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I'm reading The Drunkard's Walk: How Randomness Rules Our Lives by Leonard Mlodinow (Linkcat: The Drunkard's walk : how randomness rules our lives)
Here are a couple of nice problems:
A woman is going to have two children. Given that one of them will be a girl, what are the odds that BOTH will be girls?
You are on Monty Hall. There are three doors on the floor. Behind one door is a new IPOD Touch, behind each of the other two doors are goats. You pick door #1. Monty, who knows what is behind all the doors, opens door #2 to reveal a goat. He asks if you want to stick with door #1 or change to door #3. Should you change or not?
Answers below if you scroll down:
The odd of both being girls given that one is a girl is NOT 50%, but rather 1 in 3.
Yes, you should change doors. The host's actions skew the odds in favor of changing.
The Let's Make a Deal Applet - Give it a try.
http://montyhallproblem.com/
Monty Hall Dilemma from Interactive Mathematics Miscellany and Puzzles
Sunday, December 14, 2008
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The phrasing of the child question can be confusing. Doug has it correct. Note that it says "one of them" instead of "the first one born."
ReplyDeleteTwo children give 4 possible choices (G-G, G-B, B-G, and B-B).
If "the first one" is a girl, you eliminate 2 of the 4 choices (B-B, B-G eliminated) and the odds are now 50%.
It's only when you say "one of them" that you keep 3 choices (G-G, G-B, B-G) and the odds are 1 in 3
Exactly. If you know that the first child is a girl, then the odds on the second are 50/50 (roughly; I know births are not exactly equal by sex).
ReplyDeleteWnd what are the odds that ONE of the two will be a girl? 75%.