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 December 13th, 2011, 08:20 AM #1 Newbie   Joined: Dec 2011 Posts: 1 Thanks: 0 probability/permutations question Here is a question a friend posed to me...in 2006 the mets lost their playoff series by a count of 4 games to 3. he, being a mets fan, lamented that he had attended 3 games, and all were losses. he asked me what the odds of that happening were, knowing retrospectively that the series went 7 games. initially i thought it was as simple as 4/7 * 3/6 * 2/5 = 11.43% but then it dawned on me (perhaps incorrectly?) that this is the probability if he attended the FIRST THREE GAMES ONLY for example, if attended games 1, 2, and 7, wouldn't the probability be 4/7 * 3/6 * 1 ? what am i missing here and what would be the correct approach to this problem? the only assumption we can make is that the series went the full 7 games, and the mets lost 4 times.
December 13th, 2011, 07:42 PM   #2
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Re: probability/permutations question

Quote:
 Originally Posted by jasonbing Here is a question a friend posed to me...in 2006 the mets lost their playoff series by a count of 4 games to 3. he, being a mets fan, lamented that he had attended 3 games, and all were losses. he asked me what the odds of that happening were, knowing retrospectively that the series went 7 games. initially i thought it was as simple as 4/7 * 3/6 * 2/5 = 11.43% but then it dawned on me (perhaps incorrectly?) that this is the probability if he attended the FIRST THREE GAMES ONLY
Actually, I think this does work. But, it's not like none of us ever screw up probabilities , so never a bad idea to get a second or third or fourth opinion.

I think you're misinterpreting the "first 3 games" thing. I think it's more a matter of the 3 games he decides to attend, not which order they're in. I was trying to relate this to picking cards. Like what if you were picking 3 cards and wanted the probability they would happen to be black cards. You'd have 26/52 * 25/51 * 24/50. Same concept. (You can also do (number of ways to pick 3 black cards)/(number of ways to pick any 3 cards), which is the same answer.)

But what you said made some sense - it would be like picking the first 3 cards only. But, the probability in the case of the cards is the same regardless how they're chosen - it would be the same if you took them off the bottom or out of the middle. I suppose in a way it's like him shuffling the games and just randomly picking 3 to attend. However, I think that does have to be random for this to work. And this is also based on your initial assertion that we only know there's 7 games and they lost some random 4.

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