View Full Version : help with odds?

09-18-2003, 04:26 AM
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

09-18-2003, 05:45 AM
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

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

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

09-18-2003, 06:20 AM
10/221 - C(10,2) / C(52,4) = 4.5% = 1 in 22.2

09-18-2003, 12:39 PM
what a way to explain.. if someone who doesn't understand math reads this, stand back

09-18-2003, 01:03 PM
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?