5 ms·

Perhaps there's another formulation you meant to write, but every time I've seen it, it's been equivalent to the thing you just asked, and the only potentially

by hansvm 2y ago

Perhaps there's another formulation you meant to write, but every time I've seen it, it's been equivalent to the thing you just asked, and the only potentially ambiguous part of the question is "to your advantage". If you stay then you have a 1/3 chance of getting a car, and if you switch you have a 2/3 chance. Yes, it's also correct that you don't "know" the result, and you might prefer goats to cars (even then you should probably sell the car and buy several goats), but there's a reasonable enough interpretation of "advantage" that you shouldn't dismiss the problem outright.

raincole 2y ago
I regret opening this can of worms called Monty Hall. But anyway I'll link the relevant section from Wikipedia: https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_behaviors https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_... > The version of the Monty Hall problem published in Parade in 1990 did not specifically state that the host would always open another door, or always offer a choice to switch, or even never open the door revealing the car. However, Savant made it clear in her second follow-up column that the intended host's behavior could only be what led to the 2/3 probability she gave as her original answer. (emphasis mine) My point was that when people ask this question, they often word it like the very 1990 version did, lefting out this critical statement (which Savant considered needed as well, therefore she clarified in the follow-up column), making the question ambiguous. (Although Savant also said "Very few (out of people who said is 2/3 is wrong) raised questions about ambiguity"... so perhaps people are actually just bad at probablity...?)
- 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 ago
      I 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 ago
    The 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 ago
      Yes, 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.
      - 2y ago

by hansvm 2y ago

raincole 2y ago

I regret opening this can of worms called Monty Hall. But anyway I'll link the relevant section from Wikipedia: https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_behaviors https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_... > The version of the Monty Hall problem published in Parade in 1990 did not specifically state that the host would always open another door, or always offer a choice to switch, or even never open the door revealing the car. However, Savant made it clear in her second follow-up column that the intended host's behavior could only be what led to the 2/3 probability she gave as her original answer. (emphasis mine) My point was that when people ask this question, they often word it like the very 1990 version did, lefting out this critical statement (which Savant considered needed as well, therefore she clarified in the follow-up column), making the question ambiguous. (Although Savant also said "Very few (out of people who said is 2/3 is wrong) raised questions about ambiguity"... so perhaps people are actually just bad at probablity...?)

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 ago
    I 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 ago
  The 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 ago
    Yes, 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.
    - 2y ago