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I came across the envelopes problem (can be googled for more details) where you are given two sealed envelopes with one containing twice as much money as the other. You can pick one and then switch if you choose. I read that it makes sense to switch because expected gains go up if you do. Its quite paradoxical but when I simulated it in a python program and ran it millions of times I noticed there was no benefit to switching. I didn't see any pattern where there was a benefit to switching. So what is the way to understand this?

user3079275
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  • Hard to saw without seeing your program, but my experience has been that whenever a simulation disagreed with a well-established albeit paradoxical result, there's been some flaw in the simulation. – amd Oct 05 '15 at 17:16
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    You presumably used a particular distribution of actual money values for your simulation, such as $1$ and $2$. The paradox only works if there is not a finite expected value for either envelope – Henry Oct 05 '15 at 17:19
  • https://en.wikipedia.org/wiki/Two_envelopes_problem for those looking for a link – Henry Oct 05 '15 at 17:24
  • @Henry I didn't understand that point. What I did was create two envelopes, pick one up at random and put 100, and put 200 in the other. Then toss a coin to pick one of the two, note how much money was there and then switch and note the net gain or loss. Repeat millions of times keeping tab of how much money was gained by switching. It comes to a net negative sometimes and sometimes net positive (for each million trials). – user3079275 Oct 05 '15 at 17:35
  • @user3079275: The paradox does not exist for actual distributions with finite expectations. In your example switching could only lead to a gain or loss of $\pm100$ so the potential gain was not double the potential loss. – Henry Oct 05 '15 at 22:10
  • @Henry, so what is a real world example that would demonstrate the paradox? What is the scenario where switching actually helps? What you are saying sounds too abstract. – user3079275 Oct 05 '15 at 22:18
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    @user3079275: For example select $j$ from the positive integers with probability $\dfrac{9^{j-1}}{10^j}$ (a geometric distribution), and let one envelope have $2^j$ and the other have $2^{j+1}$. Then it is worth switching.if you look inside the first envelope: if you see $2$ then the expected value of the other envelope is $4$; otherwise if you see $2^k$ for $k\gt1$ then the expected value of the other envelope is $\frac{23}{19}2^k$. If you do not look inside the envelopes then the expectations of each are infinite. Sometimes the envelopes will have more than all the money in the world. – Henry Oct 05 '15 at 23:54
  • Realted: http://math.stackexchange.com/questions/1164157/two-envelopes-problem-what-is-the-problem-with-my-solution and http://math.stackexchange.com/questions/964381/if-you-have-two-envelopes-and and http://stats.stackexchange.com/questions/95694/two-envelope-problem-revisited – Henry Oct 06 '15 at 00:14

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