicitly state what I'm calculating. We have two players they each go out and play one game to completion. One player has the first game and the other has the second.

What is the ratio of the expected winnings of player 1 to player 2, for one complete game.

I take your last sentence to mean: After playing one game of each, calculate the expected ratio of the game 1 payout to the game 2 payout. Is that a fair interpretation?

OK, lets try to calculate the expected ratio. To find the expected ratio, we need to take a possible ratio, multiply that by the probability that it actually occurs, then sum over ALL possible ratios, correct?


Say we get m flips for Game 1, n flips for game 2. The ratio for this possible payout is:

(3^m)/[2*3^n].

The probability this will happen is:

(1/2^m)*(1/2^n).

The expected ratio is therefore:

(SUM over m)(SUM over n) {(1/2^m)*(1/2^n)} * {(3^m)/[2*3^n]}

However, this diverges to infinity. To see this, fix n to be 1, and sum over m. It should be clear that this is smaller than the expected ratio. But this is a geometric series with common ratio 3/2, so it diverges.

Therefore, the expected ratio diverges. It doesn't exist.

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[Konrad]

OK I think I see the problem. You're still thinking of independent games.

...but that's precisely an example of where it's not true!

lim (x/2x) = 1/2, but
(lim x)/(lim 2x) is undefined, (inf/inf).

But we're not talking about lim(x)/lim(2x). We are talking about lim(x/2x).

We are not comparing the total amount of money you would win in one game or the other if you played to every possible value of n. We are comparing the expected payoff ratio for one game.

To find the expected ratio, we need to take a possible ratio, multiply that by the probability that it actually occurs, then sum over ALL possible ratios, correct?

Don't know why you're doing that... The obvious formula for calculating it would be:
SUM(probability of a result)(payoff for the result for game1)
------------------------------
SUM(probability of a result)(payoff for the result game 2)
Where both sums go from 0 to n as n goes to infinity

Since the probability of getting to n flips for each game is the same we can just ignore that part. Basically it comes down to SUM(payoff for game 1)/SUM(payoff for game 2) from 0 to n as n goes to infinity

Now the reason I can say that we're summing both from 0 to n instead of summing them independently is because I'm not just talking any two random games here. I'm talking about the probable ratio which is found by that first formula. It's because those two series are both being summed by the same n that I can say they have a ratio.

It's like one person travelling a 1 meter a second and another a 2 meters a second. An infinite number of seconds later the ratio of their distances is

SUM(1)/SUM(2) from to 0 to t as t goes to infinity. Sure both sums are divergent but since both sums are linked to the same n they have a ratio. One guy obviously went twice as far as the other.

[Konrad]

Oh yeah and the flipping the coin twice per turn thing:

The expected payoff for one game would be:
SUM(1/4^n)(3^n)= SUM((3/4)^n) = 1/(1-3/4) = 4

Likewise for the second game the expected payoff would be 8.

Here's the 2:1 ratio is obvious. You may say that I've changed the game so that the payoff converges now. Yes, but I haven't changed the ratios of the payoffs. I've made the game harder but each player's winnings relative to the other guy hasn't changed.

[Cabbage]

First, for the game of flipping the coin twice per turn: The game was you flip the coin twice each time, and don't win until two heads come up, correct?

Then, for a given n, the probability of winning 3^n is:

(1/4)*(3/4)^(n-1).

You have to lose n-1 times (3/4 chance each), then win the nth time (1/4) chance. The expected value of the payout is as I said it was earlier--a divergent sum. However, as I also said before, this is a minor point, I do realize you could come up with a game with a finite expected payout, but it's irrelevant; the whole point here is that the expected payout is divergent.

I said:
To find the expected ratio, we need to take a possible ratio, multiply that by the probability that it actually occurs, then sum over ALL possible ratios, correct?

Konrad said:
Don't know why you're doing that... The obvious formula for calculating it would be:
SUM(probability of a result)(payoff for the result for game1)
------------------------------
SUM(probability of a result)(payoff for the result game 2)
Where both sums go from 0 to n as n goes to infinity

I'm treating the ratio as a random variable. Do you know what the definition of the expected value of a random variable is? It's

(SUM over ALL possible values) (Value)*(Probability of that value)

It may seem intuitively obvious to you to use only the ratios that are 1/2 (or whatever it is you're doing), but how can you possibly justify doing that? I'm merely using the definition, do you have any justification that I shouldn't be?

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[tony1234]

Two comment on a thread that has gone on much longer than I anticipated. (Longest thread of the month at least, I think.)

1. Konrad writes, "It's like one person travelling a 1 meter a second and another a 2 meters a second."

Excuse me, but my original problem is so obviously not like this profered analogy that it barely seems useful to state why. But: in Konrad's example, at every point in time, it is certain what the ratio of the distiances will be; in the "real" problem, the ratio of winnings will be jumping all over the place. Some convincing arguments have been advanced why the ratio won't settled down. (Look back to some of Cabbage's posts around post-40 in the thread).

Second (addressed primarily to Zut), I greatly doubt that the ratio of the expected returns of each game (where "expected return" is defined as the sum of all the possible returns discounted by their associated probability) has anything to do with the answer to the problem. Zut (and perhaps others) seems to think that because this ratio is the ratio of infinity/infinity and so is undefined, the expected ratio of total winnings, as the number of plays increase, is also undefined. But consider this example:

Game 1: Payoff 3^n, if n flips of fair coin.
Game 2: Payoff (3^100n!)!, if n flips of fair coin.

