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The correct answer is described here: https://www.youtube.com/watch?v=_NGPncypY68 https://www.youtube.com/watch?v=_NGPncypY68 TL;DR (Spoiler alert): the expec

by lisper 4y ago

The correct answer is described here: https://www.youtube.com/watch?v=_NGPncypY68 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.

deleted 4y ago
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hammock 4y ago
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 4y ago
  Welcome to Wikipedia. Where people who love using equation editors outnumber those predisposed to writing cogent explanations.
jonath_laurent 4y ago
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 https://news.ycombinator.com/item?id=31567251 I also made this argument separately: https://news.ycombinator.com/item?id=31569991 https://news.ycombinator.com/item?id=31569991
- Chinjut 4y ago
  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 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.

by lisper 4y ago

deleted 4y ago

[deleted]

hammock 4y ago

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 4y ago
Welcome to Wikipedia. Where people who love using equation editors outnumber those predisposed to writing cogent explanations.

jonath_laurent 4y ago

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 https://news.ycombinator.com/item?id=31567251 I also made this argument separately: https://news.ycombinator.com/item?id=31569991 https://news.ycombinator.com/item?id=31569991

Chinjut 4y ago
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 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.