[one-more-minute]

This line of reasoning isn't possible if you have specific amounts (eg £50 and £100) for the envelopes, which suggests that the variable A is being misused somehow.

`B = (2A if A=50, A/2 if A=100)`. Simplifying this to `B = 2A or A/2` loses important information: namely that when B is smaller, you expect A to be larger. Or alternatively, treating A as fixed (say A=100) conflates two different situations: one where the amounts are `100, 200` and one where they are `100, 50`. So you end up thinking B must have (200+50)/2 in it, which is incoherent. If you compute the expected B correctly then `A` is different in each branch.

It is subtle though. If this appeared in a paper, arguing something not obviously wrong, it'd be hard to convince everyone that the reasoning is bad.

[pontus]

It depends on how the envelopes are prepared. If the two envelopes have x and 2x in them and one is randomly handed to you, there's a symmetry between them that tells you there is no benefit in switching.

However, suppose x is placed in an envelope and handed to you. Then a fair coin is flipped. If it comes up heads, 2x is placed in another envelope and if it comes up tail, x/2 is placed in that envelope. Then you should switch.

It's a subtle difference in how the envelopes are prepared, but it makes all the difference.

[zmgsabst]

Oh, weird.

I’m not sure I’d have naively realized that the fixed amount in your envelope makes so much difference.

For people confused like me, think about the outcomes:

When the two are prepared together, say $2 and $4, then when you exchange your options are +2 and -2 with 50:50 odds.

When the second envelope is prepared based on $2 in your envelope, then when you exchange your options are +2 and -1 with 50:50 odds.

(I had to diagram this all out; unintuitive to me!)

[pontus]

I did a pretty deep analysis of this problem a while back here in case you're interested: https://mindbowling.wordpress.com/2020/09/14/two-envelope-pa...

[Groxx]

In some ways it's similar to the Monty Hall problem: https://en.wikipedia.org/wiki/Monty_Hall_problem

The introduction of a second "stage" based on your action changes things in a non-obvious way. Your first action is less impactful than your second one, regardless of what you did.

[bottled_poe]

This is new information which changes the definition of the problem though.

[pontus]

Yeah, I agree. This setup though is the one to which the original "naive" argument applies which is why I think it's interesting to ponder.

[stephencanon]

This is the “simple resolution” presented in the linked article. What “line of reasoning” are you responding to?

[one-more-minute]

Right. I mean the original, paradoxical line of reasoning – "the switching argument".

I see that the article also says "commonly one writer proposes a solution to the problem as stated, after which another writer shows that altering the problem slightly revives the paradox." But it doesn't seem to elaborate on where the simple resolution falls down, if it does.

[lisper]

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.

[jonath_laurent]

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.ycombinator.com/item?id=31569991

[Chinjut]

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.

[hammock]

Right away I was taken aback by the wikipedia article’s quick dive into mathematical solutions to the problem, when it seemed to me that what would be missing is consideration/framing of the initial setup.

That initial setup being, the act of starting the game with someone is(?) confers value in and of itself.

So I’m glad the real solution is wonky.

[ghaff]

Welcome to Wikipedia. Where people who love using equation editors outnumber those predisposed to writing cogent explanations.