Here to the ratio of expected returns is also infinity/infinity, but I strongly suspect that (the total winnings of Game 1 after p plays)/(the total winnings of Game 2 after p plays), as p increases, will converge to zero (and not be undefined).

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[C K Dexter Haven]

Konrad: &lt;&lt; Again, if you respond to nothing else in this post tell me why regular
algebra doesn't work in this particular case. &gt;&gt;

Regular algebra does work, but you've misstated it.

Let's start with the definition of the infinite sum: SUM(An) as n--&gt; infinity =
lim(n--&gt; inf)[PARTIAL SUM(j = 1 to n)(Aj)]

Let's use limn to mean the lim as n--&gt; inf; and let's use PSn(Aj) to mean the partial sum of the terms Aj, with j = 1 to n. So the infinite sum SUM(An) = lim(n--&gt;inf)[PSn(Aj)] ... damn the inability to write normal math notation!

OK, Konrad, when you assert SUM(2*An) = 2*SUM(An), you are asserting that lim(2*PSn(Aj)) = 2*lim(PSn(Aj))... and this is only true (epislon-delta proof) IF THE LIMITS EXIST.

[C K Dexter Haven]

OOPS! That's what I get for telling my mother how to work the TV/cable remote while I'm typing math stuff.

Please adjust the last paragraph of my comment above. It's not the SUM(2*An) = 2*SUM(An) that's the problem. The problem is that the limit of the quotient (or product) is not necessarily the quotient (or product) of the limits.

Easy example: Take n--&gt;infinity, then look at:
lim(n) = infinity, unbounded, undefined;
lim(1/n) = 0;
But lim(n*1/n) = 1.

Normal algebra doesn't always work. It doesn't work when you are dividing by zero (for instance); and it doesn't work when you are dealing with infinite quantities and treating them as if they were real numbers.

[Konrad]

Well don't have much time to respond now cause it's exam time. I'll get back to this after I'm done with my semester but my main point is that you guys are still talking about 2 unrelated games here. When we're talking about the expected ratio we take into account every possible game so the series are not seperate in the first place. It's exactly like that problem with two people walking at different speeds.

[Cabbage]

For when you get back:

It's exactly like that problem with two people walking at different speeds.

No, it's not, because their speeds are constant. That's not at all analogous to what we have here.

To set up the runner's problem the same as the original problem we would need it this way:

We have two runners, A and B.

A flips a coin until it comes up heads (for the first time, on the nth flip), then advances 3^n yards.

B flips a different coin until it comes up heads (for the first time, on the mth flip), then advances 2*3^m yards.


There's a big difference between this and having the runners run at constant rates.

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[C K Dexter Haven]

If the crux of the matter is the question of whether the games are independent or not, then I guess we're in agreement.

IF the games are dependent (for example, one coin is tossed until it comes up heads, and then a payout based on that coin is give to the Player of Game 1 and the Player of Game 2), then the ratio is clearly 1/2.

IF the games are independent (one coin is tossed for Player of Game 1 and ANOTHER coin is tossed for Player of Game 2), then there is no defined ratio. I think the OP has made clear that this is the situation he meant.


And, yes, Tony, thanks for clarifying what I meant to say. The point of the paradox is NOT that either game has an unbounded expected value; it's not hard to drum up a game with infinite expected value. The paradox is that, with this infinite expected value, "common sense" says that you wouldn't pay $3000 per round to play this game.

And, just in case there's confusion, I am not saying that lim(An)/lim(Bn) is NEVER equal to lim(An/Bn), I am saying that you can't take that for granted.

My example of multiplication was incorrect. SUM(2*An) = 2*SUM(An) isn't the problem; it's dividing SUM(An)/SUM(An) that's the problem. Like dividing by zero, dividing infinity by infinity is undefined and can lead to paradox.

And now this thread is finally done, and I'll expect Konrad's apology for saying that I was pre-Riemannian. I did have a class under Zygmund, back in the 60s, but it was the 1960s.

[zut]

Konrad, you say:

Originally posted by Konrad:
...my main point is that you guys are still talking about 2 unrelated games here. When we're talking about the expected ratio we take into account every possible game so the series are not seperate in the first place.

OK, granted. Your contention that there is a 1:2 ratio of expected winnings is certainly true if you're talking about payoffs based on the *same* coin flips, which is what (I think) you're talking about. However, the OP clarified,

Originally posted by tony1234:
With respect to Game 2, I should have designated the number of coin flips as "m" not "n." As many grasped, I did not intend to assert that the pay-off of Game 2 was based on the same number of coin flips as Game 1. More explicitly: Game 1 and Game 2 are played simultaneously *but independently.*

(The emphasis is tony's.) So most of the rest of us are talking about *independent* games, which, in my opinion, is more interesting to discuss anyway.

And, speaking of tony, he also says (to me),

Originally posted by tony1234:
I'm not convinced by [zut's] reasoning (even though I am strongly inclined to the *conclusion* that the ratio of average winnings will not converge). The problem with your reasoning is that P[2]/P[1] is the ratio of the expected winnings on a single play. I'm willing to agree that's undefined. The issue, however, is not what is the ratio of the expected winnings; it's what is the *expected ratio of the actual winnings*. Perhaps these are the same, but I don't think you have proved it.

