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Two envelopes problem

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Re: Two envelopes problem

#241

Post by JoshuaDavid »

myhf wrote: post #227
ncmncm wrote: post #176

Earlier quoted context omitted.

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.

Variant:

I tell you "These two envelopes contain money. One of the envelopes contains twice as much money as the other one. Pick one.". You pick one. I tell you the one you picked contains $60. You've now determined the amount in your selected envelope -- should you switch?

Re: Two envelopes problem

#242

Post by kmod »

baobabKoodaa wrote: post #233
kmod wrote: post #220

Earlier quoted context omitted.

If I open an envelope with $10 then I know that there is $20 in the other envelope. As soon as you assign a well-formed probability distribution to the money in the envelopes you will find that opening the first envelope is informative.

Opening the envelope is something that you added to the problem just now. It wasn't present in the original problem, and it wasn't present in your earlier post. Here's a quote from Wikipedia: "Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?" Note the part that says "before inspecting it". That means you can't open it.

Ah this is a different version of the problem. There are quite a few listed on the wikipedia page.

I believe the version you are talking about is just a simple equivocation.

Re: Two envelopes problem

#243

Post by ChrisLomont »

emerongi wrote: post #229
ChrisLomont wrote: post #221

Earlier quoted context omitted.

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.

Yeah, the solution is somewhat tricky. Here is the idea (this is somewhat sloppy, but gives the idea - if you're mathematically advanced all this can be made precise):

Suppose you have a way to pick a random number yourself, that has nonzero chance to pick any number. Suppose the person putting things in envelopes picks X Here is why it works. You pick X or Y envelope with 50/50 odds. 3 cases:

1) Suppose you picked a T 2) Suppose you picked a T with Y But, if you happened to pick X The technical details is it's minorly tricky to pick a real number out of an infinite set at random, with any number being a possible outcome. But the probability distributions above do the trick - basically any function you can draw that is everywhere positive, with the tails squeezing to zero fast enough (but never reaching it) will work.

Re: Two envelopes problem

#244

Post by baobabKoodaa »

shannifin wrote: post #240
baobabKoodaa wrote: post #236

Earlier quoted context omitted.

> No, because you're not actually wagering anything in the envelope game, so it makes no sense to calculate probabilities as though you are. If we consider the decision point where you can either keep the first envelope or switch it, you are effectively wagering the value of the first envelope. You're just muddying the waters when you're trying to make some kind of point about receiving the first envelope for free. Y…

> Yes yes, you are only gambling your "winnings" that you won earlier, it doesn't make a difference here. It makes all the difference because you don't know what those "winnings" are, and what you're potentially wagering them for depends on what they are. So ultimately you're not really wagering anything. (Again, I think this is the crux of the "paradox".) > Look, you made the claim that expected value doesn't matter…

> I never meant to claim [expected value] doesn't matter for "anything at all", just not for this envelope game.

When you say "this envelope game", I assume you are also including minor variations of this game?

If you mean specifically this exact envelope game, where the expected value for all actions is zero, then it is strictly true that it doesn't matter how we make decisions - using expected value or not - though it's not a particularly interesting point to make.

I will proceed assuming you mean also including minor variations of this game.

You wrote this earlier:

> 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".

We can make a minor variation to the game such that the probability will still remain at 1/2, but the expected value for switching can be changed. We can make the minor variation such that it will be profitable to switch, or such that it will be unprofitable to switch. We can do this while the probability remains at 1/2. According to you the probability is all that matters, and it's useless to calculate the expected value for a game like this? This would lead to making unprofitable decisions in a game with a small variation as described above.

Re: Two envelopes problem

#245

Post by JoshCole »

baobabKoodaa wrote: post #207
davidguetta wrote: post #193

Earlier quoted context omitted.

That precisely what's wrong, because A is not the same quantity in both "A" part of the equations. The right formula is changing either the 0.5A by A or the 2A by A. Try with A being a real like 100 usd numbers to see it. Basically you can't have "two realities" in the same equation. U havé to pick one.

> The right formula is changing either the 0.5A by A or the 2A by A. No it's not, that's not how you make EV calculations. As I already said in my previous post, the parts in the EV calculation do NOT refer to the pair of envelopes. Both parts refer to the same object, not to different objects. You are trying to change the EV calculation such that the different parts refer to different envelopes. This does not make a…

The person you are responding to is right actually. I drew in an ASCII art a diagram of the various game trees so it will be easier for you to see this:

On Wikipedia they created two trees one with 2A.

             -> Keep    -> 2A
          2A -> Switch  -> A
    0.5 ->
          A  -> Keep    -> A
             -> Switch  -> 2A
And another with 1/2A

                -> Keep    -> A
          A     -> Switch  -> A/2
    0.5 ->
          A/2   -> Keep    -> A/2
                -> Switch  -> A
Notice now that there is no game in which A is both 2A and A/2.

