The Two Envelopes- A Simple Yet Unsolved Paradox

Paradoxes have been confounding minds since Aristotle. It is derived from the Geek word “paradoxon” meaning contrary to expectation. Ever been confused by a Paradox? Well not surprising at all.

Paradox is a statement that contradicts itself or has the ability to be both true and untrue at the same time.  For example – ‘This Statement is A Lie’ is one of the most famous paradoxes because it is so simple. These five simple words are self- contradictory: if a statement is true, then it’s a lie, which means it’s not true. But if it’s not true, then it’s a lie, which makes it true.  Paradoxes are considered as reasonable; they might not completely make sense to a person but upon consideration that we realize there is self- defeating logic.

 

The two-envelope problem is yet again a conundrum in the decision-making theory. It has been subjected to a long-standing debate. It is a counterintuitive problem of decision making between two states in the presence of a start uncertainty. Whatever the player chooses, his payoff must be maximized in some fashion. The problem is indeed significant for it involves our in depth understanding of probability theory. 

Origin of the Paradox

The “Two Envelopes Problem”, like its better known cousin, the Monty Hall problem, is seemingly paradoxical if you don’t carefully analyse it. It arose as a variant of the necktie paradox by Maurice Kraitchik, which talks about -two men who are each given necktie by their respective wives, they wager over whose necktie is the cheaper. The terms of the bet is the man with the more expensive tie has to give it to the other man. This paradox is a rephrasing of the simplest case of the two envelopes problem’, and the explanation of "what goes wrong" is essentially the same. 

The Two Envelope Paradox

There are two envelopes, one holding twice as much money as the other (or containing pieces of papers with positive real numbers written on them where one number is twice the size of the other). The player has no knowledge of what is contained in the envelopes, apart from the fact that one envelope contains twice what the other contains. The player then randomly chooses one of the two envelopes and observes the amount of money in it. The player then deduces that the unopened envelope contains either twice the observed amount or half of it. The house then offers the player the opportunity to switch envelopes. The problem is for the player to decide whether or not to switch envelopes. By switching, the player must keep the new amount in the second envelope. If he decides not to switch, the player keeps the initial amount in the first envelope. The player is allowed to open one envelope and, having seen the contents, he may either keep that amount or opt to change the envelope. The question here is, should he swap or not?

Why Is It Counter-intuitive?

Most people would put forward the rationale that since a player has a 50-50 chance of choosing either envelope, they have an equal chance of gaining or losing money whether they decide to switch or keep the original envelope. Most people hence, believe that switching does not matter. This would be true if the player would have chosen an envelope randomly, but that is not the case. The envelope opened depends on the player’s initial choice, so the assumption of independence does not hold.

Suppose the amount in the selected envelope is denoted by A. Then, the probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2. The other envelope may contain either 2A or A/2. If A is the smaller amount, then the other envelope contains 2A. If A is the larger amount, then the other envelope contains A/2. Thus, the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2. So, the expected value of the money in the other envelope is: ½ (2A) + ½ (A/2) = 5/4. Since this amount is greater than A, so the player will gain by swapping. Now, utilizing the mathematical concept of symmetry which tells us that in a scenario when numerous components of anything is identical, then it is always possible to move the object without ultimately changing what it looks like. Similarly, in this case after the player switches, he/ she is satisfied by the decision, and hence when the player is given a second chance to switch envelopes, he will apply the same reasoning and switch again. The same thing repeats for the third, fourth and n^th time, and the player ends up in an infinite loop of switching. 

What is the correct way to think about the problem?

The main problem in this paradox lies in identifying the flaw that exists in the very convincing and what seems to be a very rational chain of reasoning mentioned above. Thus, finding the correct way to think about the problem involves finding out why the paradox isn’t resolved, by simple probabilistic thinking.

