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Two envelopes problem

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Re: Two envelopes problem

#231

Post by amalcon »

pontus wrote: post #199
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…

There actually is a paradox: Select a power of 2 according to the probability P(2^k) = 0.2 * 0.8^k i.e. p(1) = 20%, p(2) = 16%, p(4) = 12.8%, etc. Then place this amount in one envelope and twice that I'm the other envelope. Then, randomly shuffle the envelopes and select one and open it . Say you see the number 16. It could have landed there in two ways: Either 16 was generated and the other envelope has 32. This ha…

This is similar to the St. Petersburg problem. The way I satisfy myself with that answer is to note that it's kind of a spherical cow problem. Nobody in the real world can afford to run this game, since it requires an infinite amount of money -- even a currency issuer would expect to just make their currency worthless by running it, and an offering of physical resources instead of currency would violate thermodynamics in all sorts of ways.

Spherical cow problems often end in counterintuitive results.

Re: Two envelopes problem

#232

Post by shannifin »

baobabKoodaa wrote: post #224
shannifin wrote: post #141

Earlier quoted context omitted.

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?

> Just like the envelope game, right?

No, because you're not actually wagering anything in the envelope game, so it makes no sense to calculate probabilities as though you are.

An analogous coin toss example would be something like: I'll flip a coin, you call heads or tails. If you call it right, I'll give you $10. If you're wrong, I'll give you $5. (So either way you make $5, so I don't know why anyone would even offer this game!) After I flip the coin, I'll conceal it in my hand and give you a chance to change your choice of heads or tails. Well, of course there's no reason to change your choice, it's completely arbitrary! You've got a 50-50 chance of winning $10 regardless. The paradox makes you think you're wagering your unknown potential winnings, but you're not actually wagering anything.

It has nothing to do with calculating whether or not a specific wager is worth it given some set of probabilities. That's a different problem for a different sort of game.

Re: Two envelopes problem

#233

Post by baobabKoodaa »

kmod wrote: post #220
baobabKoodaa wrote: post #216

Earlier quoted context omitted.

> If it is truly the case that the second envelope has equal probability of having half or double the amount of money as the first, then you can prove that the probability distribution of money in each envelope is not well-formed (has total mass either zero or infinity). This means the premise is already contradictory... What? I can literally take 2 envelopes right now - physical envelopes - and I can literally put $…

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.

Re: Two envelopes problem

#235

Post by JoshCole »

The game looks like this in tree form:

             -> Keep    -> 2A
          2A -> Switch  -> A
    0.5 ->
          A  -> Keep    -> A
             -> Switch  -> 2A
Switch and keep both have an expectation of 3A/2 because (1/2)2A + (1/2)A. In the equation in the article they use (1/2)2A + (1/2)*A/2. They count A/2 as half of what the actual outcome should be. The mistake was abusing the A to have different meanings.

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  -> 1/2A
    0.5 ->
          1/2A  -> Keep    -> 1/2A
                -> Switch  -> A
Than they treated the chance node as selecting from these two different trees at the same time with 1/2 probability. Which is nonsensical. You can't be in an entirely different game with 1/2 probability - it isn't the same game so for analysis purposes it has 0 probability of being reached.

Re: Two envelopes problem

#236

Post by baobabKoodaa »

shannifin wrote: post #232
baobabKoodaa wrote: post #224

Earlier quoted context omitted.

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?

> Just like the envelope game, right? No, because you're not actually wagering anything in the envelope game, so it makes no sense to calculate probabilities as though you are. An analogous coin toss example would be something like: I'll flip a coin, you call heads or tails. If you call it right, I'll give you $10. If you're wrong, I'll give you $5. (So either way you make $5, so I don't know why anyone would even of…

> 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. Yes yes, you are only gambling your "winnings" that you won earlier, it doesn't make a difference here.

> An analogous coin toss example would be something like: [...] I don't know why anyone would even offer this game!

Look, you made the claim that expected value doesn't matter for anything at all, the only thing that matters is the probability of winning. I offered you a game where you have a 50% probability of winning on each coin toss and I offered to pay you $1 for each coin toss to incentivize you to play it. Can you please explain why you are refusing to play this game? Yes, you don't like the analogy, I get that, but you also don't like free money? I find that hard to believe. A more plausible explanation is that you don't actually believe the claim you are making. That's why you aren't willing to wager any money on it.

