On a table before you lie two envelopes, identical in appearance. You are told that one envelope contains twice as much money as the other, but given no indication which has more. You are then allowed to choose one of the envelopes and keep the money it contains.

You then pick one of the envelopes at random, but before you look inside you are offered the chance to exchange your envelope for the other. Should you exchange envelopes, (assuming that the more money you get, the better)?

One line of thinking is this: If the envelope you initially choose has \(x\) dollars in it, then there is a \(50\)% chance that the other envelope has \(2x\) dollars in it and a \(50\)% chance that it has \(\frac{x}{2}\) dollars in it. Thus the expected amount of money you would end up with if you chose to exchange envelopes would be

\(\frac{1}{2} * 2x + \frac{1}{2} * \frac{x}{2} = \frac{5}{4} x\) dollars.

You thus decide to exchange envelopes. But before doing so, you think to yourself, "But I could go through the same thought process with the other envelope, concluding that I should exchange my new envelope for my old one, and then go through the same thought process again, and again, ad infinitum. Now I have no idea what to do."

How can you resolve this paradox?

Edit: As mentioned in my comment below, the real "puzzle" here is to identify the flaw in the reasoning I outlined above. Or is there a flaw?

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TopNewestFirst, look at the case when the envelope is exchanged. If you pick the envelope with \(x\) dollars \((P=1/2)\) and exchange, you win \(2x\) dollars. Instead, if you pick the envelope with \(2x\) dollars \((P=1/2)\) and exchange, then you win \(x\) dollars.

\[\therefore E(\text{Money when envelope is exchanged})=\dfrac{1}{2}2x+\dfrac{1}{2}x=\dfrac{3x}{2}\]

Now, look at the case when the envelope is not exchanged. If you pick the envelope with \(x\)dollars \((P=1/2)\) and don't exchange, you win \(x\) dollars. But if you pick the envelope with \(2x\) dollars \((P=1/2)\) and don't exchange, then you win \(2x\) dollars.

\[\therefore E(\text{Money when envelope is not exchanged})=\dfrac{1}{2}2x+\dfrac{1}{2}x=\dfrac{3x}{2}\]

So it makes no difference at all whether you exchange or don't exchange your envelope. – Pratik Shastri · 1 year, 11 months ago

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But the "real" puzzle here is to find the logical miscue in the line of reasoning I laid out. It can be argued that in the two scenarios I describe, (namely that I start out with the lesser amount of money or the greater), the value \(x\) actually represents different amounts in each case, so that I can't use the same variable \(x\) for each of them, making my equation invalid.

However, suppose I did peek at the amount in the envelope before making a decision. Now this provides me with "information", but in reality it's useless since I have no clue as to whether it's the lesser or greater amount. But since we now can attach an actual value to \(x\), the mathematics I applied in my initial analysis becomes re-validated, and hence so does my "conclusion". But since, as noted, the information regarding the amount in the initially chosen envelope is in reality useless, should it matter whether we know the actual amount as far as the mathematical analysis is concerned? From this perspective, no matter what amount we see inside the initial envelope we should conclude that switching envelopes is preferable.

So the puzzle of "where's the flaw in logic?" is more a question of perspective; are we dealing with unconditional or conditional probability values, and why should this difference in perspective matter given the above discussion? A surprising amount of attention has been given to this paradox, commonly referred to as the "Two Envelope Paradox", with no universally accepted resolution as of yet. I think why it does garner such attention is that it cuts to the heart of the foundation of probability theory, and as with any foundation, the desire is that it must be as secure as possible, and if it is not, it must be either fortified or torn down.

