# Can somebody explain the two envelope paradox to me **URL:** \ **Category:** Factual Questions\ **Created:** [November 15, 2010, 8:25am UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715 "2010-11-15T08:25:52Z")\ **Posts on this page:** 20\ **Page:** 10
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [November 19, 2010, 8:42pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/181 "2010-11-19T20:42:04Z")
> [@BarryB](#): > > You aren’t using the “established formula” correctly. You are using a random variable X as a constant in a probability calculation, when the probabilities depend on different values for X. You can’t get a value for the expected value of Y without using the expected value of X, not X itself. X is one of (a,b) (let’s assume a \> b), and Y is the other one, and both a and b are fixed before you choose and can’t change. If you want to use X = a, then you also have to use Y = b, and the expected value of a constant is a constant. If you want to take both possibly outcomes into account, you can’t say that X = a in both of them, since the other outcome isn’t X = a, Y\>a, it’s X = b, Y = a. Instead you are making X fixed, but then making Y and therefore what was in both the envelopes before you started uncertain. It is uncertain to you, but it isn’t uncertain to the person who put the money in the envelopes. It isn’t uncertain to physical reality. Your calculation is along the lines of “if the envelopes contained 1 and 2, you would have a higher expected value than if the envelopes contained 1 and 0.5”. It has nothing to do with _your choice_, it has to do with how the predetermined amounts of money in the envelopes is distributed. How do you feel about the bet scenario? That’s where I have a dollar and somebody offers to bet me for my dollar. They’ll flip a coin - heads I get two dollars and tails I get fifty cents. Do you feel I can calculate the expected value of the bet in this scenario? Or is it impossible for me to do so because I don’t know what the coin will do? ---
**Author:** ![BarryB](https://avatars.discourse-cdn.com/v4/letter/b/d78d45/32.png) [@BarryB](https://boards.straightdope.com/u/BarryB)\ **Post date:** [November 19, 2010, 8:57pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/182 "2010-11-19T20:57:37Z")
I bet $1. I win. I get $2. I bet again. I can win $4, with 50% chance. I have $1 in my envelope. I swap. I get $2. Can I swap again and get $4? No, since the other envelope has 100% chance of being $1. What isn’t true: Your envelope has $1, with 50% chance of the other being bigger or smaller. Either: Your envelope has $1, with “some” chance of the other being bigger or smaller. This “some” is related to the probability that the person put in $1.50 total and the probability that the person put in $3 total, **not** that you picked the larger or smaller value, which is 50-50. This is where the infinite sum relating to E[Y | X = 1] comes in. Or: Your envelope has 50% of being larger or smaller. If $1.50 was put in ,then if you got $1 you **must** have gotten the larger value. You also have to consider that you **could** have gotten $0.50. ---
**Author:** ![Unpronounceable](https://avatars.discourse-cdn.com/v4/letter/u/9de0a6/32.png) [@Unpronounceable](https://boards.straightdope.com/u/Unpronounceable)\ **Post date:** [November 19, 2010, 9:00pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/183 "2010-11-19T21:00:03Z")
> [@Little\_Nemo](#): > > How do you feel about the bet scenario? That’s where I have a dollar and somebody offers to bet me for my dollar. They’ll flip a coin - heads I get two dollars and tails I get fifty cents. > > Do you feel I can calculate the expected value of the bet in this scenario? Or is it impossible for me to do so because I don’t know what the coin will do? Actually, in that case, it **is** always better to take the bet: you’re winning 100%, but only losing 50%, even though the odds are 50-50. ---
**Author:** ![RedSwinglineOne](https://avatars.discourse-cdn.com/v4/letter/r/a9adbd/32.png) [@RedSwinglineOne](https://boards.straightdope.com/u/RedSwinglineOne)\ **Post date:** [November 19, 2010, 9:00pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/184 "2010-11-19T21:00:27Z")
> [@Little\_Nemo](#): > > How do you feel about the bet scenario? That’s where I have a dollar and somebody offers to bet me for my dollar. They’ll flip a coin - heads I get two dollars and tails I get fifty cents. > > Do you feel I can calculate the expected value of the bet in this scenario? Or is it impossible for me to do so because I don’t know what the coin will do? This game is fundimentally different from the game described in the op, and understanding the difference is key to resolving the paradox. ---
