I recently came across a very interesting problem known as “The Monty Hall Problem.” This is a statistical puzzle named after the host of an old television show “Let’s Make a Deal” which featured a similar problem albeit a little more involved than the basic version that mathematicians use. Here is a simple description of the problem from Wikipedia:
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
If you have not come across this problem before, intuition would make you say “it just won’t matter” (if not, either you are really clever or your intuition is screwed up; I’d put my money on the later.) It does matter. But I’m not going to tell you which is a better strategy (to switch or not to switch). New York Times has a good simulation of the game, go play the game yourself a few times with different strategies and I assure you that you’ll be stunned. After you’ve played it a few times, click on “How it works” to get a decent understanding of why the best strategy is what it is. If you want a better explanation, both in words and mathematical (using Bayesian analysis), refer to the Wikipedia page linked above.
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Assume that boys and girls are equally likely to be born. Let us say that a family has two children. Given that one of them is a boy, what is the probability that the other one is a boy too? (Source: One of Scott Aaronson’s (http://scottaaronson.com/blog/) lecture notes.)
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[...] Monty Hall Problem Posted in Mathematics by Mandar on May 17th, 2008 Some time back, Sids had a post on the Monty Hall problem/game, which is as follows (Source: Wikipedia). Suppose you’re on a game [...]
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