## Sunday, October 30, 2011

### Two_envelopes_problem

1. Someone gives you two envelopes with cash inside, and tells you that one of the envelopes has twice as much as the other. You're asked to pick an envelope and open it. You pick Envelope 1 and you open it to reveal $100. Awesome! And now that person asks you if you'd like to switch to Envelope 2, for a cost of merely$5. What should you do ?
2. The basic setup: You are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead.[3]
3. The switching argument: Now suppose you reason as follows:

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7. So the expected value of the money in the other envelope is

${1 \over 2} (2A) + {1 \over 2} ({A \over 2}) = {5 \over 4}A$
8. This is greater than A, so I gain on average by swapping.
9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
10. I will conclude that the most rational thing to do is to swap back again.
11. To be rational, I will thus end up swapping envelopes indefinitely.
12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction
1. The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above