**Author:** ![BarryB](https://avatars.discourse-cdn.com/v4/letter/b/d78d45/32.png) [@BarryB](https://boards.straightdope.com/u/BarryB)\ **Post date:** [November 19, 2010, 9:05pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/185 "2010-11-19T21:05:54Z")
> [@Indistinguishable](#): > > Put another way: **BarryB** , you and many others seem to be interested in the calculations “The expected value of Y, given that one envelope contains a dollar and the other envelope contains two dollars” (which is $1.50) and “The expected value of Y, given that one envelope contains a dollar and the other envelope contains 50 cents)” (which is 75 cents). Fine, those are things, but they’re not the relevant expected values in the situation **Little Nemo** describes, because the situation they describe isn’t one where what you know is that “one envelope contains a dollar and the other envelope contains two dollars”, and it isn’t one where what you know is that “one envelope contains a dollar and the other envelope contains 50 cents”. The situation they describe is one where what you know is that “the X envelope contains a dollar”. And “the expected value of Y, given that the X envelope contains a dollar” is indeed $1.25. No, the “fact” that E[Y | X = 1] = 1.25 is useless when considering if you should swap envelopes. That is what **Little Nemo** is trying to find, if you should swap envelopes. E[Y] = E. E[Y | X = 1] depends on what was put into the envelopes to begin with. What is important in this situation is E[Y | X =1 and Little Nemo picked the envelope that would become X as X instead of the other one], which you seem to have ignored. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [November 19, 2010, 9:19pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/186 "2010-11-19T21:19:57Z")
> [@BarryB](#): > > No, the “fact” that E[Y | X = 1] = 1.25 is useless when considering if you should swap envelopes. That is what **Little Nemo** is trying to find, if you should swap envelopes. E[Y] = E. E[Y | X = 1] depends on what was put into the envelopes to begin with. What is important in this situation is E[Y | X =1 and Little Nemo picked the envelope that would become X as X instead of the other one], which you seem to have ignored. **Little Nemo** has access to the information that the first envelope contains $1. So they should condition on that information in their calculations, as they correctly did. ---
**Author:** ![glowacks](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/glowacks/32/5548_2.png) [@glowacks](https://boards.straightdope.com/u/glowacks)\ **Post date:** [November 19, 2010, 9:33pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/187 "2010-11-19T21:33:13Z")
This thread moves far too fast for me to keep up given I want to read more than just this thread in my available time. I’m satisfied I gave a complete and thorough answer, exploring the reasons why people think it’s paradoxical when it’s not. Naive probability theory is just plain wrong; one needs to axiomatize one’s thought and stop looking at things from a non-rigorous perspective. There are plenty of true things that do not have simple explanations; again, it is paradoxical that an electron by virtue of its opposite charge does not get drawn into the nucleus of an atom, and that the protons of the nucleus stay together despite holding the same charge. There have been plenty of attempts made in this thread to explain the situation, and part of the problem is that it just isn’t intuitive from first principles - you have to do the math. ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [November 19, 2010, 9:42pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/188 "2010-11-19T21:42:47Z")
If I parse out the relevant from the misleading, the overly complicated, and the pedantic nitpick, here is what I come up with: You calculate an “expected value” for the other envelope based upon a known/assumed value for your current envelope. Then you compare that expected value to the known/assumed value. It is this comparison that is invalid. You should compare the expected value against the expected value of your current envelope. When you do, you see the expected values are the same. Ergo, the expected value calculation is not helpful in determining whether to switch in this situation. The paradox comes from misusing expected value by comparing against the wrong information. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [November 19, 2010, 9:51pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/189 "2010-11-19T21:51:49Z")
