Envelopes with money (a new conundrum)

Cecil's Columns/Staff Reports
61

I must try to stop doing this - but it’s fun!

OK, I’ll take on 2 actuaries (and as I approach retirement I am grateful for your hard work).

The envelopes contain x and 2x. There are no outside clues.
Your expected win is (x+2x) / 2 = 1.5x.
Here’s where I think the misunderstanding creeps in:
If you have the ‘x’ envelope, you can win if the other envelope has ‘2x’. Total win = ‘x’.
But if the other envelope contains ‘x/2’…
stop! that’s the incorrect assumption!

The only true possibilities are:
If you have the 2x envelope (50%), then the other one contains x.
If you have the x envelope (50%), then the other contains 2x.

There is no case of x in one and x/2 in the other.

I know CKDextHavn has said this already- I’m just trying to express it differently so December is satisfied.

P.S. My nephew is an accountant - and so is his girl-friend…

62

CK – thanks for identifying your background. You inspired me to find the relevant journal article, which is from the American Statistician Volume 46 #4 November 1992

Ronald Christensen and Jessica Utts: Bayesian Resolution of the ``Exchange Paradox’’ … 274–276

The Table of Contents can be found at URL:
http://www.math.utah.edu/ftp/pub/tex/bib/toc/amstat.html#46(4):November:1992

I was unable to download this article (presumably I don’t have preper access), but my public library did so.

The authors’ view is similar to mine. Here are a few quotes:

“In this article we present a paradox that can be used to illustrate Bayesian principles in the classroom. The paradox is also resolved using a frequentist argument and illustrates how the misapplication of a symmetry argument causes problems.”

“One of the arguments in favor of using Bayesian methods is that prior assumptions are made explicit rather than being incorporated implicitly into the solution of a problem.”

“The paradox of this problem is that the rule indicating that one should always trade is intuitively unreasonable, while the method of arriving at the rule seems very reasonable.”

"The conclusion that trading envelopes is always optimal is based on the assumption that there is no information obtained by observing the contents of the envelopes. From a Bayesian perspective, the key to a successful analysis is in recognizing the potential information to be gained from the observation.

63

You probabilists always give me headaches. Below is Bayes’s Theorem

P(A|B) = P(A)P(B)/(P(A)P(B)+P(NOT A)P(B|NOT A))

Please tell me your interpretation of the values of

P(A), P(B|A), and P(A|NOT B)

For that matter, what’s A and B?

64

Mr Thin Skin:

Suppose you opened an envelope and found $8. Then the other envelope had $4 or $16.

A = the event that the original pair of envelopes had $4 and $8.

Not A = the event that the original pair of envelopes had $8 and $16

B = the event that you found $8 when you opened the envelope.

You’re seeking P(A/B) – that is, given that you found $8 in your envelope, you want to know the probability that the other envelope has $4 rather than $16.

The use of P(A) and P(NOT A) in the equation explicitly addresses the original probabilities of the two envelopes having $4 and $8 or having $8 and $16.

Prob(B/A) = 1/2 That is, if the two envelopes contained $4 and $8, then your chance of opening the one with $8 was 50%.

Prob(B/NOT A) = 1/2. That is, if the two envelopes contained $8 and $16, then your chance of opening the one with $8 was 50%.

I’m having a little trouble sorting out all the parentheses in your expression. But, in this case, since P(B/A) = P(B/NOT A) = 1/2, Bayes Rule will simplify to

P(A/B) = P(A).

65

Thanks for the link, december, but I couldn’t download it either… I’ll take a look later.

66

BTW, I trust you all understand that this game could never arise. Supposing that we must use an integer number of dollars (same argument works if we have to use an integer number of cents), and we are in charge of putting the money in the envelopes.

Note that we could never put an odd number in an envelope. If a contestant opened an envelope with $3, he would know that he must have the smaller amount, since the other envelope could not hold $1.50.

Now, what’s the smallest amount that we could put in an envelope?

(a) Well, obviously, we could not use the situation of $1 and $2; same logic. If the contestant got the envelope with $1, he’d know that he had the smaller amount, and he’d know to switch to win the maximum.

(b) We could not use $2 and $4, either. Why? Because if the contestant opened the $2 envelope, he’d know that the other envelope couldn’t contain $1 (see (a)) so it must contain $4, so he would know he had the smaller amount, and he’d know to switch and win the maximum.

© We could not use $4 and $8, because if the contestant opened the $4 envelope, he’d know the other envelope couldn’t contain $2 (see (b)), so he would know he had the smaller amount, and he’d know to switch and win the maximum.

(d) We could not use $8 and $16, because if the contestant opened the $8 envelope, he’d know the other envelope couldn’t contain $4 (see ©), so he would know…

So, in fact, the game itself is a paradox that could never be played.