Ummm... I *think* I see the disconnect. You're saying, "the expected winnings from game one is infinity; the expected winnings from game two is infinity; the ratio of infinity to infinity is undefined, and thus ratio of the expected winnings is undefined, and I can accept that. However, since any two *specific* games, once played, clearly *have* a winnings ratio, it is not clear that the expected ratio does not converge."

Is this a fair paraphrase of your argument? If so, I'd argue that the ratio does *not* converge, because any one individual game has the potential to generate an astronomical payoff (which is why the expected winnings are infinite). This astronomical payoff would result in a huge ratio, which would then severely skew the *average* ratio.

However, although I think you can't say anything about the *average* ratio, you could say something about the *distribution* of ratios. For example, by symmetry, the median expected ratio (meaning, over time, an equal number of the ratios are above and below this number) is 2:1. With some work, you could find the range within which, say, 99% or 99.9999% of the ratios fall. It's just the outliers that screw up the *average*.

You also want me to...
Originally posted by tony1234:
...consider this example:

Game 1: Payoff 3^n, if n flips of fair coin.
Game 2: Payoff (3^100n!)!, if n flips of fair coin.

Here to the ratio of expected returns is also infinity/infinity, but I strongly suspect that (the total winnings of Game 1 after p plays)/(the total winnings of Game 2 after p plays), as p increases, will converge to zero (and not be undefined).

Nope; undefined. You're confusing a really big number (in this case, (3^100n!)!) with infinity. In your original problem, the ratio is undefined because infinity over infinity is undefined; having a payoff of (3^100n!)! doesn't make it a "bigger" infinity. However, the *median* ratio will certainly be much smaller in this second case (not zero, but something pretty darn small.), as will the 99.999 percentile ratio and so forth.

[panamajack]

Okay, so I'm new to this board, but I've read almost every post so far and the discussion seems to have gotten caught up in series, sums, etc. and moved a bit off of the probability angle. So here's an explanation, without using sums (except in the background of the theory.)
:

(In this, a game represents a specific method of play, i.e. flip until heads comes up. A match will be a series of games. )
It should be clear that from a probability standpoint, the two games are exactly the same (We're not talking about payoff yet.) You flip until you get heads.

The expectation (defined, at any rate) is the mean value of the probability distribution. (1) See below for more. It represents essentially what happens in an infinite match of games, and is useful to study, among other things, likely payoff. In this case, the probablity distribution is well-known and is called the Geometric Distribution, defined as :

Waiting time T until first success of an event in Bernoulli trials (2) with probability p.

The expected value or mean of this distribution is 1/p . In other words, for this particular game, you'll have an expectation of 2 turns until you win. This isn't particularly important, but I thought you might like to know that this much of the game is a known distribution. This is just to explain that much, and to make it clear that the two games, if played infinitely without a particular payoff, will have to end up exactly the same.

Now to go to the payoff. It makes absolutely no difference to the finaly payoff if you pay off one game at a time, or just wait until the match is over, and figure the payoff. So for both of these, if you play a match of finite length, you'll get some set of numbers at the end representing winning n, and pay off. If you used the same match numbers, you'd of course get twice the payoff using Game 2. But if the numbers are different (different matches) you can't expect to get the same payoff, since you'll have a different set of numbers.

Imagine a giant board (in fact it's infinite) that has every single integer on it. Let's say you put a marker on the board every time you play a game and reach that number. So after 10,000 games, you'd have a lot of markers near the bottom, and a few out over the other numbers. And if you have Board 1 and I have Board 2, they won't necessarily look the same.
Now let's keep playing (forever, in fact). It should be clear that as this happens, my board and your board will start to look pretty similar; in fact, at inifinite time, they must be equal. (3). All the spaces will be covered by markers, and the ones that are more covered will be the same. (If you aren't seeing this, then review your concept of infinity and probability distributions). So if we pay off at infinite time (essentially what the OP was asking, since paying off after time and paying off as time passes are the same), we see that Board 2 will have twice the payoff of Board 1. Please follow up with any questions, but I have to admit I'm not that great at explaining this further.

panama jack

(1) The probability distribution is the way some random variable X looks in some range. It can be discrete or continuous, and has a value equal to the probability that X= some particular x in the range given. Ranges of infinite value are often considered. The normal distibution, or bell curve, is the most well-known probability distribution (for good reason, too, since every infinite set of distributions approaches it).

(2) This is mostly trivial, but Bernoulli trials are independent, and it must satisfy p+q=1 where q represents (not p occurs).

(3) If not, we can throw probability theory mostly out the window. The meaning of that statement, though, is that a probability distribution should work any time you use it, and represents essentially an infinite amount of events. There are in fact some who argue that we have to be careful about this, but it's more philosophical. As an example of the problem, consider a weatherman's prediction that there's a 20% chance of rain tomorrow. Does this really have a concrete distribution. After all, there's only one day that will ever be June 4, 2048 and so it's difficult to say that this 20% represents a long term or infinite number of trials of June 4, 2048.

A few more notes :

Let's look at the boards again. Imagine that instead of putting a marker on the number, we put a payoff on the number (like a casino game). It's clear that the larger numbers will have huge sums stacked on them even after one win on them, while the low numbers will rack it up slowly. Still, one win at the large number will skew the running total by quite a lot. This is why computer simulation likely won't work, and partly why some people think it doesn't converge. The thing is, it does, but only at infinity. (It's not really like a lot of things that do become bounded as time goes on.)

