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65. A Holistic Approach:
The Two Envelope Problem

The Question:

A game-show TV presenter holds two closed envelopes in her hands and tells you that one contains twice as much money as the other. She then hands you one of the envelopes. Before you open it, she informs you that if you wish you may switch envelopes. Should you do it? 
Well, you recall the Monty Hall Problem (see Chapter ///) and, therefore, yes, you will gain by switching.


The Paradox:

Let’s consider three scenarios:
Scenario 1: We designate the unknown amount that is in your closed envelope as X. By switching the envelopes, you would either gain another X, or lose half of X. Since the chances of either of these happening are half-half, the expected gain from switching is: 
½ X – ½ (X/2) = ¼ X,
This is positive! So you would expect to gain ¼ X. Therefore, SWITCH to the other envelope. 
Scenario 2: Let’s say the TV presenter keeps the envelope that contains Y and hands you the other one. It could hold with equal probabilities either ½ Y or 2Y. The expected content of the envelope you hold is
½ (½ Y) + ½ (2Y) = 5/4 Y
This is more than Y which is what you would have obtained, had you switched to the other envelope. Hence, KEEP the envelope you have.
Scenario 3: Let’s designate the lower amount as Z. Then, one envelope contains Z, the other 2Z. Now, let’s say, the TV presenter handed you the envelope containing Z. By switching you would gain another Z. Or she handed you the envelope containg 2Z. Then, by switching you would lose Z. Both scenarios occur with equal probabilities. Hence, the expected gain from switching would be:
½ Z + ½ (-Z) = 0
Since there is nothing to gain, you may as well stick with your current envelope. Or you may switch. IT SIMPLY DOES NOT MATTER.

So which is it?


The problem was first devised in 1952 by Maurice Kraitchik, a Belgian mathematican, born in Russia, whose field of research was numbers theory. But Kraitchik was mainly interested in recreational mathematicsl he organized the first congress on that subject in 1935 in Brussels. During WWII he lectured at the New School for Social Research as as associate professor for recreational mathematics. 
Naturally, the problem gained the attention of Martin Gardner who described it in 1982 in his book Aha! Gotcha. In 1989 Barry Nalebuff from Yale, subjected the problem to deeper analysis.


First of all, let us note that if scenario 1 were the correct one, one could enter a ever winning circle: after switching, one could ask oneself the same question again and come up with the same answer: switch. And again. And again. So, we can already surmise that there is something fishy…
The confusion arises because conditional probabilities are confused with unconditional probabilities. It is not correct to say, for example in scenario 1, you would either gain another X, or lose half of X with probabilities fifty-fifty. Or, in scenario 2, that you hold an envelope that contains with equal probabilities either ½ Y or 2Y. One must consider the conditions under which circumstances these gains or losses occur.
Scenario 1 considers the content of your envelope as a given, scenario 2 considers the content of the envelope the presenter keeps as given. To simply compute the expected content of the other envelope is not correct. The content of your envelope must be taken into consideration when computing the expected content of the other envelope. 
Hence, the correct way to go about it is as in Scenario 3. There are two possibilities, each with a probability of ½: your envelope holds Z, the other 2Z. Or your envelope holds 2Z, the other Z. By considering both possibilities, Scenario 3 takes the holistic approach. 
Then, the expected content of the ‘other’ envelope, conditional on what is in your envelope, is   
   ½ (Expected amount in other envelope, given that your envelope contains Z) 
+ ½ (Expected amount in other envelope, given that your envelope contans 2Z)
= ½ (2Z) + ½ Z = 3/2Z 
We see that each of the two envelopes contains, on average, the same amount, namely 3/2Z. This is, of course, what one should have expected: since the total amount contained in both envelopes is 3Z, each envelope contains, on average, half of that amount.
So there’s no point in switching.

Technical supplement:

The faulty reasoning can be seen as follows: Let’s say that a $10-bill and a $20-dollar bill are randomly stuffed into the two envelopes. You do not know what exactly was stuffed into the envelopes. For all you know, it could have been $5 and $10, or $20 and $40.
The naïve, incorrect manner to analyze the situation, that we set out in Scenario 1 above, is to implicitly assume that your envelope contains $10. Since you know nothing about the actual dollar-amounts, the other envelope could contain either $5 or $20. Or you could implicitly assume that your envelope contains the $20-bill. Then, for all you know, the other envelope could contain either $10 or $40. But neither $5 nor $40 are possibilities. That is why it is incorrect to compute expected amounts without regard of what is in your own envelope. The correct manner of analysis is the holistic approach of Scenario 3.


The Two Envelope Problem is different from the Monty Hall Problem (Chapter ////). There, the choice was between three doors, and the TV presenter knew the lucky one. If the contestant points to an unlucky door (probability 1/3), and the presenter opens a door with a goat, he collapsed the entire remaining probability of 2/3 to the lucky door. 

© George Szpiro, 2019

Comments. corrections, observations:    

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