7 ms·
Just be glad that you didn't go for the 2 envelopes problem which was her next. Vos Savant was wrong about that one, and few people like the real answer. The p
by btilly 2y ago
Just be glad that you didn't go for the 2 envelopes problem which was her next. Vos Savant was wrong about that one, and few people like the real answer.
The problem is that you're presented with two envelopes with 2 real numbers inside. You randomly select one, then look, and try to guess if you got the larger number. It doesn't seem like you can do better than even, but you can!
Unfortunately everyone hates the answer. Which is that you make up a random number and pretend it is the other one. Your odds of being right are
50% + (probability of choosing between the numbers) / 2
Which can always be strictly bigger than 50%. (Though possibly by only a little amount.)
Special case. If those numbers and yours were all independently randomly chosen from the same distribution, you'll be right 2/3 of the time.
- raincole 2y ago"Two Envelopes Problem" seems to be usaully referred to a very different problem: https://en.wikipedia.org/wiki/Two_envelopes_problem https://en.wikipedia.org/wiki/Two_envelopes_problem
- btilly 2y agoI was dealing with a more general variant of the version that Vos Savant gave. Which is that you were able to look at the contents of the envelope that you got. She claimed you didn't gain useful information. Calling it "information" is problematic, but it certainly is useful, because the stated strategy then works! This is one reason that probability theorists have learned to be more careful in stating the double money version to rule out the solution that I gave. But my version has been in the literature since, I believe the 1950s. Under, as I first encountered it, the same exact name. (I encountered it from Dr Laurie Snell at Dartmouth College in the mid 1990s.)
- deleted 2y ago[deleted]
- hansvm 2y agoThe two reals are selected via some distribution, and the only way you can do better than chance is if you have some knowledge of that distribution. The question leaves that distribution completely hidden, and your answer smuggles it back in. That feels less like a counter-intuitive math/stats question and more like a badly worded gotcha.
- raincole 2y agoYes, this two real numbers question suffers from hidden conditions as the original version of Monty Hall problem does, but even more explicitly. We can't have an uniform distribution over all real numbers, so it's quite pointless to discuss if looking into the envelope gives any new information, cause we don't even know the distribution yet.
- btilly 2y agoYou don't need a uniform distribution to get guaranteed better than even odds. And many nonuniform distributions work just fine. There are no hidden conditions. It is just a shocking result that we don't expect.
- hansvm 2y agoYou need some _known_ distribution though, and it's shocking because the distribution is ommitted from the question, and the presence of the same distribution is snuck into the answer.
- ColinWright 2y agoI think you are wrong ... see my answer here: https://news.ycombinator.com/item?id=42371563 https://news.ycombinator.com/item?id=42371563
- hansvm 2y agoNot to be too dismissive, but the title text of this xkcd [0] seems relevant. No matter how nice the explanation is, the fact that the conclusion is wrong suggests that the reasoning has a flaw. I took a stab at what I think that flaw is when I responded to your rebuttal [1]. [0] https://xkcd.com/2217/ https://xkcd.com/2217/ [1] https://news.ycombinator.com/item?id=42372285 https://news.ycombinator.com/item?id=42372285
- 2y ago