The Two Envelopes Paradox

With two envelopes provided before you one containing twice as money as other and visually identical and you get to choose which one you want to take home. But, before you get to open the envelope and find out whether you won the maximum, you have an opportunity to switch to the other envelope. Should you?

No, because you have equal probability of 50% of having the maximum or the other one. Right? Wrong!! Given the opportunity, you should switch and there is a mathematical explanation for this as well.

Origin of this problem is as confusing as the problem itself - In his book of Mathematical Recreation, Belgian mathematician Maurice Kraitchik, mentions about a different version of our problem - it is known as the Necktie paradox where two men compare the values of neckties which their wives bought them with neither of the men knowing the original prices.

There is also a similar reference using two playing cards giving credits to physicist Erwin Schrödinger - yes the same guy known for Schrödinger's cat paradox where the cat in a box is both alive and dead at the same time until you open the box!!

In our problem, the mathematics behind it is same irrespective of using playing cards, neckties or envelopes

The reason for not switching the envelope is very simple - the fact that we don't know what is inside the envelope makes 50% probability on both envelopes. But the mathematical reason behind switching is quite compelling.

Like all the mathematical proofs, lets start with some assumptions,

Let us assume that there are two envelopes e1 and e2 and let X be the amount present in envelope e1, so, the other envelope e2 can contain any of the two outcomes  2X or (1/2)*X and let e1 be the initially chosen envelope. With this initial setup,  we can now calculate the expected value for choosing envelope e2(E(e2)).

To find the expected value of the envelope we did not choose (envelope e2), we add the value of the two possibilities (2X and (1/2)*X) weighted by their probabilities.

//Expected value of choosing e2

E(e2)   = (1/2)2X +(1/2)(X/2)

= (5/4)X

= 1.25 X // 25% more than our choice

So, with this our best possible choice is to switch to the other envelope, here envelope e2.

But does that make any sense in the real world? Lets see, this solution even gets weirder. What if I say that you could switch envelopes again if you want to? Then you do the same process again and again always switching, never opening because switching gives the mathematical advantage, gaining a 25% advantage, I just proved that. So.?

Mathematician and Logician, Raymond Smullyan proposes two scenarios for the paradox,

Risk of gaining twice the amount in your envelope is worthwhile when the alternative is only losing half of it. The potential payoff is twice as high as your risk of loss and hence the expected value is higher.

But, by problem statement, one envelope (say e1) contains X and the other (e2) contains 2X. It makes sense that either we win X or lose X. In our assumptions we considered envelope e2 contains either 2X or (1/2)*X. We know both can't be true!

So, think of the problem in this way, the total sum of money in the envelopes is 3 units. So, if you get the smaller envelope then its value will be X/3 and the value in the larger envelope will be 2X/3.

When we switch from small to large we gain X/3 and when we switch from large to small, we lose X/3 (or gain - X/3) . So the expected value for switching is,

E(e2) = (1/2)*(X/3) + (1/2)*(-X/3) = 0

Same applies for switching from e2 to e1. So the expected value is 0. Now, what is the point of switching? Should we switch now? No, not necessarily! Paradox solved

This problem is more than X and 2X which makes the problem a Falsidical Paradox because there appears to be a gain from switching when we fail to consider  that the states of the two envelopes don't actually reflect the values of X and 2X.

The 2X envelope is only 2X envelope only if the X envelope is smaller, similarly in our previous consideration (1/2)*X envelope is only (1/2)*X only if the X envelope is larger. So, the envelope containing X cannot be both the larger envelope and smaller envelope. This lack of consideration made us to think that switching makes us gain 25%. Just a matter of perspective.

No matter we solve it mathematically or we try to perceive the logic, no matter what you do, when it comes to the two envelope paradox, you were right from the beginning