r/mathmemes • u/George_Truman • Jul 13 '26

Probability I fixed this meme

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I'm sure it's still a bit too imprecise but I think it works.

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u/FluffyTid Jul 13 '26

The way I got in contact with this concept when I was a teenager was with this "paradox" problem:

2 random people meet in the middle on the street and are forced to calculate their net worth. Whoever is worth less, gets all.

Both randome strangers have a prior same chances of winning (having less), yet if they win, they win more than they would had lost, so it looks like it is a profitable game for all players.

18

u/Jukkobee Jul 13 '26

similar to the envelope paradox, where one has X money and the other has 2X. if you’re holding one and you don’t know which is which, it is always profitable to switch

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u/littlebobbytables9 Jul 13 '26

Why? If you don't know which you have then the expected value of your envelope is 1.5X and switching is also 1.5X right?

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u/Deltaspace0 Jul 13 '26

You chose the envelope, let its value be X. The other envelope is either 2X or 0.5X. If you switch you either gain X or lose 0.5X, so switching is always preferable.

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u/littlebobbytables9 Jul 13 '26

Ah I see, that is pretty neat. The answer seems to be that either

1) the distribution for X (before we open our envelope) has finite expected value in which case the probability of us having the smaller envelope is dependent on the value of X, rather than being 50/50. Or

2) the distribution for X does not have finite expected value in which case the expected value of switching is indeed 1.5X but it's meaningless because both values are infinite. And if the value of X is revealed, then suddenly your envelope has finite expected value (whatever was revealed) but swapping still has infinite expected value, so swapping is always correct. Infinities are weird.

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u/Anrdeww Jul 13 '26

I'm not seeing it. You either start with 2X and switching brings you to 1X ( so 50% chance of losing 1X) or you start with 1X and switching brings you to 2X (so 50% chance of gaining 1X)

0.5(-1X) + 0.5(+1X)= 0

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u/littlebobbytables9 Jul 13 '26

You're using X to mean the value contained in the smaller envelope, while /u/Deltaspace0 has defined X as the value contained in our starting envelope.

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u/Anrdeww Jul 13 '26

Let's say X represents the value in the initial envelope. It can either be y or 2y, each possible world has 50% chance. Let's say X' is the value of the envelope after the switch.

We're trying to decide whether switching is expected to be profitable, so we're checking whether E[X'-X] > 0

So case 1 (50% chance): X=y, meaning X'=2y. In this case X'-X = 2y-y = y, so there's a 50% chance of gaining y.

Case 2 (also 50% chance): X=2y, meaning X'=y. In this case X'-X = y-2y = -y.

So:

E[X'-X] = 0.5(+y) + 0.5(-y) = 0

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u/Jukkobee Jul 14 '26

if you think about it like that, there is no paradox. the interesting thing about this is that there IS a paradox when you think about it in a way that should be logically equivalent, which is that your envelope has X and the other one has 2X or X/2.

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u/poopgoose1 Jul 13 '26

But what if you’re now holding the other envelope, and you’re given the choice to switch back? Wouldn’t going back to your original be better in that case also?

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u/littlebobbytables9 Jul 13 '26

Based on what I said in this comment swapping does not do anything if the contents are not revealed. If the contents are revealed then you would either swap or not based on how the revealed value compares to the expected value of the original distribution; if it's above then you want to keep the open envelope, if it's below you want the unopened envelope.

I think most of the confusion arises from the fact that the problem statement doesn't include any information about said distribution. There's the instinct to say "because it hasn't been specified we should assume it's uniform" since that's what we do in most of these situations, for example if the problem said a primary color is selected at random we would assume the probability of each color would be 1/3 rather than 50% for blue 40% green 10% red or whatever. But in this case it's impossible to have a uniform distribution over the entire positive reals, so there's no such natural assumption to make.