Re: Two envelopes problem

#237

Post by mike_hock »

jonath_laurent wrote: post #149
mike_hock wrote: post #64

Earlier quoted context omitted.

No, joe_the_user is exactly correct. The expectation they compute in step 7 requires that A is a constant. The formula seems to just be the regular formula for expectation: Value that B can take times probability that it takes that value, summed over all possible values. But the "possible values" they put in are A/2 and 2A, which makes no sense. A is a random variable. Don't forget that random variables are really fu…

This comment explains the fundamental reason why the reasoning is incorrect. I offer the same perspective in a different comment ( https://news.ycombinator.com/item?id=31569991 ). In my opinion, the Wikipedia article is making a disservice to readers by mentioning unnecessarily complex mathematical arguments involving bayesian reasoning and infinite distributions. I believe all of these are distractions from the more…

They also fail the article's own premise

> in particular, the puzzle is not solved by the very simple task of finding another way to calculate the probabilities that does not lead to a contradiction

Re: Two envelopes problem

#238

Post by JoshCole »

JoshCole wrote: post #235

The game looks like this in tree form: -> Keep -> 2A 2A -> Switch -> A 0.5 -> A -> Keep -> A -> Switch -> 2A Switch and keep both have an expectation of 3A/2 because (1/2) 2A + (1/2)A. In the equation in the article they use (1/2) 2A + (1/2)*A/2. They count A/2 as half of what the actual outcome should be. The mistake was abusing the A to have different meanings. They created two trees one with 2A. -> Keep -> 2A 2A -…

So why is this happening?

Some people are pointing at the error I mentioned above, but I think the problem is actually much deeper than that. Identifying this error is basically identifying the superficial reason for why they went wrong. It doesn't dig deep enough to systematically eliminate the error.

I want to point out something. Notice their logic when they call out why it is a paradox. You can switch multiple times and it would never terminate.

Mathematical formalism like markov decision processes and the bellman equations actually have the ability to capture this type of reasoning. They can even deal with infinite loops as happens here by including multiplication by a horizon term which makes the limit of the recursion equal to zero.

So the root problem is that, actually, the entire problem wasn't even using the right theoretical grounding.

To get really explicit on how it was not using the right grounding I want to point out the key error again, but stated a bit differently.

The author neglects information sets. He creates two game trees, discriminating on the basis of knowledge of which subgame he is in. This is a huge red flag. You can do that in perfect information games, but you can't in imperfect information games! You aren't playing over states in imperfect information games. You're playing over information sets.

You end up getting more insights into how this is wrong when you think about the problem in this way. For example, there might not be an error in this particular problem, but the choice to not express the action space as a probability vector is going to bite you in more complicated imperfect information decision problems.

Re: Two envelopes problem

#239

Post by [deleted] »

shannifin wrote: post #232
baobabKoodaa wrote: post #224

Earlier quoted context omitted.

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?

> Just like the envelope game, right? No, because you're not actually wagering anything in the envelope game, so it makes no sense to calculate probabilities as though you are. An analogous coin toss example would be something like: I'll flip a coin, you call heads or tails. If you call it right, I'll give you $10. If you're wrong, I'll give you $5. (So either way you make $5, so I don't know why anyone would even of…

[deleted]

Re: Two envelopes problem

#240

Post by shannifin »

baobabKoodaa wrote: post #236
shannifin wrote: post #232

Earlier quoted context omitted.

> Just like the envelope game, right? No, because you're not actually wagering anything in the envelope game, so it makes no sense to calculate probabilities as though you are. An analogous coin toss example would be something like: I'll flip a coin, you call heads or tails. If you call it right, I'll give you $10. If you're wrong, I'll give you $5. (So either way you make $5, so I don't know why anyone would even of…

> 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 for anything at all

I never meant to claim it doesn't matter for "anything at all", just not for this envelope game. (Apologies if the meaning of "at all" was ambiguous in my original post.)

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