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-   -   help with odds? (http://archives2.twoplustwo.com/showthread.php?t=45880)

 ZEN MASTER 09-18-2003 04:26 AM

help with odds?

Could anyone tell me the odds of any player having a pair of AA in a 10 handed holdem game. I know the odds of being dealt AA are 221 to 1, but I am interessted to know it in a 10 handed game. Thank you

 BruceZ 09-18-2003 05:45 AM

Re: help with odds?

The odds for a particlular player having AA in 10 handed game is 1/221. The odds that someone has AA is:

10/221 - C(10,2)*6 / C(52,4) = 4.43% or 21.6-1

 ZEN MASTER 09-18-2003 05:54 AM

Thank you bruce. Would you please tell me what the c stands for. Also could you explain your calculations so I would be able to calculate the formula for any number of players. Thank you for all your help

 BruceZ 09-18-2003 06:17 AM

I made a mistake, so I'll fix it here.

C(n,k) is the number of ways to choose k combinations of cards out of n cards, irrespective of order. You can use Excel to compute it, the function is called COMBIN.

C(n,k) = n! / [ (n-k)!*k! ] = n*(n-1)*(n-2)*...*(n-k+1)/ [ k*(k-1)*(k-2)*...*1 ]

So C(52,4) = 52*51*50*49/(4*3*2*1) = 270,725.

and C(10,2) = 10*9/(2*1) = 45.

To get the probability of someone holding AA out of 10 players, first multiply 1/221 by 10 to get 10/221. This is almost the right answer, except that it counts the cases where 2 people get AA twice, so we have to subtract the probability that 2 people get AA. There are C(10,2) ways to choose the 2 people, and there is 1 way to choose the 4 aces for those 2 people out of C(52,4) total ways to deal the 4 cards. So the probability of 2 people having aces is C(10,2)/C(52,4). So the final answer is:

10/221 - C(10,2) / C(52,4) = 4.5% = 1 in 22.2

 BruceZ 09-18-2003 06:20 AM

Correction

10/221 - C(10,2) / C(52,4) = 4.5% = 1 in 22.2

 daryn 09-18-2003 12:39 PM

what a way to explain.. if someone who doesn't understand math reads this, stand back

 BruceZ 09-18-2003 01:03 PM