The most simplified explanation for the paradox is by reevaluating the Expected Values–

Let’s label the 1st envelope as ‘A’ and the 2nd envelope as B

Now, in the case of this paradox, let’s say the player opens one envelope A to find ₹ x within, before being offered the chance to switch, then as stated earlier in this article, the expected amount of money in the other envelope i.e. envelope B is 5x/4 which is greater than x, therefore the player swaps.

The fallacy in this problem is in the player’s misunderstanding- He assumes that ₹ x, in his calculation of probability is actually representing the same values, while in reality that is not exactly the case- The value of ₹ 2x is the expected value of money in envelope B GIVEN that it is the greater one, and similarly x/2 is the amount of money in the envelope GIVEN that it is the smaller one.

One envelope contains double the money as compared to the other, hence ₹  x is considered as the lesser amount of money and ₹ 2 x is considered as the greater amount.

Total money in the scenario= ₹ 3 x

Therefore, If the Player selects Envelope A first The Expected amount of money in Envelope B GIVEN that value in B is greater than that of A,

E(B) in the envelope is-

 Simply put, calculating expected value of B considering x is small,

E(B)=   1/2 (x +2 x) = 3x/2

The same can be said for Envelope A i.e.

E(A) =  1/2 (x +2 x) = 3x/2

As, E(A) = E(B), the contradiction no longer occurs, as there is no more incentive for the player to swap the envelopes.

A Bayesian and Conditional Probability Approach

Let the lower amount of money in the be denoted by X, and the higher value is fully determined by the lower value, being twice its value. We define a prior probability distribution P(X) for the choice the host makes for the lower value in the envelopes. Now, the question here is – which envelope shall the player first choose to open – is it one that contains the lower value (denoted C=lower), or the one that contains the higher (denoted C=higher)? As the amounts are hidden from the player, he chooses entirely at random, with equal probabilities for the two options. Hence the probability P(C) is 0.5 for both cases.

 Therefore, as mentioned before whatever the amount X might be, Y is not equally likely to be x/2 or 2x. The amount in the second envelope is more likely to be greater than the amount of money in the first envelope.

Using the idea of conditional probability, we first imagine might be in the first envelope. A sensible strategy would certainly be to swap when the first envelope contains 1, as the other must then contain 2. Suppose on the other hand the first envelope contains 2. In that case there are two possibilities: the envelope pair in front of us is either {1, 2} or {2, 4}. All other pairs are impossible. The conditional probability that we are dealing with the {1, 2} pair, given that the first envelope contains 2, can be given by:

and consequently, the probability of the {2, 4} pair is 2/5, since {1,2} and {2,4} are the are the only two possibilities.

So, either the first envelope contains 1, in which case the conditional expected amount in the other envelope is 2, or the first envelope contains X > 1, and though the second envelope is more likely to be smaller than larger, its conditionally expected amount is larger: the conditionally expected amount in the second envelope will be :

3x/10 + 4x/5 = 11x/10

Since, the above value is greater than x, hence it is obvious that the player would switch to the second envelope irrespective of the amount of money in the first envelope.

This is such a paradox that although numerous solutions to it has been given by statisticians, mathematicians and even philosophers, not a single suggested explanation is generally believed to be definitive (Markosian, 2011).  Logician Raymond Smullyan has stated that this problem has nothing to do with probability at all. Thus, even after all these years, the Two Envelope Paradox remains a simple problem, yet with no definite solution.

References-

Clark, M., & Shackel, N. (2000). The Two-Envelope Paradox. Mind, 109(435), 415-442. Retrieved August 5, 2020, from www.jstor.org/stable/2659923

Markosian, N. (2011). A Simple Solution to the Two Envelope Problem. Logos & Episteme, 2(3), 347-357. doi:10.5840/logos-episteme20112318

McGrew, T., Shier, D., & Silverstein, H. (1997). The Two-Envelope Paradox Resolved. Analysis, 57(1), 28-33. Retrieved August 5, 2020, from www.jstor.org/stable/3328431

Linzer, E. (1994). The Two Envelope Paradox. The American Mathematical Monthly, 101(5), 417-419. doi:10.2307/2974901