My Math Forum Coupla pairs

 Probability and Statistics Basic Probability and Statistics Math Forum

 May 26th, 2017, 08:04 AM #1 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,398 Thanks: 829 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? Thanks from Joppy
May 26th, 2017, 05:34 PM   #2
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Quote:
 Originally Posted by Denis Skeeter has 4 decks of cards, each labelled 1 to 13.
Clearer: 4 decks of 13 cards each,

May 26th, 2017, 05:46 PM   #3
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Quote:
 Originally Posted by Denis What be the probability that both Archie and Joppy did NOT get a pair?
Are you saying neither of them has a pair?

 May 26th, 2017, 05:54 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,398 Thanks: 829 Archie did not get a pair. Joppy did not get a pair.
 May 26th, 2017, 06:29 PM #5 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,591 Thanks: 546 Math Focus: Yet to find out. Knowing Denis, the solution will be a coupla cute numbers.
May 26th, 2017, 06:39 PM   #6
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Quote:
 Originally Posted by Joppy Knowing Denis, the solution will be a coupla cute numbers.
Sorry Joppy...not this time

 May 29th, 2017, 01:28 PM #7 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 600 Thanks: 82 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.
 May 29th, 2017, 03:34 PM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,398 Thanks: 829 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...
 May 30th, 2017, 05:42 AM #9 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Hey , how come I don't get to play Thanks from Joppy
May 30th, 2017, 05:49 AM   #10
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Math Focus: Yet to find out.
Quote:
 Originally Posted by agentredlum Hey , how come I don't get to play
You can make sure skeeters shuffled correctly

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