[zut]

panamajack,

The thing you forget here is that you're discussing infinity, which is not like a regular real number. You seem to imply in your post that a) board one has an infinite number of markers on it, b) board two has an infinite number of markers on it, c) board two pays twice as much per marker, and d) (2*infinity)/infinity = 2. You can't "cancel out" infinity because 2*infinity = 3*infinity = googol*infinity = infinity. I offered a particular proof of this earlier in the thread. I realize this property of infinity is not intuitive (in fact, it's probably counter-intuitive), but that's the way it is.

[C K Dexter Haven]

Panamajack, I like your board analogy, but I think the point at which it fails is when you look at it after infinite time.

Take the two boards where you put the winnings on each number as they occur.

Forget about the "distribution" at that point of the two boards; ask what's the TOTAL? And the answer is, there's a (countable) infinite number of markers on each Board. It's like asking the question, what's the ratio of even natural numbers to ALL natural numbers -- that ratio is NOT 1/2. There are just as many even natural numbers as there are all natural numbers, even though the one is a proper subset of the other.

Again, there are situations where it is possible to take the ratio of infinite sums and get a finite sum, but I don't think this is one of them.

[tony1234]

Good. The conversation has gotten interesting again (I believe).

First, briefly, CDK. It's not really important, but you have mistaken my point about the "paradoxical" nature of the StP. Paradox. Your's is the common line on the Stp. paadox: StP is paradoxical because the exptected payoff is infinite while "common sense" says it's limited. ($20 seems to be the avearge people are actually willing to pay.) But I'm saying it's paradoxical because the expected payoff according to standard utility theory (infinite) is inconsistent with (not common sense) but what in fact must happen. The payoff for a given play must be finite even though its expected value is infinite. Usually, it's at least possible that the expected payof and the actual payof will be the same. I think it's bizarre that the expected payoff will always be more than the actual payoff and you can know this in advance. You know your expectations are wrong in effect. That's paradoxical.

Whoops. I've been notified that zut has just posted something. I'll send this, read zut's and then respond to it.

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[december]

A good many sensible posts have clarified what this problem means and why the "obvious" answer of 1/2 is not so obvious. Now several posters have gone back and asked what IS the correct answer.

Since the limit of the two ratios is a random variable, it need not always converge to the same figure. In fact, it's easy to see that certain possible specific sequences of flips could give many different numbers as a limit. But, all of these specific sequences would have probability zero.

My guess is that the ratio fails to converge, except on a set of measure zero. That is, with Probability = 1. the ratio does not converge.

However, I do not know how to demonstrate this result. Maybe a proof by contradiction:

Let F(k) be the average result of Game 1 after k trials and G(k) = the average result of Game 2 after k trials.

Suppose limit {F(k)/G(k)} = r. Then, for any epsilon, there exists a K such that F(k/G(k) is within epsilon of r for all k &gt; K

Now, I would like to show that with probability = 1, there would be some k for which F(k)/G(k) is not within epsilon of r.
Presumably this would happen because one or the other sequence would be expected to change drastically, due to a very run.

Anyone care to complete this proof or to offer a better one?

[zut]

CKDex,
I like your proof because of a corollary I didn't expect: You can rewrite the proof so that Player 3 gets the ratio of winnings of Player 2 to Player 1 (i.e., the *multiplicative inverse* of the way you wrote it). In this case, following the same logic, Player 3's expected winnings are *still* undefined and unbounded. Pretty neat.

Tony, I think you've basically summed up my position. However, in response to:

Originally posted by tony1234:
It seems to me that we have to at least ask what the likelihood is that Game 1 will produce a payoff that will screw up the ratio. If may be that the more plays there are, the greater lead that Game 2 will build up so that it becomes decreasingly likely that Game 1 will ever be able to have a payoff significantly screwing up the ratio. Of course, there is always the possibility that that it might, but if the possibility is expected to decrease, then you will get your convergence.

But the more plays there are, the more likely Game 1 is to have an *even bigger* payout, which affects the ratio even more. So, in some sense, the chance for Game 1 to "significantly screw up the ratio" keeps pace with the number of plays.

[tony1234]

Zut,

Originally posted by zut:

But the more plays there are, the more likely Game 1 is to have an *even bigger* payout, which affects the ratio even more. So, in some sense, the chance for Game 1 to "significantly screw up the ratio" keeps pace with the number of plays.

I agree that the more plays there are the more likely Game 1 is going to have an even bigger payoff, but that doesn't that the odds of it coming up are great enough to keep pace with the number of plays. It may grow, but still not keep pace. Or at least, I need I proof, like CKD's that it won't. I don't think you can simply wave infinity/infinity in the air and say the ratio is undefined. If so, CDK's long attempt at a proof is unneeded. Do you think it's unneeded?

Ok, I don't need a formal proof that it will keep pace, but I either need a proof, a convincing narrative, or consensus from the intelligent participants on the thread. (Please don't take my scepticism personally.)

Some of this message could be more clear, but I hope my point comes across,

Tony



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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[Cabbage]

Dex:

That was basically the same approach I was trying to use in my 4/15 1:56 AM post.

So the question was raised: What does this mean?