If you don't think this is the error, you're not thinking in imperfect information terms. You're thinking in perfect information terms. This sort of splitting action is allowed in perfect information games! In imperfect information games, you are in multiple subgames at the same time. A has to be the same A in both subgames, because you can't discriminate between which game you are in. So you're not allowed to do this. It is wrong!

This gets at the core of what the question is trying to make people notice. You can't do this with imperfect information. You can only do it with perfect information. You can't do it over information sets and strategy spaces. You can only do it over states and action spaces.

But they do it. And then they combine them back into the same analysis when they say 1/2(Game tree 1 switch case) + 1/2 (Game tree 2 keep case). And you can't do that, because game tree one depends on game tree ones keep case and game tree 2 depends on its switch case! You can't tell which you are in, so you are in both - not one or the other. You can't combine the two different subgames from two different trees like you could in perfect information! You have to stick with the same game tree.

And that is what the person you are mistakenly correcting is saying. Pick one or the other, you can't have both.

PS: There are other errors in the problem, like that when you solve it with an MDP or the Bellman equations you actually /do/ end up with 100% Switch having an expected value of zero. So if you spotted some other error, that doesn't mean that the person you are correcting is wrong in their correction. There are multiple issues with the way the wikipedia article approaches the conclusion.

PPS: This exact confusion that people are talking about is actually something that happened in practice in the solving of imperfect information games! IIRC the Pluribus paper author corrected a reasoning error of this type when he introduced sub game reach solving? I might be misremembering here, since its been a while since I listened to his talk, but my fuzzy memory suggests that confusion about subgames interacting with each other was the root cause of a failure in some of the poker research for subgame solving in a blueprint abstraction.

Re: Two envelopes problem

#246

Post by WhyOhWhyQ »

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.

I'm also an academic, if that matters for some reason, and I think it doesn't matter how many people didn't understand something. They must have been confusing themselves by seeking answers in the wrong direction. What matters is that the resolution I've given truly is simple. You have a 1/3 chance of being in a situation where changing doors is unfavorable, you have a 2/3 chance of being in a situation where changing doors is favorable. If you picked the door with the car, which happens with 1/3 probability, at the next stage switching is a bad option. If you picked a door with the goat, which happens with 2/3 probability, at the next stage switching is a good option. This is objectively simple, regardless of what a large number of people might think.

I also don't think changing the problem to an analogous one is as helpful as this direct and simple solution. Using 100 doors could mean multiple things. Maybe it means at the next stage you have 99 doors left. Maybe it means at the next stage you have 2 doors left. Why should I think either is the correct analog? You can just see the simple answer I've stated and be done with it.

Re: Two envelopes problem

#247

Post by ncmncm »

JoshuaDavid wrote: post #241
myhf wrote: post #227

Earlier quoted context omitted.

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.

Variant: I tell you "These two envelopes contain money. One of the envelopes contains twice as much money as the other one. Pick one.". You pick one. I tell you the one you picked contains $60. You've now determined the amount in your selected envelope -- should you switch?

Now you know x is either 30 or 60. The expectation is either 45 or 90. Switching has equal likelihood of increasing or decreasing your take. The correct calculation will show the expected value of the other envelope as 60.

Re: Two envelopes problem

#248

Post by [deleted] »

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.

[deleted]

Re: Two envelopes problem

#249

Post by [deleted] »

sorokod wrote: post #154
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.

[deleted]

Re: Two envelopes problem

#250

Post by shannifin »

baobabKoodaa wrote: post #244
shannifin wrote: post #240

Earlier quoted context omitted.

> Yes yes, you are only gambling your "winnings" that you won earlier, it doesn't make a difference here. It makes all the difference because you don't know what those "winnings" are, and what you're potentially wagering them for depends on what they are. So ultimately you're not really wagering anything. (Again, I think this is the crux of the "paradox".) > Look, you made the claim that expected value doesn't matter…

> I never meant to claim [expected value] doesn't matter for "anything at all", just not for this envelope game. When you say "this envelope game", I assume you are also including minor variations of this game? If you mean specifically this exact envelope game, where the expected value for all actions is zero, then it is strictly true that it doesn't matter how we make decisions - using expected value or not - though…

> When you say "this envelope game", I assume you are also including minor variations of this game?

No; it seems obvious to me that the argument for switching presented in the original wikipedia article is meant to apply only to the evelope game as it's presented. Of course introducing variations could easily change the meaningfulness of the article's premise. That is, it wouldn't be considered a "problem" or a "paradox" if calculating an "expected value" were actually meaningful.

> then it is strictly true that it doesn't matter how we make decisions - using expected value or not - though it's not a particularly interesting point to make

True, but my point was more of a question: if it's obvious that calculating an "expected value" is irrelevant in this specific case (as the article says, "It may seem obvious that there is no point in switching envelopes as the situation is symmetric"), why is the argument presented in the article considered compelling? That is, either it's not actually compelling, or the "expected value" being meaningless in this case is not necessarily so obvious at first... but if so, why not? (Or, to put it another way, why is the argument in the article compelling enough to warrant such a long wikipedia page with such numerous proposed "resolutions"?)

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