tag:blogger.com,1999:blog-9818962.post1431771339887973921..comments2021-07-15T18:36:28.647+05:30Comments on nanopolitan: Links ...Abihttp://www.blogger.com/profile/06790560045313883673noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-9818962.post-16820220143643225152009-07-09T07:15:14.678+05:302009-07-09T07:15:14.678+05:30At any rate, I liked some of the vadlo science car...At any rate, I liked some of the vadlo <a href="http://vadlo.com/cartoons.php?id=29" rel="nofollow">science cartoons</a>!Ligennoreply@blogger.comtag:blogger.com,1999:blog-9818962.post-6642141751298625782009-06-24T11:11:42.392+05:302009-06-24T11:11:42.392+05:30PS - an even stronger version from Gardner: someon...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.)Rahul Siddharthanhttps://www.blogger.com/profile/04809667965184094636noreply@blogger.comtag:blogger.com,1999:blog-9818962.post-9523670981394598512009-06-24T11:05:17.152+05:302009-06-24T11:05:17.152+05:30Just a note: the "original article" isn&...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?<br /><br />I 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. <br /><br />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.<br /><br />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!Rahul Siddharthanhttps://www.blogger.com/profile/04809667965184094636noreply@blogger.com