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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

Bruce further explanation please
 
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

Re: Bruce further explanation please
 
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

Re: Bruce further explanation please
 
what a way to explain.. if someone who doesn't understand math reads this, stand back

BruceZ 09-18-2003 01:03 PM

Re: Bruce further explanation please
 
It's pretty near impossible to explain the inclusion-exclusion principle to someone who doesn't understand math, and have them apply it correctly to cards. This is the simplest example of that principle there is for poker. Is it clear enough?


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