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May 26th, 2017, 09:04 AM   #1
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Coupla pairs

Skeeter has 4 decks of cards, each labelled 1 to 13.
He puts 'em all together and shuffles 'em generously.

Then he deals 2 cards to Archie.
Following that, he deals 2 cards to Joppy.

What be the probability that both Archie
and Joppy did NOT get a pair?
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May 26th, 2017, 06:34 PM   #2
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Quote:
Originally Posted by Denis View Post
Skeeter has 4 decks of cards, each labelled 1 to 13.
Clearer: 4 decks of 13 cards each,
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May 26th, 2017, 06:46 PM   #3
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Quote:
Originally Posted by Denis View Post

What be the probability that both Archie
and Joppy did NOT get a pair?
Are you saying neither of them has a pair?
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May 26th, 2017, 06:54 PM   #4
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Archie did not get a pair.
Joppy did not get a pair.
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May 26th, 2017, 07:29 PM   #5
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Math Focus: Yet to find out.
Knowing Denis, the solution will be a coupla cute numbers.
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May 26th, 2017, 07:39 PM   #6
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Quote:
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Knowing Denis, the solution will be a coupla cute numbers.
Sorry Joppy...not this time
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May 29th, 2017, 02:28 PM   #7
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Did I do this right?

There are 6 possible pairs for each number, which is 78 total. There are 52 C 2 = 1,326 ways to be given 2 cards. The probability Archie got a pair is 78/1,326 = 1/17, and the probability he didn't get a pair is 16/17. At that point, 11 numbers that Archie didn't get have 6 possible pairs, and the 2 numbers Archie got have 3 possible pairs, for a total of 72. There are 50 C 2 = 1,225 ways to be given 2 of the remaining cards. Joppy has a 72/1,225 chance of getting a pair and a 1,153/1,225 chance of not getting a pair. The probability of neither person getting a pair is (16/17)(1,153/1,225) = 18,448/20,825 = about .8859.
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May 29th, 2017, 04:34 PM   #8
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Thanks Evan.

Given answer, which I didn't agree with:
48/51 * 46/49 = ~.88355

I ran 2 simulations of 1 million each;
got 885,781 and 885, 587

You've confirmed my results...
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May 30th, 2017, 06:42 AM   #9
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Hey , how come I don't get to play
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May 30th, 2017, 06:49 AM   #10
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Hey , how come I don't get to play
You can make sure skeeters shuffled correctly
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