Imagine you are given two identical envelopes, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?
So there's no possible world in which the other envelope may with equal probability contain either two fifty or a tenner, which is the only way to make a profit by switching. — Baden
Given that there's £10 in my envelope, the probability that the other envelope contains £5 is equal to the probability that the other envelope contains £20; that probability being 1/2.
As an analogy, if I flip a coin and hide it from you, it is correct for you to say that the probability that it landed heads is equal to the probability that it landed tails; that probability being 1/2. — Michael
Then you should read this and this. — Michael
All the average return on the envelope tells us is that if we keep playing over the long term, its going to average a return greater than one. — Philosophim
That's the supposed paradox. Switching doesn't increase our expected return, but the reasoning given suggests that it does. So we need to make sense of this contradiction. — Michael
Either you were given X and the other envelope contains X/2, xor you were given X and the other envelope contains 2X. Those are separate cases. If you were given X and the other envelope contains X/2, you'd lose X/2. If you were given X/2 and the other envelope contains X, you'd gain X/2. In the second case, if you were given X and the other envelope contains 2X, you'd gain X, if you were given 2X and the other envelope contains X, you'd lose X. If you're in the first case, the expected gain of switching is 0, if you're in the second case, the expected gain of switching is 0, both cases are equally likely, so the expected gain of switching is 0. — fdrake
Whether you benefit by switching or not is dependent on what envelope you just chose, which must be Envelope X or Envelope 2X where X is a given sum of money. So, switching can only ever take you from Envelope X to Envelope 2X (+X) or from Envelope 2X to X (-X). — Baden
So, you have 10. But only one scenario applies to your choice, either the one where you lose by switching in which case X is five and 2X is 10, or you gain by switching in which X is 10 and 2X is 20. You can't be in both scenarios at the same time and X has a different value in both — Baden
We know the chosen envelope has £10 in your scenario and we also know that one of the envelopes contained an amount X and one of the envelopes contained an amount 2X, and therefore we know we can only move from X to 2X or from 2X to X by switching. Therefore we know we have a 50% chance of gaining X by switching and a 50% chance of losing X by switching and that therefore there is no point in switching.
You're using X to refer to both the X in the scenario where switching gets you less and in the scenario where switching gets you more. But given any given amount you see in front of you after already choosing an envelope, those are not the same X. — Baden
I stand by the claim that the probability that the other envelope contains twice as much as my envelope is equal to the probability that the other envelope contains half as much as my envelope, that probability being 1/2. — Michael
It isn't clear what your random variable z means in your calculation, since its sample space isn't defined. z is a particular fixed value in the first envelope, no? — fdrake
Why not?
I know that one envelope contains twice as much as the other. I pick one at random. What is the probability that I picked the smaller envelope? It seems perfectly correct to say 12
1
2
. — Michael
So by your logic we can't even talk about the probability of a coin toss landing heads being 12
1
2
? — Michael
I don't think that's at all reasonable, or even relevant when we consider puzzles like this. For the sake of puzzles like this we assume a coin toss landing heads has a probability of 12
1
2
, and we assume that my choice of envelope is truly random. — Michael
The only rational response to the two-envelopes problem as it is traditionally stated without additional assumptions, is to reply
"The probability of getting a greater or lesser prize when opening the other envelope, is between 0 and 1" — sime
That's the supposed paradox. Switching doesn't increase our expected return, but the reasoning given suggests that it does. So we need to make sense of this contradiction. — Michael
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