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Two envelopes problem

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Re: Two envelopes problem

#151

Post by ChrisLomont »

This reminds me of a fascinating problem (not sure if there's a name for it):

You play the following game (say for many rounds). Someone puts different values in two envelopes (they can pick the values arbitrarily, even maliciously, each round). You want to pick the higher valued envelope. You pick an envelope, they show you the value of the other. You can hold your value or you can switch.

It also works if the game is they show you your own value and you can choose to hold or switch. Actually, now that I think about it, you can even make the game such that you pick an envelope and tell them which envelope to show you, then you can hold or switch. More interesting :)

The questions is: is there a strategy where you get the higher value strictly more than 50% of the time?

The surprising answer is, yes, there is such a strategy. It's really neat, and I've tried to find other places to apply the reasoning, but have not found another interesting place.

And yes, this problem and strategy are mathematically correct, not cheating, not "out of the box" stuff.

Good luck :)

Re: Two envelopes problem

#152

Post by devman0 »

cosmojg wrote: post #147

Comparing this to the Monty Hall problem: Probability of choosing the correct door at time of choice: Two Envelopes: 1/2 -> 1/2 Monty Hall: 1/3 -> 1/2 ...because Monty opens one of the doors, providing information.

Monty opening the door condenses the probabilities of the two unpicked doors in to the one unpicked door. This occurs because Monty only opens a door with a goat and never the car. In the original three door version you always had a 2/3 chance of winning if you switched because you'd only lose if you correctly picked the car on the original 1/3 pick.

Re: Two envelopes problem

#153

Post by johnrob »

Sounds like a misuse of relative return (growth rate). If you gain 100% today, then lose 50% tomorrow, you are back where you started. Absolute numbers give you the correct answer: x -> 2x is a gain of x, 2x -> x is a loss of x, so expected value is zero (x is a hard number not a ratio).

Re: Two envelopes problem

#154

Post by sorokod »

kloch wrote: post #142

A simple non-mathematical resolution is that no information has been gained by selecting the first envelope. Since no information about the contents was available before selecting the first envelope switching envelopes is a neutral act. While this sounds like just a lazy intuitive explanation, it calls out the key component - a lack of information about the outcome - that can help avoid the paradox in more complex si…

This is not a resolution, a resolution is where you point out the faulty step in the reasoning that leads to the paradox.

Re: Two envelopes problem

#155

Post by ghaff »

hammock wrote: post #128
lisper wrote: post #89

Earlier quoted context omitted.

The correct answer is described here: https://www.youtube.com/watch?v=_NGPncypY68 TL;DR (Spoiler alert): the expected value of the amount of money you end up with is an infinite series whose sum changes depending on the order in which you add up the terms, and so you can choose an order that makes this value come out to be positive, negative, or zero.

Right away I was taken aback by the wikipedia article’s quick dive into mathematical solutions to the problem, when it seemed to me that what would be missing is consideration/framing of the initial setup. That initial setup being, the act of starting the game with someone is(?) confers value in and of itself. So I’m glad the real solution is wonky.

Welcome to Wikipedia. Where people who love using equation editors outnumber those predisposed to writing cogent explanations.

Re: Two envelopes problem

#156

Post by metacritic12 »

ChrisLomont wrote: post #151

This reminds me of a fascinating problem (not sure if there's a name for it): You play the following game (say for many rounds). Someone puts different values in two envelopes (they can pick the values arbitrarily, even maliciously, each round). You want to pick the higher valued envelope. You pick an envelope, they show you the value of the other. You can hold your value or you can switch. It also works if the game…

Generate a random number T from the arctan distribution. If the number in the other envelope is greater than T, switch.

For envelopes with two different numbers n, m, the above is guaranteed to give you the higher one strictly more than 50% of the time.

Re: Two envelopes problem

#157

Post by metacritic12 »

johnrob wrote: post #153

Sounds like a misuse of relative return (growth rate). If you gain 100% today, then lose 50% tomorrow, you are back where you started. Absolute numbers give you the correct answer: x -> 2x is a gain of x, 2x -> x is a loss of x, so expected value is zero (x is a hard number not a ratio).

Exactly! The logic of "the average of +100% and -50% is positive" only applies if the denominator is the same. In this case it isn't.

The relative percentage "rule of thumb" doesn't even feel like a primitive in math. It seems wrong to reason from that.

Re: Two envelopes problem

#158

Post by Nihmie »

IshKebab wrote: post #70
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…

I think that's an ill-posed question without knowing (or making assumptions I guess) how the distribution was chosen isn't it?

I'm inclined to agree with you here. The solutions suggested here seem to depend on heuristics that depend on the type of distribution.

A tangential question: real numbers are uncountably infinite. Are probability distributions over the real numbers likewise uncountably infinite, or do they form a higher infinity?

Re: Two envelopes problem

#159

Post by platz »

metacritic12 wrote: post #157
johnrob wrote: post #153

Sounds like a misuse of relative return (growth rate). If you gain 100% today, then lose 50% tomorrow, you are back where you started. Absolute numbers give you the correct answer: x -> 2x is a gain of x, 2x -> x is a loss of x, so expected value is zero (x is a hard number not a ratio).

Exactly! The logic of "the average of +100% and -50% is positive" only applies if the denominator is the same. In this case it isn't. The relative percentage "rule of thumb" doesn't even feel like a primitive in math. It seems wrong to reason from that.

why isn't this listed in wikipedia as an answer.

Re: Two envelopes problem

#160

Post by jonath_laurent »

lisper wrote: post #89
one-more-minute wrote: post #87

This line of reasoning isn't possible if you have specific amounts (eg £50 and £100) for the envelopes, which suggests that the variable A is being misused somehow. `B = (2A if A=50, A/2 if A=100)`. Simplifying this to `B = 2A or A/2` loses important information: namely that when B is smaller, you expect A to be larger. Or alternatively, treating A as fixed (say A=100) conflates two different situations: one where th…

The correct answer is described here: https://www.youtube.com/watch?v=_NGPncypY68 TL;DR (Spoiler alert): the expected value of the amount of money you end up with is an infinite series whose sum changes depending on the order in which you add up the terms, and so you can choose an order that makes this value come out to be positive, negative, or zero.

I do not like this explanation. In my opinion, the fancy mathematical argument involving infinite series is an unnecessary distraction from a much more fundamental and mundane mistake (i.e. incorrectly using a variable out of its scope).

I believe mike_hock provides the clearest and simplest explanation in this thread: https://news.ycombinator.com/item?id=31567251

I also made this argument separately: https://news.ycombinator.com/item?id=31569991

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