Challenging Probability Question

Factual Questions
1

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.)

2

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

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

3

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

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

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…

4

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

5

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.

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

6

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.

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

7

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:=mp13; mp2:=mp23;
{ 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>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.




------------------
Wrong thinking is punished, right thinking is just as swiftly rewarded. You'll find it an effective combination.
8

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.

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

9

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.

10

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.

11

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.

12

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

13

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 :smiley: 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 will be of some interest to you on a related matter…

14

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

15

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).

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

16

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?

17

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.

18

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.

19

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

20

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.