The Two Envelopes Fallacy

During his Hamilton lecture in Dublin recently, Fields medalist Martin Hairer made a passing mention of the “Two Envelopes Paradox”. This is a well-known problem in probability theory that has led to much misunderstanding. It was originally developed in 1912 by the leading German number theorist Edmund Landau (see Gorroochurn, 2012). It is frequently discussed on the web, with much misunderstanding and confusion. I will try to avoid adding to that.

Two-Envelopes

The Problem

Anne puts money into two envelopes and asks Bill to choose one of them. The only information she gives him is that one envelope contains twice the amount in the other. Being something of a genius, Bill concludes that one envelope contains half the amount in the other!

But how to choose? Since he has no further information, Bill chooses an envelope at random. Then, before he opens it, Anne gives him the option to change his mind. Is there any advantage in switching envelopes? It would seem not, but Bill reasons as follows:

Suppose I have chosen an envelop with x euros. Given the absence of any further data, there are equal chances that the other envelope has a greater or lesser sum. If it contains 2x euros, my gain is x. If it contains x/2 euros, my loss is x/2. Averaging these, the gain if a switch is made is:

Gain = ½ ( x ) + ½ ( –½ x ) = ¼ x             [1]

So, I stand to gain by ¼ x just by switching.”

But suppose he does switch! Anne, being of a generous nature, gives him the option of switching again. Surely, on the same basis as before, he can increase his gain by switching once more, and this process may be continued forever.

What is wrong with Bill’s reasoning? Think about this before reading on.

* * *

Suppose that Anne started with a total amount of money 3y euros. She divided this into two amounts, y and 2y. She puts y euros into one envelope and 2y euros into the other. Then she computes as follows:

We assume that Bill chooses one envelope at random. His expected amount is

½ ( y ) + ½ ( 2y ) = ½ ( 3 y ) ,

half the total amount, as we might anticipate. Now, what happens if Bill switches? If he has y he will get 2y, gaining y. If he has 2y, he will get y, loosing y. Since both are equally probable, his expectation is

Gain = ½ ( y ) + ½ ( –y ) = 0             [2]

Poor old Bill: his expectation does not change whether he chooses to switch or to stick.”

Slippery Variables and Sticky Variables

Where did Bill go wrong in the first calculation? The problem is that he used x to represent the amount in the envelope first chosen, and then used this to represent two different quantities in his formula [1] above.

If he has chosen the envelope with the smaller amount, then x = y. If he has chosen that with the greater amount, then x = 2y. But he uses x for both these quantities in his equation [1] and, as a result, his calculation gives a meaningless result.

We could refer to x as a “slippery variable”, since it represents different quantities in different cases. In contrast, we might call y a “sticky variable” as, once chosen, it sticks to its initial value.

Conclusion

The Wikipedia page on the two envelopes problem discusses many interpretations and also several variants that have generated interest. The value of these problems is that they caution us to examine lines of argument that appear reasonable but that contain logical flaws. The problems are of little interest in themselves; the benefits lie in the sharpening our reasoning capacity.

Sources

Gorroochurn, Prakash, 2012: Classic Problems of Probability. Wiley, 314pp. ISBN: 978-1-118-06325-5.

Wikipedia article: Two Envelopes Problem.


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