Here is an accessible discussion on the "search for the flaw" and why many find it so intriguing. – Brian Charlesworth · 1 year, 11 months ago

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Here's another question: Are we then told that "hey, you chose the lesser or greater amount of money" after we've made our choice? I think not. So the probability will not even depend on outcome. For all intents and purposes, we need not even bother with the maths of 'x' and '2x'; and we might as well adopt an approach with x and y since the relationship between the monies simply do not matter. I might as well be a choice between winning the lottery and detonating an atomic bomb. – Olawale Olayemi · 1 year, 7 months ago

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As for the choice between winning the lottery and detonating an atomic bomb, well, I think I would just walk away from the table; too many lives other than my own would be at stake without them having any say in the matter. However, if it were a choice between my eternal health, wealth and happiness on the one hand and my immediate extinction on the other, I would give it a bit more thought. I would probably still walk away, since it would feel like I'm making some kind of deal with the Devil, but it does make for an interesting psychological experiment. This is an entirely different question than the previous "paradox" we were dealing with; with the money in the two envelopes, no matter what choice we made we would be richer than before. With this new question, there is the possibility of a terrible outcome, so the first choice becomes whether to engage at all, and if we do then why do we think it is worth such a high risk of calamity. The more we would have to lose in our present life the less likely we would engage, but if we did not have much to live for then the risk might be worth it. However, as I said before, only the Devil would make such a proposition, so anyone, no matter what their circumstances, would be wise to just walk away. Sorry, I do ramble on, but i always do that when a good philosophical question happens by. :) – Brian Charlesworth · 1 year, 7 months ago

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– Olawale Olayemi · 1 year, 7 months ago

Hahaha.I do love to read you rambling on. It's more engaging than many other types of conversations. Your ending pretty much wrapped up my premise: which is that, the paradox seems to rely on the probability of making a more money outcome which is identical to that of making a lesser money. I do have some philosophical questions if you're interested, I'd love to read your views. I'm relatively new on this site so is it OK if I create a note and invite you to give your thoughts on it?Log in to reply

– Brian Charlesworth · 1 year, 7 months ago

Of course; I look forward to seeing what the topic is. :)Log in to reply

Here are two notes I have posted. I invite you to share your thoughts on them: https://brilliant.org/discussions/thread/number-bases-in-a-different-world/# https://brilliant.org/discussions/thread/the-infiniteness-of-light/?ref_id=654556

Hi @Calvin Lin. Do you think it's a good idea to find a way to incorporate Philosophy (Logic and Reasoning) on Brilliant? I think it is a very fascinating pseudoscience and it would appeal to many older people and others who are not strictly geeks and nerds like the rest of us. Just a thought – Olawale Olayemi · 1 year, 7 months ago

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I'm reading Brian's posted scenario, and it seems to me that if it's known that one envelope contains twice the other, then the expected value in choosing either is simply \(\dfrac { 1 }{ 2 } x+\dfrac { 1 }{ 2 } 2x=\dfrac { 3 }{ 2 } x\) either way, as pointed out by Pratik. It's 50% percent chance that it's twice the one you've chosen, and 50% percent chance it's half that--but you can't assume the same \(x\) dollars in your chosen envelope in both cases! That was the flaw. Maybe Brian is trying to alluding to the problem of trying to determine probabilities when information is incomplete, which is what led to Bayesian probability theory. Let me offer a slightly different scenario. You are given \(2\) envelopes. You are told that one of them may or may not contain a valuable stock certificate. You open one up and do not find it inside. What's the probability the other one does? Extend that to \(100\) such envelopes. This is actually what happens when you are looking for gold in the ground in a certain area when you don't have much information to go on. After so much digging, at what point in time do you give up? Are more likely to find gold, because you've already almost eliminated the places where it can be, or are you more less likely to find gold, because you've already tried to find it and failed? A lot of science can be like that. – Michael Mendrin · 1 year, 7 months ago

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I'm fascinated this "blind" probability theory. It is a little bit anti cognitive. Which reminds me of something else which defies our flawed cognizance. Consider flipping a coin and getting heads five times in a row. Flipping it a sixth time seems as though there is a higher probability to get tails. In fact, I'd bet that if you did a survey of ten mathematics professors, a higher percentage of them would choose tails for that sixth coin toss except they are just fighting against their human nature. But sometimes I think about it and really wonder if the law of averages is that subjective. If you mix two liquids together, they mix together perfectly EVERY TIME if they are of similar viscosity. I'm rambling. I have even forgotten what the point I'm trying to make is – Olawale Olayemi · 1 year, 7 months ago

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Likewise, a lot of technology was preceded by science fiction. Smart phones? Didn't Star Trek popularize that idea? – Michael Mendrin · 1 year, 7 months ago

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