At Coding Horror Jeff Atwood revisits the Monty Hall problem; he also has some links to several other pieces on this very interesting puzzle that even mathematicians are known to have fumbled on.
One of the links is to the original article by Marilyn vos Savant, published in 1991 [Update: Check out Rahul's comment, below, about a version of this problem discussed by Martin Gardner in 1959 !] . If you click through [to von Savant's site], don't forget to browse through the adverse comments on her (correct) solution that came from so many academics!
A Mystery link (cartoon).
Interdisciplinary wars: the economics vs. sociology edition: Bryan Caplan, Fabio Rojas, Henry Farrell.
Michael Nielsen: How to read mathematics and physics.
Just when you thought Elsevier can't stoop any lower, you get this: Elsevier offered to pay reviewers for posting good reviews -- 5 stars -- on Amazon.com and Barnes&Noble.
An interesting research finding on open access publications: Is freely available literature better disseminated?
Empirically, we find that articles that received good evaluations on F1000 biology (a website where experts post evaluations of recently published papers in biology) were more likely to be in open access.
Wednesday, June 24, 2009
Links ...
2 comments:
Would you like to comment on this post (or, in response to one of the comments)? If so, please note:
1. This blog does not allow anonymous comments (any more), so please use an open-id account to comment.
2. Comments on posts older than 15 days go into a moderation queue, and may take some time to appear.
Thank you for joining the conversation. Have your say:
Just a note: the "original article" isn't from Marilyn vos Savant. Martin Gardner described an equivalent problem, involving prisoners, in 1959, reprinted in one of his collections. Gardner's version was: of three prisoners, A, B and C, condemned to death, one will be pardoned. Prisoner A asks the warder who, and the warder refuses to tell him, but agrees to tell him that B will be executed. What are A's chances of survival now?
ReplyDeleteI remember being confused by it for about a day (I was in high school) and Gardner's explanation that "the warder can always name a prisoner who will die, so A's chances are unaffected and remain 1/3, while C's become 2/3" was rather hard to grasp.
But in an addendum, he supplied a reader-contributed argument that anyone familiar with elementary probability theory should grasp: Imagine a large number, N, of such situations. In N/3 B is pardoned and the warder names C; in N/3, C is pardoned and the warder names B; in N/3 A is pardoned, and in half of these (N/6) the warder names B. So, in the N/2 cases where the warder names B, C is pardoned in N/3 cases and A is pardoned only in N/6 cases, i.e. 1/3 of the time, just as before.
That convinced me. Today I'd express it more concisely with conditional probabilities. I'm astonished that many mathematicians, including Erdos, not only failed to get the right answer but actually attacked vos Savant for her answer!
PS - an even stronger version from Gardner: someone places 52 cards face-down in front of you and asks you to pick the ace of spades. You pick a card. The probability that you are right is 1/52. The other person knows where the real ace is, and turns over 50 cards that are not the ace, leaving you with the card you picked and another face-down card. What is now the probability that the card you picked is the ace? The answer is it remains 1/52 and the other has a 51/52 probability of being the ace. (If 50 cards were turned over randomly, and none proved to be the ace, then of course it's 50-50 for the remaining cards.)
ReplyDelete