Re: Probability of AA,KK,QQ and JJ.....
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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.
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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
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