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[icon] Probability puzzle redux - Patti
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Subject:Probability puzzle redux
Time:01:28 am
I've really enjoyed watching people noodle on the probability puzzle I posted earlier. One of the things I like about it is that even for mathematically literate people, the answer that they find intuitively is usually wrong. Mine was. I've seen a lot of really smart people get this one wrong, though they usually get to the right answer eventually.

If you want to noodle on it yourself, but are finding it too hard, here's a non-spoiler hint.

It's not really a 52-card problem-- it's a 5-card problem. Think about it with a deck that contains five cards-- four aces and the two of clubs-- and it starts to look easier to solve.
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ronsrants
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Time:2009-08-20 01:27 pm (UTC)
As long as we both agree that I'm right, I won't have to prove it with math and numbers and what not.

-R
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tom_bayes
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Time:2009-08-20 02:23 pm (UTC)
Dude, NO CREDIT if you can't show your work.
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ronsrants
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Time:2009-08-20 02:40 pm (UTC)
I can easily proof pi(e). 3.124cherry.


-R
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whipartist
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Time:2009-08-20 05:27 pm (UTC)
We are looking for cases where the first ace is immediately followed by the As or 2c, and attempting to calculate the relative probability of it being one of those two cards.
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whipartist
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Time:2009-08-20 07:00 pm (UTC)
How would you restate it to make it clear?
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evwhore
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Time:2009-08-20 07:11 pm (UTC)
I rephrased it in e-mail to some friends as:

Suppose we have a standard deck of 52 cards, with no shenanigans, and I tell you I am going to follow the following procedure:

1. I will thoroughly shuffle the deck in a fair/balanced/random manner.

2. I will then turn over cards one at a time until I turn over an ace of any suit.

3. I will then turn over the next card following that first ace; call this new card X.

But before I actually carry out this procedure, I ask you to predict: is card X more likely to be the ace of spades, or the 2 of clubs, or are they equally likely?
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[icon] Probability puzzle redux - Patti
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