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Two envelopes problem

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Re: Two envelopes problem

#221

Post by ChrisLomont »

metacritic12 wrote: post #156
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.

Yep - any distribution that is nonzero over the set of choices works, so over the reals, anything that is everywhere nonzero works.

For example, exp^{-x^2} with normalization factor is a simple choice.

Re: Two envelopes problem

#222

Post by kmod »

pontus wrote: post #218
kmod wrote: post #214

Earlier quoted context omitted.

If you disallow negative values of k then you break the symmetry argument, and you do indeed get quite a bit of information by opening an envelope. I assume you meant to use a exponential base <0.5 so that the mean is finite

The point I'm making is that if you open the envelope you will always conclude that it's advantageous to switch regardless of what you see in that envelope. In the case where you see $1, you're guaranteed that switching is a good idea. If you see anything other than $1 you come to the same conclusion but you have to rely on a probabilistic argument using expected values. Now, if you will decide to switch regardless o…

Every envelope you open will have a higher expected value than the previous one. This is not paradoxical, it is a standard property of probability distributions with infinite means.

I believe the problem you are trying to elicit is that you can almost show that either envelope is a priori better than the other. The problem with this is that the means are infinite so the math isn't sound. So there's not an actual contradiction; the strategy is to open either envelope and then switch to the other one. It's unintuitive but only because we're not used to reasoning about distributions with infinite means.

Re: Two envelopes problem

#223

Post by baobabKoodaa »

shannifin wrote: post #79

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?

> 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".

I'd be happy to take you up on that offer. Let's play some dice or card games that are set up such that you'll always have 9/10 probability of winning. You'll play, regardless of the expected value, right?

Re: Two envelopes problem

#224

Post by baobabKoodaa »

shannifin wrote: post #141
wccrawford wrote: post #101

Earlier quoted context omitted.

Right, the difference here is that the calculation that determines the expected payout is the same as the calculation that determines how much it costs to switch. With the coin, the cost to play is set and is lower than the expected outcome.

I was thinking more along the lines of there being no need to actually calculate anything at all because the only relevant information is the 50-50 probability of choosing the higher envelope. And since you gain no new information after choosing, there's no need to calculate anything afterward either. The "paradox" is in making one think there's something to be calculated beyond the 50-50 chance.

Let's play a game where I wager $10 on each coin toss and you're the casino. Heads, I pay you $5. Tails, you pay me my $10. Just like the envelope game, right? "The only relevant information is the 50-50 probability", right? So if I pay you $1 per toss, you'd be happy to play this game and take my money, right?

Re: Two envelopes problem

#225

Post by pontus »

kmod wrote: post #222
pontus wrote: post #218

Earlier quoted context omitted.

The point I'm making is that if you open the envelope you will always conclude that it's advantageous to switch regardless of what you see in that envelope. In the case where you see $1, you're guaranteed that switching is a good idea. If you see anything other than $1 you come to the same conclusion but you have to rely on a probabilistic argument using expected values. Now, if you will decide to switch regardless o…

Every envelope you open will have a higher expected value than the previous one. This is not paradoxical, it is a standard property of probability distributions with infinite means. I believe the problem you are trying to elicit is that you can almost show that either envelope is a priori better than the other. The problem with this is that the means are infinite so the math isn't sound. So there's not an actual cont…

Right, I mentioned above that the resolution to this is that the expected value of both envelopes is infinite before you open anything and thus the difference is undefined. Once you open one, the difference becomes well defined.

I still think it's quite unintuitive that one needs to open an envelope in order to regularize things this way. The fact that you will always decide to switch, regardless of what you see in the first envelope, yet you still have to open it is what I find paradoxical.

Re: Two envelopes problem

#226

Post by orthoxerox »

entropicgravity wrote: post #201
WhyOhWhyQ wrote: post #200

The Monty Hall problem has a simple resolution. At the beginning you had a 1/3 chance of picking the good door, and 2/3 chance of picking a bad door, meaning you probably picked a bad door at the start, so you should switch.

You can say 'simple' but a whole pantheon of academics famously got it wrong. If you really want to understand the problem use 100 doors instead of three and then it becomes very clear whats going on.

Or a million scratch-off tickets.

Re: Two envelopes problem

#227

Post by myhf »

ncmncm wrote: post #176
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…

The goalposts here are different from a normal "paradox". The actual solution is explained right there on the page, the expectation for both envelopes (one has x, one has 2x) is x*3/2, so switching changes nothing. The "solution" demanded is supposed to be pointing out the specific logical error in the erroneous calculation that says switching should win. Citing the correct logic is not considered sufficient.

It seems like this "solution" is saying that the error is in step 1:

> Denote by A the amount in the player's selected envelope.

which is information that we have not actually gained.

Re: Two envelopes problem

#228

Post by tom-thistime »

davidguetta wrote: post #198
oh_my_goodness wrote: post #192

Earlier quoted context omitted.

This may be progress, but I don't see how it eliminates the "paradox". The claim in the "paradox" answer is that when you switch envelopes, "A" is not one of the possible outcomes. If "A" isn't a possible outcome then it shouldn't show up in the expectation value.

Because A has just not the same value in both terms of the equation. In the left it means "the lowest" and in the other "the highest". It's two versions of the reality in the same equation, which is wrong. Use 100 usd / 200 usd instead of A and 2A and you'll see. If you never switch and if you always switch in both case the equation is 100×0.5 + 200x0.5

I agree that's true once we know that the two envelopes have 100 and 200 dollars. For me the tricky part is, how do we know it's 100 and 200? Maybe we have two envelopes with 100 and 50 dollars instead.

Re: Two envelopes problem

#229

Post by emerongi »

ChrisLomont wrote: post #221
metacritic12 wrote: post #156

Earlier quoted context omitted.

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.

Yep - any distribution that is nonzero over the set of choices works, so over the reals, anything that is everywhere nonzero works. For example, exp^{-x^2} with normalization factor is a simple choice.

Is there a complete explanation for this somewhere?

Asking for a friend, I obviously fully comprehend what you're talking about here.

Re: Two envelopes problem

#230

Post by scapp »

Nihmie wrote: post #158
IshKebab wrote: post #70

Earlier quoted context omitted.

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?

First, a point of terminology: uncountably infinite refers to any infinite cardinality other than countably infinite, so if the number of probability distributions is at least the cardinality of the real numbers, then it's already uncountable.

As for what uncountable cardinality they form, it's the same as the real numbers [0]. Roughly, a probability distribution is determined by the countable collection of real numbers P(X [0] https://math.stackexchange.com/questions/3698864/why-does-th...

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