## Monday, 5 February 2018

### Revisiting a Classic Probability Puzzle: the Two Envelopes Problem

Many people have heard about the Monty Hall problem. A similar (but less well known and more mathematically interesting) problem is the two envelopes problem, which Wikipedia describes as follows:
“You are given two indistinguishable envelopes, each containing money, one contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?”
The problem has been around in various forms since 1953 and has been extensively discussed (see, for example Gerville-Réache for a comprehensive analysis and set of references)  although I was not aware of this until recently.

We actually gave this problem (using boxes instead of envelopes) as an exercise in the supplementary material for our Book, after Prof John Barrow of University of Cambridge first alerted us to it. The ‘standard solution’ (as in the Monty Hall problem) says that you should always switch. This is based on the following argument:
If the envelope you choose contains \$100 then there is an evens chance the other envelope contains \$50 and an evens chance it contains \$200. If you do not switch you have won \$100. If you do switch you are just as likely to decrease the amount you win as increase it. However, if you win the amount increases by \$100 and if you lose it only decreases by \$50. So your expected gain is positive (rather than neutral). Formally, if the envelope contains S then the expected amount in the other envelope is 5/4 times X (i.e. 25% more).
In fact (as pointed out by a reader Hugh Panton), the problem with the above argument is that it equally applies to the ‘other envelope’ thereby suggesting we have a genuine paradox. In fact, it turns out that the above argument only really works if you actually open the first envelope (which was explicitly not allowed in the problem statement) and discover it contains S. As Gerville-Réache shows, if the first envelope is not opened, the only probabilistic reasoning that does not use supplementary information leads to estimating expectations as infinite amounts of each envelope. Bayesian reasoning can be used to show that there is no benefit in switching, but that is not what I want to describe here.

What I found interesting is that I could not find - in any of the discussions about the problem - a solution for the case where we assume there is a finite maximum prize, even if we allow that maximum to be as large as we like. With this assumption it turns out that we can prove (without dispute) that there is no benefit to be gained if you stick or switch. See this short paper for the details:
Fenton N E, "Revisiting a Classic Probability Puzzle: the Two Envelopes Problem" 2018, DOI10.13140/RG.2.2.24641.04960

#### 1 comment:

1. The argument for switching is incorrect (well, incomplete) whether or not you look in an envelope. Two random variables are required: one representing some measure of a dollar amount, and one representing the relative value of your envelope to the other.

Let me illustrate with examples. Say I fill two pairs of envelopes with {\$5,\$10} and {\$10,\$20}, respectively. I pick a pair at random and give it to you, and ask you to pick one from the pair. All of the conditions of the problem, as you stated it, are satisfied. If you open your envelope and see \$10, then the expected value of the other is indeed \$12.50, but only because of the even distribution of these pairs.

Now say I fill nine pairs with {\$5,\$10} and one with {\$10,\$20}. This time if you see \$10, the expectation for the other is only \$6.50.

If you don't look, but name your value X, then the correct expectation Y for the other envelope is:

Y = (X/2)*Pr(pair={X/2,X})*Pr(high} + (2X)*Pr(pair={X,2X})*Pr(low}
Y = (X/4)*Pr(pair={X/2,X}) + X*Pr(pair={X,2X})