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View Full Version : Probability of AA,KK,QQ and JJ.....

01-11-2005, 03:45 AM
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

gaming_mouse
01-11-2005, 03:58 AM
(6/(52 choose 2))*(6/(50 choose 2))*(6/(48 choose 2))*(6/(46 choose 2))*4!=1.64016364 × 10-08.

gm

Cobra
01-11-2005, 09:42 AM
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

gaming_mouse
01-11-2005, 09:55 AM
[ 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

gaming_mouse
01-11-2005, 09:59 AM
[ 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.

Cobra
01-11-2005, 10:10 AM
Thanks for the quick reply. That does make sense now.

Cobra

aloiz
01-11-2005, 02:40 PM
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