In an earlier puzzle, you were handed two envelopes, one of which contained twice as much money as the other. After opening one, you were given the chance to swap. At first glance, it appeared that the your chance of getting more money could only increase each time the envelopes were swapped, but clearly this was nonsense: since there is no probability distribution which allows all real numbers to have the same probability, some values would have to have been more likely than others.

Suppose instead envelopes contain the non-negative integer sums 2ⁿ and 2ⁿ⁺¹ with probability q(1 − q)ⁿ for some fixed q < 1/2

Now of course if the envelope you open contains a 1, you know the other must contain 2, so you ought to swap.

But you can do even better than this. Suppose you open an envelope and find an amount of money 2^k

What would the expected value of the second envelope be?

Does this lead to the same paradox?

Well yes, apparently Sam has cleaned up this problem and produced the same paradox.  Josh has concluded that after opening the first envelope one should always opt for the second one.  Clearly, this implies that before opening one envelope one should always opt for the other.  And clearly, this does not make sense, because the process has no end. 

I'll give some thought to this and post again.

Posted by Steve Herman on 2005-08-13 02:41:44