Two envelopes problem • Hacker News Board

Re: Two envelopes problem

#11

Post by RHSeeger » Tue, May 31, 2022, 5:02 AM UTC

NamTaf wrote: post #7

Schroedingersat wrote: post #5

Earlier quoted context omitted.

Step 3 (and thus also 1 and 2 on reflection) stands out to me immediately as describing two different situations for the entire game. The case where switching gives you 2A and the case where switching gives you A/2 describe 2 completely different universes, not two different actions in one.

Agreed. Step 6 is where it breaks down, because it's redefined A as being one of the two envelopes to being the halfway point between those two envelopes. Until step 6, it's describing two parallel views of the universe (depending on which envelope you first get - hence steps 4 and 5), and then mashes them together in a way that doesn't work. Take the practical example of $100 and $200: A is either $100 or $200 depen…

Even after looking at those numbers, it still felt wrong to me. It eventually occurred to me that this was because my intuition was telling me the total should add up to A. And it doesn't; it adds up to 9/8 A. It took me a while to realize that my intuition was wrong, and that there's no reason the total should be A.

The thing that helped me reconcile this was realizing that the A in the two equations are different values; then replacing them.

    (0.5⋅(0.5⋅A + 0.5⋅2.A)) + (0.5⋅(0.5⋅A + 0.5⋅0.5⋅A)) = 9/8 A

^ but the A in the left grouping (where it's 100) is different than the A in the right grouping (where it's 200).

Replacing the As with their actual values

    0.5⋅(0.5⋅100+0.5⋅2⋅100)+0.5⋅(0.5⋅200+0.5⋅0.5⋅200) = 150

And, since one envelope has 100 and the other has 200, an expected outcome across both envelopes is, as calculated, 150.

Re: Two envelopes problem

#12

Post by antognini » Tue, May 31, 2022, 5:02 AM UTC

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 paper inside. You then need to guess whether or not the number in the other envelope is larger or smaller than the one you are looking at. Can you guess with better than 50% chance?

Re: Two envelopes problem

#13

Post by RHSeeger » Tue, May 31, 2022, 5:05 AM UTC

scrapheap wrote: post #9

jdkee wrote: post #2

Notably different from the Monty Haul problem where additional information is given to the player of the game prior to making the decision to switch. See https://en.wikipedia.org/wiki/Monty_Hall_problem

Personally I don't think you are given any new information in the Monty Hall problem that's worth anything. You always knew there was a 100% chance that one of the doors in that set has a goat behind it, which door that is is pretty much irrelevant. Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stic…

> switch and take the best result from the other doors

That is, effectively, what the choice boils down to. But that being said, since the problem is stated as only getting to pick one "other" door, the "extra information" received is knowing which of the other 2 doors has the best result.

Re: Two envelopes problem

#14

Post by arnarbi » Tue, May 31, 2022, 5:22 AM UTC

scrapheap wrote: post #9

jdkee wrote: post #2

Notably different from the Monty Haul problem where additional information is given to the player of the game prior to making the decision to switch. See https://en.wikipedia.org/wiki/Monty_Hall_problem

Personally I don't think you are given any new information in the Monty Hall problem that's worth anything. You always knew there was a 100% chance that one of the doors in that set has a goat behind it, which door that is is pretty much irrelevant. Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stic…

Try thinking of it this way: The information is in which door the host does not open.

2/3rds of the time, your initial choice will be a goat. In those cases the host deliberately avoids opening the remaining door with a car - thereby telling you where it is.

If you chose the car door initially, then indeed you get no new information. But that only happens 1/3rd of the time.

I am curious if that changes your opinion.

Re: Two envelopes problem

#15

Post by dllthomas » Tue, May 31, 2022, 5:23 AM UTC

scrapheap wrote: post #9

jdkee wrote: post #2

Notably different from the Monty Haul problem where additional information is given to the player of the game prior to making the decision to switch. See https://en.wikipedia.org/wiki/Monty_Hall_problem

Personally I don't think you are given any new information in the Monty Hall problem that's worth anything. You always knew there was a 100% chance that one of the doors in that set has a goat behind it, which door that is is pretty much irrelevant. Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stic…

When Monty reveals a goat, your odds increase from 1/3 to 1/2 if Monty is picking randomly and just happened not to reveal a goat (and if he revealed a car we start from the top, or you automatically lose, or you automatically win, or...), but to 2/3 if Monty will reliably reveal a goat. I think it's hard to say that you have no extra information in that latter case.

It's true that we can add other options (pick both doors) of equal likelihood, but it's because of our increased information that the reduced option of picking just one door isn't actually worse off.

Re: Two envelopes problem

#16

Post by kazinator » Tue, May 31, 2022, 5:26 AM UTC

It is very simple.

Let's represent the poorer envelope by the value 0.5 and the more bountiful one by 1.0.

The expected value from the envelope draw is 0.75: 0.5 drawn with a 50% probability contributing 0.25 to the expected value, and 1.0 drawn with a 50% probability contributing another 0.5 to the expected value for a total of 0.75.

If you were to do a large number N of these draws, the value you will obtain will be from a distribution that is centered on 0.75 N. So far so good?

So the way to look at a draw is this: you're going to be getting the expected/average value of 0.75, together with either a 0.25 bonus if you pick the better envelope, or a else -0.25 penalty if you pick the poorer envelope.

So you see, those two situations are equal and opposite. By switching, you either turn a 0.25 bonus into a -0.25 penalty (- 0.5) or vice versa (+ 0.5) with equal probability.

If you need any more complex analysis than this, you need to be thwacked on the head.

The rhetoric about doubling versus halving the money is a distracting red herring. If you switch from the 1.0 envelope to the 0.5 envelope, you're down 0.5, and if you switch the other way, you're up 0.5. That's it.

Accounting is based on debits credits not halving and doubling. You would never update a ledger by crediting double some amount to one account, and debiting half the amount from another account. It's all purely additive.

You have to look at what you're potentially gaining or losing on its own not as a fraction of some guaranteed fixed portion that you're getting from either envelope.

Re: Two envelopes problem

#17

Post by dllthomas » Tue, May 31, 2022, 5:39 AM UTC

joe_the_user wrote: post #8

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.

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.

I think that subtly misses the point. The problem is that you're implicitly using a distribution that... isn't a distribution. And with this particular not-a-distribution, whether you should switch or not does not depend on the value of A. But with any actual distribution (... I think?) it does, at which point... no paradox. It's true that it's not clear (at least to me, but perhaps more generally) what distribution we should assume, but as long as we avoid treating something that isn't a distribution as if it were one we avoid the worst of it.

Re: Two envelopes problem

#18

Post by Nihmie » Tue, May 31, 2022, 5:40 AM UTC

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…

Is there actually some way to guess with better than 50% chance?

I'm pretty sure this could be reduced to the secretary problem where there's a pool of 2 candidates, which yields an optimal hire with probability 50%.

Maybe you play word games and say, "I guess the other number is not higher," or "I guess the other number is not lower." Since the number you observed was produced once, there is some non-zero probability that it was produced twice, in which case the other number is neither nigher nor lower.