Two Envelopes Problem: Am I just dumb?

It seems the recent craze about statistician being a profession of choice in the future gains steam. In future where we will be surrounded by quality BigData, capable computers and bug-free open source software including OpenDremel. Well the last one I made up… but the rest seems to be the current situation. Acknowledging this I was checking what is the state of open-source statistics software and who are the guys behind it and etc.. But it is not my today topic, it is the topic of one of my next posts. Today I want to talk about one of the strangest problems/paradoxes on the internet I have ever seen. The story is, that I encountered right now “the two envelops problem” or “paradox” as some put it. Having worked with math guys for a long time (having math mastery that I’ll never reach in my life) I immediately recognized that problem as I was teased by it few times a long time a go. However, I was never told that this is such a big deal of a problem. Wikipedia lists it under “Unsolved problems in statistics”. Heh? And I never understood what is so paradoxical or hard or even interesting in it? For me it seemed high-school grade problem at most. So I put “two envelopes problem” into Google and found tens of blogs trying to explain it and propose over-engineered over-complicated and long solutions to such a simple problem. I have a very strange feeling that I’m either totally dumb or a genius and I know I’m not a genius In some sources it is mentioned that only Bayesian subjectivists suffer from this, however in large majority of other sources it is presented as an universal problem… Well enough talking lets dive in into simplest solution on internet (or I will be embarrassed by someone pointing my mistake or similar solution published elsewhere).

The problem description for those who never heard about it:

You are approached by a guy that shows you two identical envelopes. Both envelopes contain money. You are allowed to pick and keep any one of them for yourself. After you pick one, the guy makes you an offer to swap envelopes. The question is if one should swap. For me it is as clear as sunny day that it doesn’t matter if you swap and it is easily provable by simple math. Somehow most folks (some Ph.D. level!) get into very hairy calculations that suggest that one should swap and than even more hairy ones why one should not. Some mention subjectivity but most don’t.

The simplest solution on internet (joking… but seriously I haven’t found such simple one):

Let’s denote the smaller sum in one of envelopes as X and therefore the larger sum in the other will be 2X. Then expected value of current envelope selection before swap consideration is 1.5X. How I got to it? Very simply we have 0.5 probability of holding the envelope with larger sum that is 2X and 0.5 probability with holding an envelope with smaller sum witch is X. So:

0.5*2X+0.5*X = X+0.5X=1.5X

So far so good… let’s now calculate the expected value if we swap. If we swap, we will have same 0.5 probability of holding larger sum and same 0.5 probability of holding the smaller sum. Needless to repeat the calculation, you will get exactly same 1.5X as an expected value, meaning that the swap doesn’t matter. Or if time has any value it doesn’t make sense to waste it by swapping envelopes.

Do you see it as hard problem? I bet 10-year old will do fine with it, especially if offered some reward.

How come others get lost here?

The answer is that some try to apply Bayesian subjectivism probability theory and then innocent folks follow it and gets lost as well.

If you look to Wikipedia article for example you will find a classic wrong solution that allegedly is “obvious” and then a link outside Wikipedia to a “correct” solution. The correct solution seems a long post with a lot of formulae and usage of Bayes theorem that at the end came to correct answer.

Well… I see clearly a flaw with the solution published in wikipedia. That solution really looks artificial, but according to the number of followers it should be obvious for many. The blunder is in the third line:

The other envelope may contain either 2A or A/2

By A they denoted the sum in the envelope they are holding. The mistake is in “either 2A or A/2″, it should be “either 2A or A”, Then everything will be ok and no “paradox” will emerge in the end. The mistake stems from the fallacy of using same name for two separate variables that are dependent but not equal! And then repeatedly confusing them since they have same name. Here is a “patch” to be applied to wikipedia published reasoning:

1. I denote by A the amount in my selected envelope. => FINE
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2. => FINE
3. The other envelope may contain either 2A or A/2. =>INCORRECT variable A denotes different values so it is highly confusing to write it this way.

let’s explicitly consider two cases here instead of implicit “either..or…”

in first case let’s assume we are holding the smaller sum then the other envelope contains 2A
in the second case let’s assume we have holding the larger sum then the other envelope contains A/2. However, the A is different from A of first case so let’s write it as _A/2

Moreover we know that _A is not just different from A but is exactly twice the other so
_A = 2A
So the expression “either 2A or A/2″ must be written as “either 2A or _A/2″ or substituting _A=2A as “either 2A or A”.

Then for calculating expected value you also substitute A instead of A/2 and get same expected value than before swap.

That said, I saw many people feeling so “enlightened” by reading a complicated “correct” solution that they erroneously think and argue that one should not accept the following offer thinking it is equivalent to the above problem (well not exactly this but I rephrased it for clarity):

One guy comes to you and says there are three envelopes. You are allowed to pick one and keep it. One envelope is red and two are white. All three contain money. One of white envelopes contain twice as many as red one. Another white one contains half of red one. The white envelopes are identical and there is no way to know which one contains double and which one contains half. The question is which envelope you should choose: the red one or one of the white ones. And the answer is that you should pick one of white envelopes! In fact the calculation errorneously applied to the two-envelopes problem is 100% correct to the three-envelopes-problem. And on average you will win choosing one of white envelopes rather than a red one.

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