What does it mean that the expected value for the ratio is unbounded?

I think it means that if you play the two games, take the ratio of the payoffs, and do this again and again, you will get a sequence of ratios whose average value will continually increase. In other words:

[(p1/q1) + (p2/q2) + ... (pn/qn)] / n

goes to infinity as n goes to infinity (pi is the payoff for the ith game of Game 1, qi is the payoff for the ith game of Game 2).

The question that occurs to me, though, is this. How does this relate to the ratio of the Total Game 1 payoff to the Total Game 2 Payoff. In other words, how does the above expression relate (if at all) to:

(p1 + p2 + ... + pn)
--------------------------
(q1 + q2 + ... + qn)

as n goes to infinity?

I think the latter expression seems more akin to the nature of the OP.

I haven't tried working with it, but my conjecture is: The fact that the former expression goes to infinity probably implies that the latter expression doesn't converge. (Doesn't necessarily go to infinity, but just bounces around).

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[C K Dexter Haven]

Since all the terms in this series are positive, I think that "unbounded" and "converge to infinity" and "undefined" are all the same. I have been using "unbounded" in the mathematical sense of "increasing without bound" because the phrase "converges to infinity" is like chalk on a blackboard to me. OK?

Thus, I am using "unbounded" in the sense of "summing to countable infinity" or "to distinguish from "divergent but bounded" like the alternating (+ 1 - 1) example.

Cabbage, today is kinda rushed, but I think your question is about lim(SUM(An)/SUM(Bn)) compared to lim(SUM(An))/lim(SUM(Bn)) and I think in this example that both diverge (OK, OK, "converge to countable infinity" or "are unbounded.")

As you say, Tony, an interesting and challenging set up. Part of the problem here I think has been understanding what the "ratio" of winnings means, when we're dealing with an infinite sequence of rounds. However, I think I demonstrated that the expected ratio (expressed as an infinite series) has a sub-series that is unbounded ("converges to infinity," Lordy, I hate that term) and (since all elements are positive) so it must also be unbounded.

Anyway, this topic now has almost an unbounded number of posts.

[tony1234]

Originally posted by CKDextHavn:
Part of the problem here I think has been understanding what the "ratio" of winnings means, when we're dealing with an infinite sequence of rounds. However, I think I demonstrated that the expected ratio (expressed as an infinite series) has a sub-series that is unbounded ("converges to infinity," Lordy, I hate that term) and (since all elements are positive) so it must also be unbounded.
[/B]

CKD,

Not to be too picky, but just because a series has an unbounded subseries embedded within it, it doesn't mean the series is unbounded. Its limit may be undefined. For example, the series 1, 0, 2, 0, 4, 0, 8, 0, 16, 0 . . . . has an unbounded subseries, but the series does not converge (even at infinity). The limit of the series is undefined. So we are still stuck between "increases without limit" and "undefined" in the "indeterminate" sense.

If this is just a quibble, nevermind. But it make be relevant to the bottom-line of your analysis.

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[Cabbage]

Dex:
Cabbage, today is kinda rushed, but I think your question is about lim(SUM(An)/SUM(Bn)) compared to lim(SUM(An))/lim(SUM(Bn)) and I think in this example that both diverge (OK, OK, "converge to countable infinity" or "are unbounded.")

No, actually, I was conjecturing that, if the expected ratio is unbounded, then the total ratio (ratio of the sums of the payoffs) is divergent. In other words: If

lim [SUM(An/Bn)] / n

is unbounded, does that imply

lim {[SUM(An)] / [SUM(Bn)]}

is divergent? (The first comes from the fact that the individual ratios per game will have an unbounded average, which is what you proved. The second is what the OP asked--Will the total winnings per game have converging ratios?)

I agree that both statements are true in this case, but I was wondering about the general case. If it is true in the general case, then we would also have a proof that, for example, Tony's variation on the game involving the factorial payoffs would also involve divergent ratios. Anyway, I'll try to look into it.

Tony:
Not to be too picky, but just because a series has an unbounded subseries embedded within it, it doesn't mean the series is unbounded. Its limit may be undefined. For
example, the series 1, 0, 2, 0, 4, 0, 8, 0, 16, 0 . . . . has an unbounded subseries, but
the series does not converge (even at infinity).

I believe you're confusing "sequences" with "series". As a sequence, what you said is true--the sequence does not diverge to infinity. As a series (sum of terms in the sequence), it's not true (when we're dealing with all positive terms, anyway). In other words:

The "subseries"

1+2+4+8+... is unbounded,

and so is the original series:

1+0+2+0+4+0+8+...

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[tony1234]

Originally posted by Cabbage:
Dex:
I believe you're confusing "sequences" with "series". As a sequence, what you said is true--the sequence does not diverge to infinity. As a series (sum of terms in the sequence), it's not true (when we're dealing with all positive terms, anyway). In other words:

The "subseries"

1+2+4+8+... is unbounded,

and so is the original series:

1+0+2+0+4+0+8+...



Yup, you're right.

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[tony1234]

Panama Jack,

Welcome to the conversation.

