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Two envelopes problem

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Re: Two envelopes problem

#71

Post by wardedVibe »

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…

If we take any finite interval [0,a], there's a unique uninformed prior of the uniform distribution, and we get a (a-b)/a probability of gaining by switching under that distribution. If we drag a to infinity, we get with probability 1 that switching is better. Numbers are big yo

Re: Two envelopes problem

#72

Post by djmips »

mountainwalker wrote: post #6

change your mind as many times as you like, there will only ever be one actual choice

It's a simple as that. If you know you pick A and then you will switch to B then it's the same as if you had just decided to pick B in the first place.

Re: Two envelopes problem

#73

Post by maest »

thekiptxt wrote: post #25
sjburt wrote: post #3

I love this problem because it is so simple, and the false line of reasoning is so compelling that it would hardly raise an eyebrow if you saw it in an academic paper and yet the conclusion is so obviously wrong. Decision problems are tricky and in non intuitive ways.

This reminds me of how chess tactics feel so obvious when I’m on a tactics trainer, yet I can’t identify them in games where they actually happen. Or code that “obviously” has a bug only after it caused an issue in production. Would be nice if someone warned me before I do a PR: “ok, fyi there is a subtle but devastating bug in the new_feature.cpp”

This is a class of "hard to find, easy to verify" solutions.

Re: Two envelopes problem

#74

Post by paganel »

And what is the solution? I would just take the money at point A, i.e. at the very beginning, you never know when the experiment supervisor decides that he has enough of potentially giving even more free money (in case I'm smarter compared to him/her). Plus, free money at moment t is better than potentially double free money at moment t+t1

Re: Two envelopes problem

#75

Post by IshKebab »

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…

This is a far better explanation than the confused hand waving in the Wikipedia article.

Re: Two envelopes problem

#76

Post by smitty1e »

thekiptxt wrote: post #25
sjburt wrote: post #3

I love this problem because it is so simple, and the false line of reasoning is so compelling that it would hardly raise an eyebrow if you saw it in an academic paper and yet the conclusion is so obviously wrong. Decision problems are tricky and in non intuitive ways.

This reminds me of how chess tactics feel so obvious when I’m on a tactics trainer, yet I can’t identify them in games where they actually happen. Or code that “obviously” has a bug only after it caused an issue in production. Would be nice if someone warned me before I do a PR: “ok, fyi there is a subtle but devastating bug in the new_feature.cpp”

> Or code that “obviously” has a bug only after it caused an issue in production.

Cognitively, the mind filters out quite a bit of information for performance reasons.

Are we biased in favor of reasoning that "sounds plausible"?

My "hot dice" say "yes".

Re: Two envelopes problem

#77

Post by billpg »

Back when I made videos for YouTube, I did a little video on this paradox. I used a pair of cash boxes instead of envelopes because I thought opening boxes looked better on screen than opening an envelope. I know I didn't explain the issue very well because a lot of commenters thought I was discussing the TV game show "Deal Or No Deal".

https://www.youtube.com/watch?v=0yjVv_iB1h0 (Interestingly, a lot of those DoND comments have gone. Did they delete them?)

Re: Two envelopes problem

#78

Post by robocat »

> Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?

Normally we would assume there is someone who owns the envelopes of money. Now their incentive could be to offer to switch only if the envelope you chose contains more money (so they have a chance of keeping more). More like Monty Hall.

Even without that, after switching, should you switch back? The equation given argues you should switch back, so obviously the equation must be wrong.

It is easier to think: if I switch I have a 50% chance of making $X and a 50% chance of losing $X. Switching doesn’t change expected $.

I actually think the later problem in the Wikipedia article is more clever:

  Two people, equally rich, meet to compare the contents of their wallets. Each is ignorant of the contents of the two wallets. The game is as follows: whoever has the least money receives the contents of the wallet of the other (in the case where the amounts are equal, nothing happens). One of the two men can reason: "I have the amount A in my wallet. That's the maximum that I could lose. If I win (probability 0.5), the amount that I'll have in my possession at the end of the game will be more than 2A. Therefore the game is favourable to me." The other man can reason in exactly the same way. In fact, by symmetry, the game is fair. Where is the mistake in the reasoning of each man? — "Martin Gardner: Aha! Gotcha"

Re: Two envelopes problem

#79

Post by shannifin »

I must admit I don't find the "compelling line of reasoning" all that compelling; to me it seems the "paradox" has more to do with the meaning of probability than with its calculation.

That is, the probability is 1/2. That's all that matters to the decision making. Calculating an "expected value" at all is completely useless, whether or not you do it "correctly".

Am I missing something?

Re: Two envelopes problem

#80

Post by dotancohen »

davidguetta wrote: post #52
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.

Yeah, that's obvious when they say "one envelope contains 2*A and the other A/2". Well no, one doesn't contain one fourth of the other.

I don't think that this is so obvious to the layman, when A has already been defined as "the amount of money in the envelope I am holding".
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