The Two Envelopes
Date: 09/08/97 at 07:56:43 From: Ofri Becker Subject: The two envelopes probability problem (Maybe a paradox) Dear Dr. Math, I am an Aeronautics graduate student in the Technion-Israel Institute of Technology, but in my free time I love mathematical riddles, paradoxes, etc. Here is one whose solution I don't exactly know to explain: Let's say you are given two sealed envelopes, and told that in both there is money, while in one of them (unknown which), there is twice as much as in the other. You are told to choose one. After you open it and see how much there is inside it, you are asked whether you want to regret and take the other. Should you regret or not? At first glance it sounds very simple - there's no difference, BUT - let's look at the odds: 50 percent to get double the sum = 50 percent * 2 = 1 50 percent to get half the sum = 50 percent *.5 = .25 All together = 1.25 times the sum. So - it brings us to the conclusion that you should regret! BUT - you didn't have to see the sum inside the first envelope in order to know that it's better to regret - so why not "regret" from the beginning and choose the other envelope? But then you have to regret again, etc. etc.... I tried to explain why the odds are not 50/50, but 2/3 for the lower sum and 1/3 for the higher sum, which would make the average of regretting or not equal, but I couldn't. Please help me explain the solution of this interesting problem.
Date: 09/08/97 at 12:33:37 From: Doctor Anthony Subject: Re: The two envelopes probability problem (Maybe a paradox) Let the two sums of money be M and 2M, in envelopes A and B. Suppose the strategy is always to swap. The possible sequences are: First choose A with prob 1/2. Swap to B. Expectation = (1/2) x 2M = M First choose B with prob 1/2. Swap to A. Expectation = (1/2) x M = .5M Total expectation = 1.5M If the strategy is never to swap, then the Expectation will again work out at 1.5M. So there is no advantage in swapping. The mistake in your reasoning is to consider the expectation based on a variable starting point, leading to a factor of 4 in the different sums of money, 2 times and 1/2 times. Dealing with sums that have been seen, as you do, the full calculation is: Suppose you are looking at M, then if you swap, the expectation is 1 x 2M = 2M Suppose you are looking at 2M, then if you swap, the expectation is 1 x M = M The probability is 1/2 for each situation, so total expectation is (1/2) x 2M + (1/2) x M = 1.5 M as we got before. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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