I believe I see your argument. In a nutshell, I think you're saying that after an infinite number of games, the distribtution of payoffs would be the "expected" distribution and therefore we get the 1:2 ratio. That's Konrad's argument too, I think. But it's impossible to play an infinite amount of games. We can only play more and ask about limits. To be concrete (my math is very basic), after a billion plays, the ratio of 3's in game 1 to 6's in game 2 will be very close to 1:1. We can be pretty sure of that. Granted, that's a good foundation for thinking the ratio of total winnings will be 1:2. But what about the possible 3^billion and 2*3^billion payoffs? Even after a billion plays there will always be payoffs that are (1) large enough to throw off the ratio, (2) likely enough to be a worry, and (3) not so likely that we can say anything about their frequency with confidence. That's why the ratios will always bounce around.

That's the argument, even if it is not quite a proof.

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[tony1234]

Zut,

Your's is the point I am really interested. If I understand you, you believe that whenever Game 1 and Game 1 have infinite expected utilities, the exptected ratio of the winnings will be undefined. An you beleive so because infinity/infinity is undefined. So if doesn't matter if Game 2's payoff is (10^10n!)! or whatever (as long as Game 2's payoff is not dependent of Game 1's or vice-versa). Infinity over infinity is undefined. This is your position, right?

Do others agree?

It seems to me that we have to at least ask what the likelihood is that Game 1 will produce a payoff that will screw up the ratio. If may be that the more plays there are, the greater lead that Game 2 will build up so that it becomes decreasingly likely that Game 1 will ever be able to have a payoff significantly screwing up the ratio. Of course, there is always the possibility that that it might, but if the possibility is expected to decrease, then you will get your convergence.

I am willing to admit that if the payoff of Game 2 is t*3^n, then for any t, there will be no convergence, ie, the ratio will be undefined. Undoubedly the class of nonconvergence payoff functions is lrager than that. But I would need more argument to accept no-derfined-ratio for *any* payoff function with infinite expected utility.

Tony


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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[C K Dexter Haven]

Tony: &lt;&lt; The payoff for a given play must be finite even though its expected value is infinite. Usually, it's at least possible that the expected payof and the actual payof will be the same. I think it's bizarre that the expected payoff will always be more than the actual payoff and you can know this in advance. You know your expectations are wrong in effect. That's paradoxical.&gt;&gt;

Yes, agreed. It's a fascinating problem.

I think I have finally come up with a proof that will satisfy everyone. Warning: it's kinda tedious and not at all elegant.

OK, the set up, from several hundred posts ago:
Player 1 plays Game 1, and gets 3^n if the first heads is on the nth toss.
Player 2 plays Game 2, and gets 2*(3^m) if the first heads is on the mth tosses.

NOW... here's my new insight.

Let's add a new Player 3, where Player 3 gets the ratio of winnings of Player 2 to Player 1. Got it?

EXAMPLE: SUPPOSE first round, Player 1 flips 2 tails and a head (TTH), gets 3^3 = 27. Player 2 flips TH, gets 2*3^2 = 18. So Player 3 gets 18/27 that round.

Every one clear? Now, we can reframe the ratio question by asking: what's the expected value of the game for Player 3?

Obviously, if the same coin-sequence is used (n = m), then on each round, Player 3 gets $2. That's the dependent games issue, that I think we've already dealt with.

OK, so now we need to solve for the expected value of Player 3 when n and m are independent. I contend this is also infinite (unbounded.) I prove this by enumerating all the outcomes:

Game 1: H
Game 2: H, TH, TTH, TTTH, ....

Game 1: TH
Game 2: H, TH, TTH, TTTH, ...

Game 1: TTH
Game 2: H, TH, TTH, TTTH, ....

....

OK? If you're with me so far, then let's calculate the expected value for each outcome (probability x payoff for Player 3), and the expected value for Player 3 (which will be the ratio of Game 2/Game 1) will be the SUM of the expected values for each outcome. Ready?

Game 1: H (Prob = 1/2) Pays $3
Game 2: H (Prob = 1/2) Pays $6
Game 3: 1/4 X payoff $2 = $1/2

Game 1:H (Prob = 1/2) Pays $3
Game 2:TH (Prob = 1/4) Pays $18
Game 3:1/8 X payoff $6 = 3/4


Game 1:H (Prob = 1/2) Pays $3
Game 2:TTH (Prob: 1/8) Pays $54
Game 3: 1/16 X payoff $18 = 9/8

Game 1:H (Prob 1/2) Pays $3
Game 2:TTTH (PRob 1/16) Pays $162
Game 3:1/32 x payoff $54 = 27/16

Etc when Game 1 = H. Then eventually, we list all the outcomes when Game 1 = TH.
Game 1:TH (Prob 1/4) Pays $9
Game 2:H (Prob 1/2) Pays $6
Game 3: 1/8 x payoff 2/3 = $1/12

Game 1:TH (Prob 1/4) Pays $9
Game 2:TH (Prob 1/4) Pays $18
Game 3: 1/16 x $2 = $1/8

Etc. And then, eventually, Game 1 = TTH, etc.

In this way, we can enumerate all the outcomes, their probability of occurring, and the payoff for Player 3.

But it's clear from the first set of outcomes (Player 1 gets H on first flip) that the expected value for Player 3 is already unbounded, hence the expected value for Game 3 is also unbounded. QED.

Thus, the ratio of Game 2 to Game 1 is, in fact, undefined (and even unbounded!)

Does this do it for y'all??

