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Two envelopes problem

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Re: Two envelopes problem

#41

Post by eru »

Nihmie wrote: post #18
antognini wrote: post #12

There is a related puzzle which I've also heard called the Two Envelopes Problem. The premise is that you are handed two envelopes. Inside each envelope is a piece of paper with a number written on it. I have written down these numbers by sampling from some probability distribution, but you don't know what it is. All you know is that the two numbers are different. You get to pick one envelope and then look at the pap…

Is there actually some way to guess with better than 50% chance? I'm pretty sure this could be reduced to the secretary problem where there's a pool of 2 candidates, which yields an optimal hire with probability 50%. Maybe you play word games and say, "I guess the other number is not higher," or "I guess the other number is not lower." Since the number you observed was produced once, there is some non-zero probabilit…

Yes, there is a way to guess better than 50%. But it's a bit weird:

You pick an arbitrary threshold, if your envelope is below the threshold, you switch.

(You can add randomness as necessary.)

Re: Two envelopes problem

#42

Post by CamperBob2 »

scrapheap wrote: post #9
jdkee wrote: post #2

Notably different from the Monty Haul problem where additional information is given to the player of the game prior to making the decision to switch. See https://en.wikipedia.org/wiki/Monty_Hall_problem

Personally I don't think you are given any new information in the Monty Hall problem that's worth anything. You always knew there was a 100% chance that one of the doors in that set has a goat behind it, which door that is is pretty much irrelevant. Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stic…

The Monty Hall problem is easily resolved by imagining there are a million doors instead of just three. The host opens 9,999,998 doors, revealing goats behind each. You'd obviously switch doors to the one the host left closed, and the reasoning you followed will be just as valid for any number of doors including 3.

This problem is completely different, since the number of degrees of freedom remains fixed throughout the decision process. As someone else pointed out, it is trivially resolvable by realizing that no matter how many times you switch envelopes, you're really only making one choice, and doing so with 50-50 odds of picking the more profitable envelope. The offer to switch amounts to meaningless magician's patter, whether the offer is made once or a million times.

Re: Two envelopes problem

#43

Post by kurthr »

dllthomas wrote: post #17
joe_the_user wrote: post #8

Earlier quoted context omitted.

Well, the simple way to put is that "A" isn't fixed. The "expected value" argument in steps 6-7 is acting like "A" is a single value when "A" will be larger or smaller depending on what envelope you picked.

I think that subtly misses the point. The problem is that you're implicitly using a distribution that... isn't a distribution. And with this particular not-a-distribution, whether you should switch or not does not depend on the value of A . But with any actual distribution (... I think?) it does , at which point... no paradox. It's true that it's not clear (at least to me , but perhaps more generally) what distributi…

Well taking the mean of products doesn’t make much sense anyway… but if you take a log and average then you end up with the typical solution. If you take a geometric mean then you end up with 1, which implies there’s no difference either.

Re: Two envelopes problem

#44

Post by beanlog »

What an interesting trick. A starting point to intuit a resolution: yes you have a 50% chance of doubling and a 50% of halving. But the doubling only happens if you have x dollars, and the halving only happens if you have 2x. So you can see you either gain x or lose x, so your expected return is 0.

When the amount you multiply your money by depends on the amount of money you currently have, you have to factor in your initial money to each case to compute expected return.

Re: Two envelopes problem

#45

Post by accountLost »

My take :

let M be the maximum amount of money that a envelope can contain (at worst it can be the total amount of money in the world). You don't know the value of this upper bound M.

Most of the time, you get 2*x or x/2 with 50% chance if you choose the other envelope (which on average is a win)

But when x=M you get M/2 instead of M and therefore you loose M/2 with 100% chance. And when it happens you have lost much more than in the other cases (because exponentiel...)

If you do the math, you can see that what you win for 1 2 4 8 16 ... M/2 is equal to what you loose for M.

And so on average what win from switching is 0.

Re: Two envelopes problem

#47

Post by cyborgx7 »

beanlog wrote: post #44

What an interesting trick. A starting point to intuit a resolution: yes you have a 50% chance of doubling and a 50% of halving. But the doubling only happens if you have x dollars, and the halving only happens if you have 2x. So you can see you either gain x or lose x, so your expected return is 0. When the amount you multiply your money by depends on the amount of money you currently have, you have to factor in your…

Oh, that actually really helped me getting a grasp on the problem. I feel like other comments and sections of the article were trying to get at that idea, but your phrasing of it made it finally click for me. Thanks.

Re: Two envelopes problem

#48

Post by ossopite »

chii wrote: post #35

https://www.youtube.com/watch?v=_NGPncypY68 this video explains the paradox properly, with a very reasonable explanation.

that's an excellent treatment. I'm curious: this video picks a particular example of a distribution of the amounts and shows that the expected profit from switching is not defined. Can one somehow prove that for all possible distributions, either the expected profit from switching is 0 or it is undefined?

Re: Two envelopes problem

#49

Post by Elte »

scrapheap wrote: post #9
jdkee wrote: post #2

Notably different from the Monty Haul problem where additional information is given to the player of the game prior to making the decision to switch. See https://en.wikipedia.org/wiki/Monty_Hall_problem

Personally I don't think you are given any new information in the Monty Hall problem that's worth anything. You always knew there was a 100% chance that one of the doors in that set has a goat behind it, which door that is is pretty much irrelevant. Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stic…

When I was initially exposed to Monty Hall, what confused me the most is that I didn't know that the host would always open a door with a goat, rather than just picking one of the remaining doors at random. In the latter case (i.e. the case where the host can pick a car), even provided that you are in the universe where he picked a goat, switching and staying both yield a 50% chance of success.

Re: Two envelopes problem

#50

Post by pooper »

CaptainNegative wrote: post #36
Nihmie wrote: post #18

Earlier quoted context omitted.

Is there actually some way to guess with better than 50% chance? I'm pretty sure this could be reduced to the secretary problem where there's a pool of 2 candidates, which yields an optimal hire with probability 50%. Maybe you play word games and say, "I guess the other number is not higher," or "I guess the other number is not lower." Since the number you observed was produced once, there is some non-zero probabilit…

The reduction is leaky in that it requires you to assume that there is nothing to learn from looking at the first number. Here's a hint that is not a full solution. If you knew that the numbers in both envelopes were independently sampled from the same Gaussian (but not necessarily which Gaussian), then there is a simple strategy that wins more than 50% of the time: pick an envelope to peek at uniformly at random and…

I didn't even think of the possibility of negative numbers. I can't ever imagine playing this x, 2x envelope game if there was any chance at all that the numbers would be negative. Like you open an envelope and it says I owe them either one million or two million dollars. Yeah, screw that game.
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