December – don’t tell them! Let them figure it out for themselves!

67

CDextHavn, very interesting point.

Now let’s assume they are checks and they will be rounded to the nearest cent.

68

CK, I like that reasoning. It reminds me of the story (that you certainly know, but I’m telling it for the others) of a prisoner who was sentenced to death. The judge said he would be executed in the following week (Mon - Fri). He wouldn’t be told the exact day in advance (to make it less cruel), but the execution would take place at noon on the respective day. The prisoner thought:

(a) “They can’t execute me on Friday, because if they don’t come for me Thursday noon I’ll know the day in advance which they said I wouldn’t.”

(b) “They can’t execute me on Thursday, because if they don’t come for me Wednesday noon, I’ll know it’s Thursday (Friday being impossible due to (a)), and they said I wouldn’t know in advance.”

© “They can’t execute me on Wednesday due to (a) and (b).”

(d) “They can’t execute me on Tuesday due to (a), (b), and ©.”

(e) “They can’t execute me on Monday due to (a), (b), © and (d).”

(f) “So they can’t execute me that week, but they said they would!”

With the verdict logically flawed, the prisoner thought he had found a chance to file an appeal and challenge the sentence. Maybe he could get a new trial, a better lawyer! Maybe he’d be free soon!!

Sadly, his appeal was dismissed. The prisoner was executed at noon on Tuesday, and he didn’t know the day in advance, just like the judge said.

69

Yes, Holg, exactly the same paradox (even though no actual doctors are involved in the statement of the problem.)

70

How would you know, CK? Actually, I’m working on my PhD in CS. Can I play too?

71

CK – good example. It shows that one cannot assume that all pairs of form X, 2X were equally likely. Furthrermore, it provides some sort of a basis to guess at the probabilites before an envelope was opened.

72

december, sorry, but you’re wrong. Switching envelopes does not change the outcome in any way.

I just wrote a little computer program that simulates this problem. It uses the following algorithm:

  1. Randomly pick a number of dollars, and put that amount into envelope one.
  2. Put twice as much money into envelope two.
  3. Randomly select one of those envelopes to be given to the (AI) player. Call this envelope A; call the other one B.
  4. If the player is following the “always kee p” rule, add the contents of A to his score. If he is following the “always switch” rule, add the contents of B to his score.
  5. Repeat.

When the two strategies is compared, the totals are very close. I’ve just done a few runs of several hundred thousand trials apiece, with a maximum dollar amount of $200; in every case, the difference between the totals is less than $2 per trial. Neither strategy is consistently higher.

Whether or not you ever switch envelopes, you’ll end up with about the same total over the long term.

(And just think, I didn’t even need a PhD to figure this one out.)

I’m not a warlock.
I’m a witch with a Y chromosome.

73

A correction of my last post:
In each run of more than 100,000 trials, each strategy’s total was within $100,000 of the other. This is true even for my most recent run, consisting of 1,000,000 trials. As the numbers get higher, the percentage difference between the totals is getting smaller, as is to be expected.

The “$2 per trial” line that I wrote above doesn’t make sense, sorry. Chalk it up to lack of sleep.

74

So, AuraSeer, what have you proved? That without opening the envelope, switching doesn’t make any difference? I thought we had agreed on that.

Try changing your program so that the player opens the envelope. If he finds $1, he switches; if he finds $200 (or whatever the upper limit), he keeps the envelope; in between, he may always switch or always keep.

Average winnings will increase because if you have any knowledge (however vague) of the range of possible amounts, you can exclude some possible combinations. My rule refers to the limits of this range; CK’s example concerned other limitations such as the amount always having full dollars (so that $99 can never be the “2X” envelope).

75

My limitations using whole dollars was arbitrary, you’d get the same result using whole cents (If you open an envelope with $.33, you know it must be the low.)

If you say everything is rounded to the nearest cent, you’d still know that if you open an envelope with $0.01, you should switch (the other envelope would either also have $0.01 or would have $0.02), so the same idea works, but cent by cent rather than dollar by dollar.

Aura, the question in your program is partly whether the contestant KNOWS the upper and lower bounds. If you know the upper bound is $200, and you open an enevelope with $102 in it, you know it must be the maximum – the other envelope must contain $51, it can’t contain $204.

76

AuraSeer – Holg is right. Your example can be used to illustrate the Bayesian case. The recipient’s probabilities and expected value depend on your chosen original probability distribution. When you “randomly” picked a number of dollars, you actually made that selection from your particular pre-determined set of possible values. Had you begun with a different set of values, the answer would have been different. The recipient can improve her odds to the degree that she can guess the shape of your original distribution. She can use her judgment and do better than 50-50.

