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Two envelopes problem

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Re: Two envelopes problem

#51

Post by kazinator »

kazinator wrote: post #38
thematrixturtle wrote: post #21

Earlier quoted context omitted.

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/…

Because the variable A refers to an amount, has to have a consistent meaning across the two cases, and so it has to be:

1. We have A; the other envelope has 2A; or else

2. We have 2A; and the other envelope has A.

In case 1, if we switch we gain A.

In case 2, if we switch, we lose A.

We have no idea which case we are in.

We cannot use the variable A to denote our initially selected envelope in both cases, because then A simultaneously represents the smaller amount and the larger amount, and we are absurdly mixing the quantities A/2 and 2A in the same calculation. No quantities in the actual problem are related by a factor of four!

People might be seduced into the A equivocation perhaps because they are initially thinking of A as being a label for the envelope, which is arbitrary and innocent enough. We have some envelope A and it has some money in it. Oh, but the other envelope then has either A/2 or 2A. At that point, A is no longer an envelope name, but a name for an amount. Then it has to pin down a specific number: it has to denote either the smaller value, or the larger one, and that value is found in different envelopes in the different cases; A cannot both be "name of initially chosen envelope" and "amount in that envelope".

Re: Two envelopes problem

#52

Post by davidguetta »

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.

Yeah, that's obvious when they say "one envelope contains 2*A and the other A/2". Well no, one doesn't contain one fourth of the other.

Re: Two envelopes problem

#54

Post by junon »

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

A fascinating line that shocked me, tucked away at the end of a paragraph with no further additional context aside from its source:

> Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans.

Re: Two envelopes problem

#55

Post by junon »

thekiptxt wrote: post #24
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…

I didn’t question the math, but what made me intuitively believe the Monty Hall problem was this: Me: I am thinking of a number between 1 and a million. If you guess correctly, you get a prize. you make your guess Me: Ok, I will reveal that the number is either $YOUR_GUESS or 452,871. You may keep your original guess or switch to 452,871. Of course, you would switch, because the chances you guessed correctly original…

Wow. I don't think it's ever been explained to me this well before.

Re: Two envelopes problem

#56

Post by mike_hock »

m463 wrote: post #46

This reminds me of the Monty Hall Problem: https://en.wikipedia.org/wiki/Monty_Hall_problem (except that problem has a more definite solution)

In fact, the fake/paradoxical solution probably plays on the reader's familiarity with the Monty Hall problem. ("Ah yeah I remember this. Switching is better!")

Re: Two envelopes problem

#57

Post by xcambar »

jansan wrote: post #20

A related problem, but actually interesting, because there is a beneficial strategy, is the Monty Hall Problem. https://en.wikipedia.org/wiki/Monty_Hall_problem

Also, it is a good opportunity to remind that the Monty Hall Problem was mentioned in a Brooklyn 99 episode, in case someone missed on Brooklyn 99.

https://m.imdb.com/title/tt6026160/movieconnections

Re: Two envelopes problem

#58

Post by xmprt »

davidguetta wrote: post #53

The bug is obvious when they say "one envelope contains 2*A and the other A/2". Well no, one doesn't contain one fourth of the other.

That's not what it's saying. It's saying one envelope contains A and the other contains either 2*A or A/2.

Re: Two envelopes problem

#59

Post by mike_hock »

davidguetta wrote: post #53

The bug is obvious when they say "one envelope contains 2*A and the other A/2". Well no, one doesn't contain one fourth of the other.

Basically the sleight of hand trick is using a random variable A as though it was a constant.

Yes, B = A/2 or B = 2A with probability .5 each, but not for the same value of A.

Re: Two envelopes problem

#60

Post by xmprt »

Schroedingersat wrote: post #5
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.

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.

The breakdown for me was in step 4-5.

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

These are conditional probabilities so you can't simply add them up and compute an expected value of switching like they show in the example.

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