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July 11, 2003
MONTY HALL
They say the hardest thing to say is 'sorry', but actually it is 'I was wrong'. However, having read Keith Devlin's article, I must admit I was wrong about Monty Hall. Colour my face red! Just as well I didn't take that bet... I console myself with the thought that some of the people who got it right seem to have cheated by searching the web for the answer, as shown by the uncanny similarities between their comments and Devlin's article. Or maybe it's another of those billiontoone coincidences... DAVID BURBRIDGE
Posted by David B at
12:27 PM
It's all good Dave :), Devlin notes that all kinds of brainiacs and number jacks are fooled by this problem. I'm just glad that the dreadful puzzle didn't tear this blog asunder, like the proverbial Charles Murray dinner party. Posted by: Jason Malloy at July 11, 2003 01:15 PM"i was wrong" is one of the greatest assertions one can make in science ;) Posted by: razib at July 11, 2003 01:43 PMAnd the utterer is marked as a genuine seeker of truth. When I first encountered the problem the guy who recounted it to me mentioned that he had argued long and hard with a statistics professor who would not accept the correct answer... Posted by: bbartlog at July 11, 2003 07:23 PMThanks for the understanding words. I think one reason for the difficulty of accepting the right answer is a kind of intellectual snobbery that many of us (me anyway) are guilty of. We think 'the naive maninthestreet would accept the offer to switch, but we are smarter than that, we know there are lots of fallacies about probability, and this looks like one one them'. But in this case the rules of the puzzle have been devised so that it is not a fallacy after all. It's a kind of doublebluff, and it is very, very, very annoying to fall for it! Posted by: David B at July 12, 2003 03:02 AMgc  The outcome table described above: [O] /X/ X is incorrect because in line 1 there are actually 2 possible outcomes, not 1. The other possible outcome is: [O] X /X/ Adding this to the table of possible events gives: [O] X /X/ which on first glance suggests that the probability of winning on switching is 1/2! But all is not lost because the probability of the sequence of events described in lines 1 and 2 is the product of their individual probablities, since they are independent: 1/3 (the probility of chosing door 1) X 1/2 (the probility of the host choosing 1 of the remaining 2 doors) = 1/6. The sum probility of lines 1 and 2 is 1/3. The probability of lines 3 and 4 are calculated similarly: 1/3 (the probility of chosing door 1) X 1 (the probility that the host will choose the only option available) = 1/3. So the odds of success on switching are indeed 2/3. This arguement rests on the two events in lines 3 and 4 being "independent". I think they are but am not quite good enough at probability theory to justify this assumption. Posted by: Neil at July 15, 2003 04:52 PM 