[tony1234]

CKD,

Great attempt. Very helpful. Certainly it establishes the proper framewok for analysis. But small point of clarification:

You repeatedly equate unbounded and undefined. As I sais many posts ago, these are not the same concepts. Unbounded means greater than any finite number. The series 1, 2, 3, . . . . is unbounded and so is the sum of the terms in the series. But it's not undefined in the sense that 1/0 or (1 + -1 + 1 + -1 . . . ) is undefined. Here we cannot say anything about the quantity, not that it's greater than 2, or greater than any finite unnumber or anything.

So exactly what are you saying about the ratio and Game 1 and 2?

Are you sasying that "thinking about ratio of winnings does not give us any reason to prefer G2 because the ratio is undefined" or "thinking about ratios give us a reason to prefer G2 before the ratio is greater than any finite number."

Of course, I could study your proof, but it's easier to just ask.

Tony

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Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[panamajack]

Thanks all for the replies. Upon reviewing my position, I've seen I was looking at the wrong distributions and I'm slowly coming around to the idea that it involves a ratio of infinite sums that doesn't converge. (I'd be coming around faster if I hadn't sold my calculus book a few years back.) The points made are generally clear, I'd just like to point out the right formula if someone cares to do it ... I think it may not matter, but the expectation of the ration actually comes out something like this (althogh CKDex's piecewise definition is probably good enough, since it proved the point) :

If X1 is the random variable of flipping coins until heads shows up, the distribution is
P(X1=n) = p^n where p is 1/2 for {n=1,2,3,...}

Call the payoff for game 1 G1, and for game 2 G2, and for game 3 G3

P(G1=n) = p^n for n = 3^n*p and 0 everywhere else
P(G2=n) = p^n for n= 2*3^n*p and 0 everywhere else

The expected value E(G1) for G1 is
SUM (n=1 to inf) : n * P(G1=n)
= SUM (n=1 to inf) : 3^n * p^(n+1)

and E(G2) is clearly 2*E(G1)

Since G1 and G2 are independent, E(G1/G2) = E(G1)/E(G2)
This is E(G3), the expected value of the ratio of the two payoffs. So it will equal 1/2 only if the two sums cancel, and I'm not really sure they do. In fact, i'm inclined to think they don't.

Another, perhaps deeper, perhaps not if this isn't what's behind this, is the concept of infinity as a limit or as an extant thing. In other words, does it make any sense to speak of playing for an infinite amount of time, and have an infinite sequence that comes out to something or just some large number ratios bouncing around a lot that just keeps going, and we could look in on the game at any time. Here's a thought about this : It is imaginable that given a fair (and admittedly immutable and indestructible as well) coin, there is a game that is infinite, i.e. you could keep flipping the coin and it will never come up heads. So in an infinite amount of time, this happens, yes, but how to account for it? Mathematics assigns the probability of this to be the limit, which is in fact 0. I always found both infinite and infinitesimal events to be interesting mathematical concepts (and I used to get really scared as a child that some day, while playing a simple game, I might end up having to roll the dice for all eternity until a tie was resolved, or I hit the winners' circle on exact count, or whatever).

It's not as important a question as whether the universe is infinite, but it might be interesting.

panama jack

[C K Dexter Haven]

Tony says: &lt;&lt; Originally posted by Cabbage:
Dex:
I believe you're confusing "sequences" with "series". As a sequence, what you said is true--the sequence does not diverge to infinity. &gt;&gt;

Just to be clear, *I* wasn't confusing sequences with series. A series (all positive terms) with an unbounded subseries must itself be unbounded.

For sequences, a sequence with an unbounded subsequence cannot converge (to a finite sum), but the divergence itself could be undefined or could be unbounded. (Repeating, I *hate* the term "converges to infinity.")

[Cabbage]

Dex:

I don't think your proof is quite sufficient to answer the question originally raised by the OP (not that I disagree with the conclusion, however), which is why I brought up my conjecture a few posts ago (which turned out to be false).

For example, consider the following two sequences (sorry for the sloppy notation, but it should be clear):

An = 1,4,1,9,1,16,1,25,...
Bn = 4,1,9,1,16,1,25,1,...

(Both sequences alternating between the constant 1 and the squares, shifted by 1 from each other).

If we look at the average of the individual ratios (Ai/Bi), that is unbounded. In other words:

[(A1/B1) + (A2/B2) + ... + (An/Bn)] / n

is unbounded as n goes to infinity.

However, the ratio of the sum of the first n Ai's to the sum of the first n (same n here) Bi's converges to 1:

A1 + A2 + A3 + ... + An
---------------------------
B1 + B2 + B3 + ... + Bn

goes to 1 as n goes to infinity.

The point is that your proof shows that the individual ratios, (Ai/Bi), have unbounded expected value. That's not enough to demonstrate that

A1 + A2 + A3 + ... + An
---------------------------
B1 + B2 + B3 + ... + Bn

doesn't converge as n goes to infinity, which is what the OP was asking.

I can see a couple of ways around this. One would be to show that the set of pairs of sequences {An,Bn} where such a counterexample to my conjecture occurs has probability 0.

Another way would be to show that the expected value of:

A1 + A2 + A3 + ... + An
---------------------------
B1 + B2 + B3 + ... + Bn

doesn't exist for n=1,2,3,... (Your proof took care of the case n=1).

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[C K Dexter Haven]

Cabbage, I'm not entirely following you (on the other hand, it's early, and last night was a late night.)

