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Two envelopes problem

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Re: Two envelopes problem

#191

Post by canjobear »

jonath_laurent wrote: post #183

I would argue that this Wikipedia article is misleading and that it confuses more than it clarifies when it comes to resolving the paradox. In particular, I am disputing the fact that "no proposed solution is widely accepted as definitive" (exact quote). Indeed, the switching argument (or at least the argument as it is presented in the Wikipedia article) makes a clear and precise mistake that I claim any trained math…

The claim is that no solution is accepted as definitive, which is an empirically true fact about the academic literature. To respond to this with "no, my favored solution is definitive" is uninteresting.

Re: Two envelopes problem

#192

Post by oh_my_goodness »

davidguetta wrote: post #137
dotancohen wrote: post #80

Earlier quoted context omitted.

I don't think that this is so obvious to the layman, when A has already been defined as "the amount of money in the envelope I am holding".

Of course it's not obvious, that's why it's a "problem". But if only look at the formulat at bullet point 7 there "2A" and "0.5A" as amounts but not "A". It just doesn't describe the current state of reality not trying to brag, just reformulating the simplest way for everyone to see

This may be progress, but I don't see how it eliminates the "paradox". The claim in the "paradox" answer is that when you switch envelopes, "A" is not one of the possible outcomes. If "A" isn't a possible outcome then it shouldn't show up in the expectation value.

Re: Two envelopes problem

#193

Post by davidguetta »

baobabKoodaa wrote: post #186
davidguetta wrote: post #136

Earlier quoted context omitted.

the "writing" doesn't say it but the written formula at bullet point 7 says it. That's the mistake, the formula is wrong and does not describe reality.

This is incorrect. Bullet point 7 in the written formula does not refer to a corresponding pair of envelopes, as you imply. Instead, it refers to "the other" envelope only: 1/2 chance that "the other" envelope contains 2A, and 1/2 chance that "the other" envelope contains A/2. Notice that both references reference the same object.

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.

Re: Two envelopes problem

#194

Post by [deleted] »

johnrob wrote: post #153

Sounds like a misuse of relative return (growth rate). If you gain 100% today, then lose 50% tomorrow, you are back where you started. Absolute numbers give you the correct answer: x -> 2x is a gain of x, 2x -> x is a loss of x, so expected value is zero (x is a hard number not a ratio).

[deleted]

Re: Two envelopes problem

#195

Post by GeorgeKangas »

Why is no one looking at the chance-giver, and considering his motive? One would reasonably assume he'd like to minimize the payout -- so he'd offer you the chance to switch only if you had chosen the richer envelope. The description explicitly allows this possibility:

"Having chosen... you are given the chance to switch"

Nowhere does it say, you'll be offered the switch no matter which envelope you choose.

Same idea applies, to the very similar Monty Hall problem.

Re: Two envelopes problem

#196

Post by mathogre »

Take 10 minutes to model it and you'll find there is no benefit in changing. As a couple others have pointed out, it isn't the Monte Hall problem.

Below I did two runs with a million cases. One envelope has 1 unit; the other has 2 units. I shuffle the envelopes, and the "player" chooses an envelope. "noChange" means this is the payoff if the player doesn't switch envelopes. "yaChange" means this is the payoff if the player switches envelopes.

$ python3 ./twoEnvelopes_00.py 1000000 noChange = 1500023 yaChange = 1499977

$ python3 ./twoEnvelopes_00.py 1000000 noChange = 1499984 yaChange = 1500016

When modeling Monte Hall, the payoff for changing is obvious.

Re: Two envelopes problem

#197

Post by jonath_laurent »

Chinjut wrote: post #185
jonath_laurent wrote: post #183

I would argue that this Wikipedia article is misleading and that it confuses more than it clarifies when it comes to resolving the paradox. In particular, I am disputing the fact that "no proposed solution is widely accepted as definitive" (exact quote). Indeed, the switching argument (or at least the argument as it is presented in the Wikipedia article) makes a clear and precise mistake that I claim any trained math…

This is not the issue. When expected value is well-defined, it is perfectly fair to compute E[X] = E[E[X | A]], where E[X | A] is a function of random variable A ("conditional expectation"). If E[X | A] > A for every particular value of A, then E[X] > E[A], whenever the expected values E[X], E[A], and E[X - A] are all well-defined. The issue in this case is that E[X - A] is not well-defined. The expected profit from…

This is a good comment and I believe your justification is equally valid.

The reason for the apparent disagreement is that in order to point out a logical flaw in an argument, one must make the argument formal and explicit enough first. There are several plausible ways that the Wikipedia argument can be translated into a machine-checkable formal proof (and this ambiguity is at the core of the paradox). When I make such an attempt in my head, I am not interpreting the expectation as a conditional one and the problematic step is about conflating a random variable with a fixed constant. In your attempt to formalize the Wikipedia argument, you try and use the iterated expectation theorem and the problematic step becomes different. Seeing from the comments, several people agree with my interpretation but clearly there is an ambiguity here that is part of the paradox.

Thanks for prompting me to add nuance to my comment.

Re: Two envelopes problem

#198

Post by davidguetta »

oh_my_goodness wrote: post #192
davidguetta wrote: post #137

Earlier quoted context omitted.

Of course it's not obvious, that's why it's a "problem". But if only look at the formulat at bullet point 7 there "2A" and "0.5A" as amounts but not "A". It just doesn't describe the current state of reality not trying to brag, just reformulating the simplest way for everyone to see

This may be progress, but I don't see how it eliminates the "paradox". The claim in the "paradox" answer is that when you switch envelopes, "A" is not one of the possible outcomes. If "A" isn't a possible outcome then it shouldn't show up in the expectation value.

Because A has just not the same value in both terms of the equation. In the left it means "the lowest" and in the other "the highest".

It's two versions of the reality in the same equation, which is wrong.

Use 100 usd / 200 usd instead of A and 2A and you'll see. If you never switch and if you always switch in both case the equation is

100×0.5 + 200x0.5

Re: Two envelopes problem

#199

Post by pontus »

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 happens with probability 0.2*0.8^4 ~ 8.192%. Alternatively 8 was generated and the envelope you have is already the largest one. This happens with probability 0.2*0.8^3 ~ 10.24%. Given that you see 16 already, the relative probability becomes 4/9 vs 5/9. In other words the chance that you have the largest envelope already is 5/9. The expected value of the other envelope is then 5/9*8 + 4/9*32 = 18.67 > 16 so you should switch.

In fact, try this for different initial envelopes and you'll see that you should always switch.

But, if you should always switch, why even look inside the envelope?

But, then you're back at the symmetry / no information argument.

Re: Two envelopes problem

#200

Post by WhyOhWhyQ »

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.
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