Quote:
Originally Posted by OnceAgain
Hi,
I'm trying to figure out how to determine the number of hand combinations in 2-7 TD under various circumstances. For example, in holdem we know that there are 16 ways to make any unpaired hand.
16 is just the number of ways to choose the suits. Multiply by the C(13,2) ways to choose the ranks to get the number of unpaired hands:
C(13,2)*16 = 13*12/2 * 16 = 1248
Quote:
In TD, if I'm doing this correct, there are 1024 ways to make any unpaired five card hand.
1024 is the number of ways to choose the suits (4
5). Multiply by the C(13,5) ways to choose the ranks to get the number of unpaired hands:
C(13,5)*1024 = 13*12*11*10*9/(5*4*3*2*1) * 1024 = 1,317,888
Quote:
Assuming this is correct, can anyone help me with how I would figure out stuff like ...how many ways there are to make:
1. A 5 card hand containing 3 cards between a deuce and 7
Exactly 3:
C(24,3)*C(28,2) = 765,072
3 or more:
C(24,3)*C(28,2) + C(24,4)*28 + C(24,5)
= 1,105,104
Note that the 3 cards 2-7 don't have to be different ranks. See next post if they must be different ranks.
Quote:
2. A 5 card hand containing 3 cards with specific value (ex. 2,3,7, x, x)
4
3*C(40,2) +
3*C(4,2)*4
2*40 +
C(3,2)*C(4,2)*C(4,2)*4 +
3*C(4,3)*4
2
= 62,064
The first term is for exactly 1 of each 2,3,7. The second term is for 2 of one of them. The third term is for 2 of two of them. The last term is for 3 of one of them.
C(n,k) = n!/[(n-k)!*k!]
Last edited by BruceZ; 12-03-2010 at 05:09 PM.
Reason: different rank comment