#1
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Probability of AA,KK,QQ and JJ.....
being dealt to 4 players on the same hand playing 4 handed. This happened in a live tourney. I was holding the queens and
ended up laying them down preflop. I will hang up and listen! Thanks |
#2
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Re: Probability of AA,KK,QQ and JJ.....
(6/(52 choose 2))*(6/(50 choose 2))*(6/(48 choose 2))*(6/(46 choose 2))*4!=1.64016364 × 10-08.
About 1 in 61 million. gm |
#3
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Re: Probability of AA,KK,QQ and JJ.....
GM
This is a neet way of doing it but I am sorry to say I don't understand it. I would have done it with inclusion-exclusion which gets complicated. If you have time could you quickly explain your terms. Thanks, Cobra |
#4
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Re: Probability of AA,KK,QQ and JJ.....
[ QUOTE ]
This is a neet way of doing it but I am sorry to say I don't understand it. I would have done it with inclusion-exclusion which gets complicated. If you have time could you quickly explain your terms. [/ QUOTE ] If you ignore the "4!" in the calculation, that gives you the chance that, eg: Seat 1 has AA Seat 2 has KK Seat 3 has QQ Seat 4 has JJ But this is obviously the same chance as, eg: Seat 1 has KK Seat 2 has AA Seat 3 has QQ Seat 4 has JJ There are 4! possible matchups of seats to hands. Each different matchup represent a mutually exclusive event. The entire matchup as a unit is the event -- the above two examples are mutually exclusive even though Seat 3 and Seat 4 have the same cards in each example. Thus we just multiply by 4!. gm |
#5
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Re: Probability of AA,KK,QQ and JJ.....
[ QUOTE ]
This is a neet way of doing it but I am sorry to say I don't understand it. I would have done it with inclusion-exclusion which gets complicated. If you have time could you quickly explain your terms. [/ QUOTE ] Also, in case this is not clear: 6/(52 choose 2) comes from the fact that there are 6 possible AA hands out of a total of (52 choose 2) hands. Now there are only 50 cards left, which explains the chancing denominators. |
#6
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Re: Probability of AA,KK,QQ and JJ.....
Thanks for the quick reply. That does make sense now.
Cobra |
#7
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Re: Probability of AA,KK,QQ and JJ.....
Actually you don't need the inclusion exclusion principle because we're only calculating the odds of four players having one of four hands, so we don't need to account for the possibility of double counting. Throw in a couple more players and ask the same question and the problem becomes pretty difficult.
aloiz |
#8
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Re: Probability of AA,KK,QQ and JJ.....
Thanks gaming_mouse!
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