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Two envelopes problem

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Re: Two envelopes problem

#31

Post by tsimionescu »

Schroedingersat wrote: post #23
scrapheap wrote: post #9

Earlier quoted context omitted.

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…

It becomes a lot clearer if you increase the number of goats. I have a deck of a billion cards, one entitles you to the car, the others have a picture of a goat. You pick one at random but do not get to see it. Of the remaining 999,999,999 cards, I discard 999,999,998 goats. Do you want your initial card, or the remaining card in my deck? Or to frame it another way, do you want the car iff you picked it first or iff…

The GP is not contesting this, they just think that the whole discarding/revealing of the goats is an unnecessarily complicated explanation for why this works.

The way they put it, the strategy becomes obvious if you instead reformulate the problem like so: there are three doors with three different prizes. You get to pick one door, but are then given a choice: stick with your original choice, or choose to get the best prize from the other two doors.

The solution (switch!) is more obvious in the second formulation. I have no idea if it's easier to convince someone else of the reasoning in the first formulation, or to convince them that the second formulation is equivalent to the original.

Re: Two envelopes problem

#32

Post by tsimionescu »

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…

Your first paragraph makes it seem like you don't accept the common solution to the Monty Hall problem, which is why people have responded explaining it to you, even though the second paragraph makes it clear that you do, and just dislike the way it is explained.

Personally, I'm not sure that it would be easier to convince someone that the second formulation is equivalent to the original problem, but next time I find a dis-believer I will give it a shot!

Re: Two envelopes problem

#33

Post by na85 »

thekiptxt wrote: post #25
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.

This reminds me of how chess tactics feel so obvious when I’m on a tactics trainer, yet I can’t identify them in games where they actually happen. Or code that “obviously” has a bug only after it caused an issue in production. Would be nice if someone warned me before I do a PR: “ok, fyi there is a subtle but devastating bug in the new_feature.cpp”

Well if debugging is the process of removing bugs from code, then logically the act of writing the code must be called "bugging" and the probability of having a bug in any new piece of code asymptotically approaches 1 with increasing SLOC ;^)

Re: Two envelopes problem

#34

Post by tsimionescu »

arnarbi wrote: post #14
scrapheap wrote: post #9

Earlier quoted context omitted.

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.

Read their second paragraph. That is exactly how they think of it, and they do think it makes it obvious you should switch.

Re: Two envelopes problem

#36

Post by CaptainNegative »

Nihmie wrote: post #18
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 probabilit…

The reduction is leaky in that it requires you to assume that there is nothing to learn from looking at the first number.

Here's a hint that is not a full solution. If you knew that the numbers in both envelopes were independently sampled from the same Gaussian (but not necessarily which Gaussian), then there is a simple strategy that wins more than 50% of the time: pick an envelope to peek at uniformly at random and guess that it is the larger one if and only if it shows a positive number.

Why does this work? If both numbers are positive, your strategy is equivalent to "pick a random envelope" and you're left with a uniformly random choice of envelope. If both numbers are negative, you're also left with a uniformly random choice of envelope. But if one is positive and the other is negative, you win 100% of the time.

What are the odds of the third case? Well every Gaussian has at least some mass on positive numbers and some mass on negative numbers. So there is some nonzero chance, even if you don't know what it is. So while the strategy might not always do better than a coin toss, some positive percent of the time you win every time. And so you get a distinct advantage over random guessing.

The problem is now how would one transfer such an approach to the original problem without the Gaussian assumption. Somehow the two problems are less different than they initially seem.

Re: Two envelopes problem

#37

Post by cygaril »

CJefferson wrote: post #28
kazinator wrote: post #16

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 b…

I'm not sure it's that simple. let's imagine the first envelope contained between 1 andand 1000 dollars, the second double the first. if you open your envelope and it contains a whole number of dollars between 1 and 1000, then you should swap, id take "250 or 750" over 500.

Only if each value is equally likely. If you see $1,000 but figure the envelope-filler is a lot more likely to have been willing to put $1,500 in than $3,000 then you should stick.

Re: Two envelopes problem

#38

Post by kazinator »

thematrixturtle wrote: post #21
kazinator wrote: post #16

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 b…

You are missing the point. The problem of whether you should actually switch is obvious and trivial (you should not). The challenge here is explaining why the plausible 'proof' that you should switch is wrong.

I read that in the Wikipedia page. There is some obvious argument involving repeatedly switching sides and whatnot, and it has some subtle logical flaw. Why bother with that at all.

OK, but let's look at it:

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

2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.

3. The other envelope may contain either 2A or A/2.

OK, here is where this argument got seduced by the multiplicative thinking. The correct reasoning is: let's call the smaller base amount A and the larger one 2A. One envelope contains A, the other one 2A. That's it.

3. If A is the smaller amount, then the other envelope contains 2A.

4. If A is the larger amount, then the other envelope contains A/2.

Here, the argument is splitting into cases because of the IF, and in these two cases, A has a different meaning.

5. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.

6. So the expected value of the money in the other envelope is:

This is wonky. The overall two-envelope situation has an expected value, not the individual envelope.

The correct view is that if this is the richer envelope, it contains X more money. If it is the less endowed one, then it contains X less.

    1        1           5
    - 2A +   - (A/2) =   - A
    2        2           4
7. This is greater than A so, on average, the person reasons that they stand to gain by swapping.

That calculation is wrong though, by itself. That issue should be attacked directly.

Separate calculations have to be performed for the two cases: A is the smaller amount and A is the bigger amount. For each of these cases, the expected value from both envelopes has to be calculated.

The way the above formula is doing the expected value calculation is off because it's equivocating on A.

It ends up working with two unrelated quantities: 2A and A/2. 2A is four times bigger than A/2. A is simultaneously denoting the smaller amount and the larger one.

8. After the switch, denote that content by B and reason in exactly the same manner as above.

Yes, and if you do, you will end up with 5/4B. So that can't be right and the whole remaining argument about infinite switching being irrational compared to just opening an envelope is superfluous.

Re: Two envelopes problem

#39

Post by kazinator »

CJefferson wrote: post #28
kazinator wrote: post #16

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 b…

I'm not sure it's that simple. let's imagine the first envelope contained between 1 andand 1000 dollars, the second double the first. if you open your envelope and it contains a whole number of dollars between 1 and 1000, then you should swap, id take "250 or 750" over 500.

But the problem is explicitly about swapping before opening any envelope.

Re: Two envelopes problem

#40

Post by eru »

dllthomas wrote: post #17
joe_the_user wrote: post #8

Earlier quoted context omitted.

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 distributi…

Yes.

The big problem is that you can't have a uniform distribution on the natural numbers.

(And by extension, you can't have a uniform distribution on the rounded-to-integer version of a distribution on the real numbers.)

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