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Two envelopes problem

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Re: Two envelopes problem

#181

Post by jstanley »

ChrisLomont wrote: post #151

This reminds me of a fascinating problem (not sure if there's a name for it): You play the following game (say for many rounds). Someone puts different values in two envelopes (they can pick the values arbitrarily, even maliciously, each round). You want to pick the higher valued envelope. You pick an envelope, they show you the value of the other. You can hold your value or you can switch. It also works if the game…

FWIW, I think all 3 versions of your game are completely equivalent. In all 3 versions you first get to decide which envelope you look inside, and then you get to decide which envelope you open.

Re: Two envelopes problem

#182

Post by twinkletwinkle_ »

not2b wrote: post #177
camgunz wrote: post #164

Earlier quoted context omitted.

Yeah I was looking for someone saying this to double down. I think the "paradox" comes down to information loss. The tricky bit is this framing: "when you're holding an envelope, the other envelope contains >= $MONEY, so there's no reason not to switch". But this omits (loses) the information of " both envelopes contain >= $MONEY". That is, as soon as you switch to the other envelope, the situation is still true, the…

The right resolution is: suppose the two amounts are $50 and $100. If the envelope you have contains $50 and you switch, your gain is $50. If it contains $100 and you switch, your gain is -$50. So your expected gain from switching is (0.5)(50) + (0.5)(-50) which is zero. The way that they fool you is by saying that you either double your money or you halve it, leading you to think that your expected ratio is 2.5/2. B…

I think you nailed it. If the question were posed as "Two envelopes with $A and $(A+1)" then there would be no talk of ratios and no "paradox" would ever emerge.

Re: Two envelopes problem

#183

Post by jonath_laurent »

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 mathematician will immediately agree on once pointed to it.

This mistake is explained in the following comments: https://news.ycombinator.com/item?id=31566226, https://news.ycombinator.com/item?id=31567251 and https://news.ycombinator.com/item?id=31569991.

Once you have identified it, this mistake is all but subtle. It is not about infinite series or Bayesian reasoning. It does not require a 30 minutes rebuttal talk. It is simply about misusing the very definition of an expected value in a way that could be qualified as a typing error (see linked comments). This error can only go undetected because of the ambiguity of the English language. Most of the fancy mathematical discussions I've been reading are distractions. This paradox is about language and not about any deep mathematical fact of probability theory.

Re: Two envelopes problem

#184

Post by MrBlueIncognito »

whatshisface wrote: post #172
platz wrote: post #159

Earlier quoted context omitted.

why isn't this listed in wikipedia as an answer.

Because if the larger envelope is changed to contain four times as many dollars, then /2 vs. *4 points back towards swapping indefinitely.

If one envelope contains 4 times the money of the other envelope, then by swapping you either gain 3x or lose 3x, where x is the amount of money in the smaller envelope. Even in this case, the average return is +3x -3x = 0.

Re: Two envelopes problem

#185

Post by Chinjut »

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 switching is given by an infinite series whose positive and negative components are each of infinite total magnitude, thus yielding different sums when bracketed differently ("conditional convergence"; just a coincidence that the word "conditional" comes up here again).

Re: Two envelopes problem

#186

Post by baobabKoodaa »

davidguetta wrote: post #136
wccrawford wrote: post #100

Earlier quoted context omitted.

It doesn't actually say that. It says the other envelope will either contain 2A or A/2. The envelope you're holding is A. The other envelop is either twice that, or half that.

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.

Re: Two envelopes problem

#187

Post by t_mann »

zeroonetwothree wrote: post #166
t_mann wrote: post #127

Earlier quoted context omitted.

You'd never win, because you can't have more than the other guy. A draw is not winning

You win if you have less

My bad - then your probability of losing would be zero with zero cash, yes. But we still don't have a proof for the constant probability of 0.5 for winning.

Re: Two envelopes problem

#188

Post by dllthomas »

eru wrote: post #40
dllthomas wrote: post #17

Earlier quoted context omitted.

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.)

We're not dealing with the natural numbers, right? Otherwise, you obviously want to open the envelope and always switch when you see an odd number.

Of course, the lack of a uniform distribution on the reals or rationals presents the same problem.

Re: Two envelopes problem

#189

Post by [deleted] »

sorokod wrote: post #154
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…

This is not a resolution, a resolution is where you point out the faulty step in the reasoning that leads to the paradox.

[deleted]

Re: Two envelopes problem

#190

Post by Chinjut »

jonath_laurent wrote: post #160
lisper wrote: post #89

Earlier quoted context omitted.

The correct answer is described here: https://www.youtube.com/watch?v=_NGPncypY68 TL;DR (Spoiler alert): the expected value of the amount of money you end up with is an infinite series whose sum changes depending on the order in which you add up the terms, and so you can choose an order that makes this value come out to be positive, negative, or zero.

I do not like this explanation. In my opinion, the fancy mathematical argument involving infinite series is an unnecessary distraction from a much more fundamental and mundane mistake (i.e. incorrectly using a variable out of its scope). I believe mike_hock provides the clearest and simplest explanation in this thread: https://news.ycombinator.com/item?id=31567251 I also made this argument separately: https://news.yc…

There's nothing wrong with conditional expectation. E[X] = E[E[X | A]], where E[X | A] is a function of the random variable A. See https://en.wikipedia.org/wiki/Conditional_expectation.

When E[X], E[A], and E[X - A] are all well-defined, it is indeed the case that if E[X | A = a] > a for every particular value a, then E[X] > E[A].

What goes awry in this case is that E[X - A] is not well-defined (where X is the value in the unselected envelope and A is the value in the selected envelope). It is given by a conditionally convergent series, as noted.

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