andrewducker (
andrewducker) wrote2011-08-17 04:15 pm
Monty Hall
[Poll #1770413]
Explanation
I have known what the answer was for ages, but for some reason it only "clicked" in my head today. You can blame
sarahs_muse for triggering it.
Explanation
I have known what the answer was for ages, but for some reason it only "clicked" in my head today. You can blame
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no subject
One of the correct answers is here.
And the short version is Bayes' Theorem and the sum of an infinite series.
(rather than sum the infinite series, a more elegant solution: Consider the two die rolls as simultaneous. If the first is a 6 (6/36 of the time), Bob loses. If the first is not a 6 and the second is a 6 (5/36) of the time, Bob wins. If neither is a 6, the test repeats.
Therefore, you can view this as an infinite loop with 11 possible exits, 6 of which are Sue wins and 5 are Bob wins. Given that SOMEONE wins, we see that Bob wins 5/11 of all games.)
If you don't like that, then yank out your geometric progression rules and sum the infinite series the hard way. I'll wait, you'll get 5/11.
Now: You've already shown that, for any game, the odds of Bob winning in turn 2 are 5/6*5/6*5/6*1/6: not-6, not-6, not-6, 6. That's 125/1296.
So, the odds of him winning in ANY game in turn 2, divided by the percentage of games he wins, gives you (125/1296) / (5/11).
Which is 275/1296.
275/1296 of Bob's wins will come in turn 2, which means that, for any given game where Bob wins, the odds of his win happening in turn 2 are about 21%.
Mainly as at this stage I think you are probably trolling rather than just being stupid.
You're demonstrably, aggressively wrong about the dino babies, the pub coins, and Bob and Sue, and you're calling *me* stupid because of it. You have no leg to stand on with accusations of "trolling"
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