> [@Irishman](#): > > If I parse out the relevant from the misleading, the overly complicated, and the pedantic nitpick, here is what I come up with: > > You calculate an “expected value” for the other envelope based upon a known/assumed value for your current envelope. Then you compare that expected value to the known/assumed value. It is this comparison that is invalid. You should compare the expected value against the expected value of your current envelope. When you do, you see the expected values are the same. Ergo, the expected value calculation is not helpful in determining whether to switch in this situation. > > The paradox comes from misusing expected value by comparing against the wrong information. This is definitely not true. For one thing if you’ve opened your envelope, you’re not dealing with unknown values. You know the value of your envelope and you know the two possible values of the other envelope. And the expected values are not the same - if that were the case the other envelope would be “double or nothing” the value of your envelope when it’s actually “double or half”. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [November 19, 2010, 9:57pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/190 "2010-11-19T21:57:41Z")
> [@](#): > > How do you feel about the bet scenario? That’s where I have a dollar and somebody offers to bet me for my dollar. They’ll flip a coin - heads I get two dollars and tails I get fifty cents. > > Do you feel I can calculate the expected value of the bet in this scenario? Or is it impossible for me to do so because I don’t know what the coin will do? > [@](#): > > This game is fundimentally different from the game described in the op, and understanding the difference is key to resolving the paradox. Take it to the next step. You have an open envelope with a dollar in it. You’re offered an opportunity to trade it for a sealed envelope that has an equal chance of contained either fifty cents or two dollars. Do you trade envelopes? ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [November 19, 2010, 10:09pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/191 "2010-11-19T22:09:48Z")
> [@BarryB](#): > > What isn’t true: Your envelope has $1, with 50% chance of the other being bigger or smaller. > Either: Your envelope has $1, with “some” chance of the other being bigger or smaller. This “some” is related to the probability that the person put in $1.50 total and the probability that the person put in $3 total, **not** that you picked the larger or smaller value, which is 50-50. This is where the infinite sum relating to E[Y | X = 1] comes in. > Or: Your envelope has 50% of being larger or smaller. If $1.50 was put in ,then if you got $1 you **must** have gotten the larger value. You also have to consider that you **could** have gotten $0.50. That’s only because you’re starting with the assumption that you somehow know what the total amount is. You don’t - no scenario that’s been discussed has offered you that information. What you know is that you opened an envelope with a dollar in it. And you know there’s a 50% chance that you picked the larger envelope, in which case the total amount is $1.50 and the other envelope contains fifty cents, and there’s a 50% chance that you picked the smaller envelope, in which case the total amount is $3.00 and the other envelope contains two dollars. You seem to be saying that the chance that the total amount is $1.50 is somehow more or less than the chance that the total amount is $3.00. That may be abstractly true in the sense that there’s some minute number of people in the world who have $1.50 but don’t have $3.00. But the actual effect is infintesimal - you might calculate the odds of it being $1.50 as .500000000000000000000001 and the odds of it being $3.00 as .499999999999999999999999. ---
**Author:** ![BarryB](https://avatars.discourse-cdn.com/v4/letter/b/d78d45/32.png) [@BarryB](https://boards.straightdope.com/u/BarryB)\ **Post date:** [November 19, 2010, 10:36pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/192 "2010-11-19T22:36:54Z")
> [@Little\_Nemo](#): > > Take it to the next step. You have an open envelope with a dollar in it. You’re offered an opportunity to trade it for a sealed envelope that has an equal chance of contained either fifty cents or two dollars. Do you trade envelopes? Why do they have an equal chance? ---
**Author:** ![Unpronounceable](https://avatars.discourse-cdn.com/v4/letter/u/9de0a6/32.png) [@Unpronounceable](https://boards.straightdope.com/u/Unpronounceable)\ **Post date:** [November 19, 2010, 10:40pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/193 "2010-11-19T22:40:42Z")
> [@BarryB](#): > > Why do they have an equal chance? They don’t really - one has a 100% chance, the other has a 0% chance. You simply have no way of knowing which is which. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [November 19, 2010, 10:50pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/194 "2010-11-19T22:50:54Z")
> [@BarryB](#): > > Why do they have an equal chance? Because you made your first pick from two identically appearing envelopes, one of which contained a larger amount and one of which contained a smaller amount. So you have an equal chance of picking the larger or smaller envelope. Which means there’s an equal chance that the unpicked envelope is smaller of larger. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [November 19, 2010, 10:53pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/195 "2010-11-19T22:53:55Z")