We have two games. Define a "round" to be one play of Game 1 (that is, flipping the coin until it comes up heads and getting paid accordingly) and one play of Game 2.

I'm not showing the ratios of each round to be unbounded, I'm showing the EXPECTED VALUE of the ratios of a round to be unbounded. The expected value of the ratios is obtained by enumerating all the possible outcomes of a single round, multiplying each outcome by the probability of that outcome arising, and summing.

So, I'm not looking at a particular sequence of outcomes, I'm looking at an enumeration of ALL outcomes, and then summing the expected value of each one... and the sum in unbounded.

Not sure whether that's connecting, but I'm not sure where your example of partial sums is leading. I'm not looking at the average of the ratios, I'm looking at the sum of the expected values.

[Cabbage]

Dex:

Basically, my point is this: Your proof shows that the expected value of (A1/B1) is unbounded--the expected value of the ratio after each game has been played once.

But what's the expected value of the ratio after the games have been played, say, twice--the expected value of (A1+A2)/(B1+B2)? or, more generally, the expected value of the ratios after the games have been played n times (not just once): (A1 + A2 + ... + An)/(B1 + B2 + ... + Bn)? I don't think it necessarily follows that, since the expected value of (A1/B1) is unbounded, then the others are unbounded as well. (It may actually follow, but I don't think it's obvious).

So, for n=2, we would have:

Game 1 payoffs: $3, $3
Game 2 payoffs: $6, $6
Ratio: 1/2
Probability of this: 1/16
Add to the expected value: (1/2)*(1/16)

Game 1 payoffs: $3, $9
Game 2 payoffs: $6, $6
Ratio: 1
Probability: 1/32
Add to the expected value: (1) * (1/32)

Game 1 payoffs: $9, $3
Game 2 payoffs: $6, $6
Ratio: 1
Probability: 1/32
Add to the expected value: (1) * (1/32)

Game 1 payoffs: $3, $27
Game 2 payoffs: $6, $6
Ratio: 30/12
Probability: 1/64
Add to the expected value: (30/12) * (1/64)

And so on, through all the possibilities, and then sum up the expected value of the ratio of two payoffs of Game 1 divided by two payoffs of Game 2.

In general, it seems at least possible to me that this expected value (or the expected value of the ratios of n games, for some n) could exist, even if the expected value of the ratio of the payoffs after a single game of each doesn't exist.

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[Cabbage]

PS--Just for the record, I do think the expected value of (total payoff after n games of Game 1) / (total payoff after n games of Game 2) probably doesn't exist for any n in this case, I'm just claiming it's not obvious based only on the fact that it doesn't exist for n=1.

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[C K Dexter Haven]

OK, I see where you're going, Cabbage, thanks. You're saying, play Game (1) and Game (2) exactly n times (where n is some fixed integer, yet to be determined) and then sum all the winnings from both games, and Player 3 gets the ratio of the sum of the winnings... is it possible that Game (3) has a determinable (finite) expected value?

I don't want to think that hard.

But... suppose there is such a fixed n.
We enumerate all possible outcomes for Game 3 (the ratio of the sums), and the probability for each outcome, as you have started. Then we sum all the expected values.

I'm going to try to find an unbounded sub-series of the sum, as follows:
One set of outcomes is obviously:
Game 1: Payout $3, $3, $3..., $3
Sum of payouts = 3*n
Game 2: Payout $6, $6, $6,...,2*3^m where m ranges over all the integers m = 1, 2, ....
Sum of payouts = 3*(n-1)+2*3^m = 3^(n-1)[1 + 2*3^(m-n+1)]

For each m, (3) = Ratio =(2)/(1) = [1 + 2*3^(m-n+1)]/3

Now we need to multiply this by the probability of that outcome, for each m... but the probability which is something like (1/2)^n or perhaps a sum of 1/2^n, I presume. My point is that (I think) the probability doesnt depend on m, just depends on n.

So multiplying the ratio of an outcome by the probability of that outcome may introduce a large denominator with 2^n, but n is fixed, so that denominator is finite, and eventually 3^m in the numerator will overtake it.

Hence, the sum of ratios has an unbounded subseries. Since everything in sight is positive, this means that the series itself must be unbounded, QED.

I think I'm basically trying to do this by induction. We've shown it's true for n = 1, and assume it's true for fixed n, then adding one more play won't change the unboundedness of the expected value series.

Man oh man. Now I got a headache.

[Cabbage]

Actually, I think I've found a fairly simple way around it. Instead of trying to find the expected value of the ratio after n rounds (A1 + ... + An)/(B1 + ... + Bn), find the expected ratio of just one Game 1 payoff, to the total of n Game 2 payoffs: A1 / (B1 + ... + Bn). If this is unbounded, it seems fairly clear to me that the former is unbounded as well. Now it's pretty easy to get an unbounded subseries:

Game 1 payoff: $3
Total Game 2 payoffs: $6n
Ratio: 3/6n
Probability of this: 1/2^(n+1)
Add to the expected value: (3/6n)*{1/2^(n+1)}

Game 1 payoff: $3^2
Total Game 2 payoffs: $6n
Ratio: 3 * (3/6n)
Probability of this: (1/2) * {1/2^(n+1)}
Add to the expected value: (3/2) * (3/6n)*{1/2^(n+1)}

And so on, increasing the Game 1 payoff each time, where I've highlighted the fact that this is a geometric series with ratio 3/2.

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