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?
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.
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.
This game is fundimentally different from the game described in the op, and understanding the difference is key to resolving the paradox.
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.
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.
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”.
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?
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.
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.
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’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.
Lemme try…
EV (other envelope) .5(2x)+.5(x)=1.25x
EV (your envelope) .5(x)+ .5(.5x)=.75x
EV of trading=.51.25+.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.
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?
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.
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.’
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.