Sunday, October 01, 2006

The two envelopes problem

Many times the investors switch stocks with fallacious reasoning created by a flawed argument. I would show an example of one simple puzzle which leads into complex error of judgment.

Let's say 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're offered the possibility to take the other envelope instead.

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's the larger 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

V = ½ * 2A + ½ * A/2 = 5/4 *A

  1. This is greater than A, so I gain on average by swapping

Source- wikipedia

Your intuition would most probably realize that there is something wrong but you may find it difficult to spot flaw in the reasoning given above.

When you are calculating expected values using the equation in step 7, you are assuming that 50% of the time the other envelop would contain 2A and 50% of the time it would contain A/2. Its like saying that 50% of the time the envelops would be (A/2, A) and rest 50% of the time these would be (A, 2A). This is incorrect because the problem statement doesn’t say that. I can play the same game by giving (100$,200$) envelops repeatedly.

What you should see is that for any given set of envelops (A,2A), you would pick A 50% of the time and 2A, 50% of the time. If you are holding the 2A then you have conditional probability of 100% to lose by a swap. When your are holding A then you have conditional probability of 100% to gain by a swap. It is important to note that both his gain and loss are same as the smaller sum and the expected gain by swapping is zero.

V = 1/2 * 100% * A + 1/2 * 100% * (-A) = 0

When you are offered a chance to swap, you haven’t gained any extra information about the contents of another envelop, hence the probability of choosing the higher value envelop remains same. I’m as good as with the chosen envelop as I’m by swapping.

This situation can have real life equivalents with an additional twist. A transaction cost involved in swapping which makes swapping a losing proposition because the expected value is negative.

Another variation of the problem where the following 2 arguments lead to conflicting conclusions:

  1. Let the amount in the envelope you chose be A. Then by swapping, if you gain you gain A but if you lose you lose A/2. So the amount you might gain is strictly greater than the amount you might lose.
  2. Let the amounts in the envelopes be Y and 2Y. Now by swapping, if you gain you gain Y but if you lose you also lose Y. So the amount you might gain is equal to the amount you might lose.

Solution:

Argument 2 has already been explained above so let us focus on argument one. When I lose I’m holding a sum twice as large as the sum I’m holding when I gain by swapping. So the A/2 in losing swap transaction is equal to A in the gaining swap transaction because A in both are not same.

2 comments:

Tad Boniecki said...

Hi,

I read your solution to the two envelope problem but I think it does not solve the paradox. What you have done is to assimilate the argument of 2) into 1). You have not shown at what point the argument of 1) actually goes wrong, as the argument of 1) the value A is the same in the two cases being examined ie A, A/2 vs A, 2A.

I have spent a lot of time on solving the paradox and if you are interested my solution appears here:
http://soler7.com/IFAQ/two_envelope_paradox_solution.htm

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