> [@Unpronounceable](#): > > They don’t really - one has a 100% chance, the other has a 0% chance. You simply have no way of knowing which is which. If you’re going to flip a coin, you don’t say I have a 100% chance of it being heads and a 0% chance of it being tails if it comes up heads and a 100% chance of it being tails and a 0% chance of it being heads if it comes up tails. You say I have a 50% chance of it being heads and a 50% chance of it being tails. ---
**Author:** ![UltraVires](https://avatars.discourse-cdn.com/v4/letter/u/ecccb3/32.png) [@UltraVires](https://boards.straightdope.com/u/UltraVires)\ **Post date:** [November 19, 2010, 10:56pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/196 "2010-11-19T22:56:56Z")
Lemme try… EV (other envelope) .5(2x)+.5(x)=1.25x EV (your envelope) .5(x)+ .5(.5x)=.75x EV of trading=.5_1.25+.5_.75=1 The other envelope EV might be worth 1.25, but you have to factor in the .75 that you are giving up. ---
**Author:** ![BarryB](https://avatars.discourse-cdn.com/v4/letter/b/d78d45/32.png) [@BarryB](https://boards.straightdope.com/u/BarryB)\ **Post date:** [November 19, 2010, 11:04pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/197 "2010-11-19T23:04:00Z")
> [@Little\_Nemo](#): > > Because you made your first pick from two identically appearing envelopes, one of which contained a larger amount and one of which contained a smaller amount. So you have an equal chance of picking the larger or smaller envelope. Which means there’s an equal chance that the unpicked envelope is smaller of larger. If you had picked the smaller envelope, and opened it, would you have seen the same value as if you had picked the larger envelope and opened it? ---
**Author:** ![RedSwinglineOne](https://avatars.discourse-cdn.com/v4/letter/r/a9adbd/32.png) [@RedSwinglineOne](https://boards.straightdope.com/u/RedSwinglineOne)\ **Post date:** [November 19, 2010, 11:10pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/198 "2010-11-19T23:10:50Z")
> [@Little\_Nemo](#): > > Take it to the next step. You have an open envelope with a dollar in it. You’re offered an opportunity to trade it for a sealed envelope that has an equal chance of contained either fifty cents or two dollars. Do you trade envelopes? This is again, not the same game as the op. You absolutely must understand the difference between the two situations to resolve the paradox. The difference is subtle, but significant. ---
**Author:** ![Unpronounceable](https://avatars.discourse-cdn.com/v4/letter/u/9de0a6/32.png) [@Unpronounceable](https://boards.straightdope.com/u/Unpronounceable)\ **Post date:** [November 19, 2010, 11:13pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/199 "2010-11-19T23:13:20Z")
> [@Little\_Nemo](#): > > If you’re going to flip a coin, you don’t say I have a 100% chance of it being heads and a 0% chance of it being tails if it comes up heads and a 100% chance of it being tails and a 0% chance of it being heads if it comes up tails. You say I have a 50% chance of it being heads and a 50% chance of it being tails. On any given toss, yes, the odds are 100% and 0% _ **after-the-fact** _. On future tosses, the odds are 50/50. Once you’ve taken an envelope, then you either have x or 2x. The probability has collapsed. The other envelope has a 100% chance of being the one not in your hand. It doesn’t matter that you don’t know which you have. The whole 1/2x thing is utterly unnecessary. And the source of the ‘paradox.’ ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [November 19, 2010, 11:25pm UTC](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715/200 "2010-11-19T23:25:20Z")
Man, everyone here is wrong except **Little Nemo**. Imagine the following game is played, over and over again: I pick a random number, put that number in one envelope, put twice that number in another envelope. Then I flip a coin and hand the envelope to you, while keeping the other envelope for myself. For any particular k, out of all the times when you get k in your envelope, I, on average, get 1.25 k in mine. That genuinely is true. It’s not wrong. It’s correct. What do you think it is instead? **Little Nemo** ’s calculation that E[Y | X] = 1.25 X is correct. E[Y | X] = E[Y | X, Y = 2X] \* P(Y = 2X | X) + E[Y | X, Y = X/2] \* P(Y = X/2 | X) = 2X \* 1/2 + X/2 \* 1/2 = 1.25 \* X. Given that P(Y = 2X | X) = P(Y = X/2 | X), then that calculation is entirely correct. Those of you attacking **Little Nemo** ’s calculation are missing the salient aspect of the paradox. It’s not that the calculation went wrong. [Previous page](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715.md?page=9) [Next page](https://boards.straightdope.com/t/can-somebody-explain-the-two-envelope-paradox-to-me/560715.md?page=11)