Paradox Involving Swapping Dollars
Date: 08/01/2005 at 09:27:00 From: Dan Subject: Mathmatical Paradox ? This is a mathematical paradox that really boggles me. Suppose two people (Jack and Jill) make a bet with each other; whoever has MORE money in their wallet must give their money to the person who has LESS money in his/her wallet. (We can assume that both are carrying a random amount of dollars, say between $1-$100.) Jack says to himself, "If I have more than Jill, she'll win only what I have, but if she has more than me, I'll win MORE than what I have, and my odds of winning are 50/50, so the bet is in my favor!" Now of course Jill says the same exact thing for herself, concluding that the bet is in HER favor. I can't find a way to disagree with Jack OR Jill, but yet the bet cannot mathematically be in both Jack AND Jill's favor. Where has their logic gone wrong??? I'm usually very good with mathematical trick questions, but this one has me baffled. Please help! :)
Date: 08/02/2005 at 05:29:03 From: Doctor Jacques Subject: Re: Mathmatical Paradox ? Hi Dan, The short answer is that your probability of winning depends on the sum you have in your wallet. The more money you have, the more likely you are to lose it, and your expected gain is negative. Contrary to what you say, your odds of winning are not 50/50 for a fixed amount in your wallet. When you average everything, it turns out that the expected gain of both players is 0, because for each combination of amounts in the wallets, there is a symmetrical combination with the opposite gain. This assumes that the amounts of money are distributed in both wallets with the same probability. In fact, this is the key point in such problems. For the problem to make sense, you must specify the probability distribution for the money (the probability that a given wallet contains a given amount of money). Let us look at a simpler example: each wallet contains either $1, $2, or $3, each with probability 1/3. We also assume that, in the case where both wallets contain the same amount, no transfer of money takes place. Consider now what happens from Jack's point of view. 1. Jack's wallet contains $1. ---------------------------- If Jill's wallet contains $1, Jack's gain is $0. If Jill's wallet contains $2, Jack's gain is $2. If Jill's wallet contains $3, Jack's gain is $3. As each of the three cases happens with probability 1/3, Jack's expected gain is, in this case: (1/3) * ($ 0 + $ 2 + $ 3) = $ 5/3. 2. Jack's wallet contains $2. ---------------------------- In this case, we have the following gains for Jack: Jill's wallet | Jack's gain --------------+------------ $1 | -$2 $2 | $0 $3 | $3 and Jack's average gain is: (1/3) * ( -$ 2 + $ 0 + $ 3) = $ 1/3. 3. Jack's wallet contains $3. ---------------------------- In this case, we have the following gains for Jack: Jill's wallet | Jack's gain --------------+------------ $1 | -$3 $2 | -$3 $3 | $0 and Jack's average gain is: (1/3) * ( -$ 3 - $ 3 + $ 0) = -$ 2. Now, each of the three cases (for Jack's wallet) happens with probability 1/3, so Jack's average gain is: (1/3) * ($ 5/3 + $ 1/3 - $ 2) = $ 0 The third case is a good illustration of the fallacy. In that case, Jack cannot win more than he can lose, because it is impossible for Jill to have more money than he does. It turns out that the average gain will always be 0, whatever the probability distribution is, provided that it is the same for both players (imagine playing that game with Bill Gates...). You might ask: what if the amount of money is unlimited? In that case, the answer will still be the same, but there are restrictions on the probability distribution. For example, you can no longer assume that any amount is equally likely (you cannot have a uniform distribution). Indeed, if we assume that the amounts are integer numbers of dollars (for simplicity), and that every amount has the same probability, say p, the total of the probabilities (the probability to have _any_ amount of money) must equal 1. But the total number of possibilities is infinite, and this means that the total probability will be p * (infinity), which cannot be 1. There are many other interesting paradoxes related to that idea. For example, if you search the Internet for the phrase "two envelopes paradox", you will find lots of articles about another similar paradox (also known as the "St Petersburg's paradox"). Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
Date: 08/02/2005 at 12:57:34 From: Dan Subject: Thank you (Mathmatical Paradox ?) Thank you, Dr. Jacques! It all seems so clear and simple now that you explain it that way. I am ashamed not to have figured that out myself. You guys are AWESOME! Thanks again! Dan
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