Speaking of 50-50, here’s a quote from GOTCHA by Martin Gardner:

"The “principle of insufficient reason” which the economist John Maynard Keynes renamed the “principle of indifference” in his famous Treatise of Probability, can be stated as follows: If we have no good reasons for supposing something to be true or false, we assign even odds to the probability of each truth value.

"The principle has had a long and notorious history… [Gardner gives several examples of mis-use of this principle.]

“The principle has legitimate application in probability, but only when the symmetries of a situation provide objective grounds for assuming probabilities to be equal. For example, [flipping] a penny…”

77

Okay, let me see if I have this right. Are you saying that the player knows the upper and lower limits of the money he could be offered? I didn’t think this was part of the setup.

Perhaps I’ve missed something important in the discussion here. I thought the whole argument about expected values and so forth was on whether it was advantageous to always switch once you’ve seen the money in the envelope.

A player who knows the possible range of dollar amounts will of course end up with more money than an ignorant player. That much is so obvious that it’s barely worth mentioning. If that’s what this whole thread has been about up to this point, I’ll be very disappointed.

I’m not a warlock.
I’m a witch with a Y chromosome.

78

<< A player who knows the possible range of dollar amounts will of course end up with more money than an ignorant player. That much is so obvious that it’s barely worth mentioning. >>

Yes, but I think that’s really the question deep down, although it’s more complex than that.

  • If the game includes all possible X and 2X in the real numbers, then it’s a probability theory game (50/50) and the contestant has no new information from opening the first envelope.

  • If there is some slight edging – for example, if the number of dollars is an integer – then some combinations of X and 2X will give the contestant new information from opening the first envelope. For instance, as noted, if the first envelope contains an odd number of dollars, then it must be the smaller amount. Or if an envelope contains $10,000, you know it must be the larger amount. Thus, there are some situation where opening the first envelope gives information that may improve the odds from 50/50.

I haven’t read up on Baysian probabilities in quite a while, but I have to think that in this situation, the number of such modificiations is miniscule.

In any expected value problem, one must look at a large number of trials (my earlier comment to this effect was not irrelevant, december). Most of the simulations have assumed one contestant plays the game many times, with X and 2X varying. My approach was different: I assumed that there were many, many participants playing independently; and that X and 2X were always fixed (such as, always $8 and $4, for every contestant.)

Viewing the problem this way eliminates the subjective aspect but leaves the essential “paradox” that the true expected value is 1.5X, but that the contestant on opening the first envelope must estimate the expected value at 1.25 * (content of first envelope).

79

AuraSeer wrote:

The setup is not “standardized”. This seems to be part of the problem because people on this thread have been talking about slightly different things. The favorite setup is a gameshow scenario in which I think you would have some idea of the possible range of prizes.

Always switching after seeing the money would have the same effect as switching without ever seeing the money.

Personally, I think the opposite case is obvious: Always switching (with or without seeing the money) cannot give you any advantage because it’s just like having picked the other envelope in the first place, and assuming there is no bias in the first choice, the expected values must be the same.

I suppose both cases are pretty obvious to common sense. The point of the problem is that the calculation of expected seems to show that switching is advantageous. In fact, you are led to believe that you should not just switch once for every pair of envelopes; you should switch the same pair of envelopes back and forth again and again (if you’re allowed to) to increase your expected winnings each time! This is obviously wrong - the question is: Why? It’s a bloody paradox!

So was Cecil (posted 08-08-1999). Me, I was at least surprised at the longevity of the thread. I had contributed my thoughts in May. I suppose this paradox is just very hard to see through.

80

Said succinctly, Holg, the nature of the paradox is:
(a) the expected value is clearly the halfway point between the two amounts.
(b) the contestant believes the expected value to be 1.25 x (the amount in the first envelope that he selected and opened), because he has insufficient information about the contents of the envelope (that is, about the distribution of amounts in the envelopes.)

The whole point of the relevant bits of debate hinges around that question, and that leads to the question of what he should assume to be the possible amounts in the second envelope.

  • The paradox hinges around the contestant, finding amount M in the first envelope, thinks that 0.5M and 2M are equally likely to be in the second envelope. The true distribution, of course, is that one of those has zero chance and the other has 100% chance of being in the second envelope.
  • The question then becomes, what distribution should the contestant assume? And that’s where knowing something about the limits will help eliminate some possibilities and help him get a better estimate of the true distribution. For example, if he knows that $1000 is an upper bound, and he opens an envelope with $700, then he has good reason to think that the other envelope contains $350 and not $1400… thus, helping him assess the true probabilities.

So you are incorrect, Holg, in saying that opening the first envelope doesn’t give information. It could give information, if there are limits on the payout amounts. That’s the hinge of this debate. To the quizshow master, who knows the amounts in both envelopes, the expected outcome is clear. To the contestant, the expected outcome is not clear because he doesn’t have enough information; he therefore assumes a distribution as the best estimate he can make, and proceeds from there.