[PairTheBoard]

"I'll stay out of the meaning of probability discussion, but I have to defend the envelope paradox. It is considerably deeper than suggested here. Like all good paradoxes, the point is not to explain it, but to peel away the layers."

--AaronBrown

I realize the Two Envelope problem as usually stated stirs most discusion around the amounts in the envelopes being "randomly chosen". I am eliminating that part of the discussion by defining the problem more precisely and specifying the repeated experiment. Here is the version of the Two Envelope Problem I am talking about.

There are Two Envelopes, Env1 and Env2. A certain amount of money has been placed in each. We have no idea what the amounts are but we do know that one amount is twice as much as the other. In the repeated experiment the exact same amounts are used over and over. But we have numerous players of the game, none of which know the amounts in the envelopes. We also assume that if an envelope is opened and it's contents are revealed, it gives no information to a player as to whether the amount is the larger or smaller amount.

The envelopes are randomly shuffled and presented to a player. At this point there is a 50% probability that Env2 has twice the amount as Env1 and a 50% probability that Env2 has half the amount as Env1. Now, Env1 is opened and the amount X dollars is revealed to the player. The player is now asked to either choose the X dollars she sees or switch to Env2.

The player says to herself, I still have no idea whether the larger amount is in Env2 or not. But I now know there is X dollars in Env1. So I can calculate the expected value of the amount in Env2 as follows:

50%(.5X) + 50%(2X) = 1.25X

The player figures it's a nobrainer and takes the amount in Env2. In fact, every player makes the same calculation and if the calculation is correct the players on average must come out 25% better by always switching. What's wrong with this picture? Clearly it can't be right so what's wrong with the calculation?

What's wrong with the calculation is exactly the same mistake being made by people who insist after seeing the Twin with the Tony that there is still a 50% probability that the older Twin is behind door #2. Maybe a Philosophical Baysian can use that language but it sure doesn't help when it comes time to actually calculate the expected value in the Two Envelope problem.

Here is the correct way to calculate the expected value of the amount in Env2. Let the amounts in the envelopes be A and 2A and let the amount in Env1 which is revealed to a random player be Z. Then the conditional probabilty that Env2 contains 2Z given Z=A is 100%. The conditional probabilty that Env2 contains .5Z given Z=A is 0%. Just like the Twin with the Tony. It's either 100% or 0%, we don't know which. Similarly when Z=2A. For a random player the amount Z is equally likely to be A or 2A. So the expected value for Env2 is:

50%(100%(2Z) + 0%(.5Z) : when Z=A) +
50%(0%(2Z) + 100%(.5Z) : when Z=2A) =

50%(2A) + 50%(A) = 1.5A

While the expected value of Z for a random player is also

50%(2A) + 50%(A) = 1.5A


For all of you who have been saying mean things to me, I accept your apologies unspoken.

PairTheBoard