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Old 12-02-2010, 11:21 PM   #1
OnceAgain
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2-7 Triple Draw Hand Combinations

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. In TD, if I'm doing this correct, there are 1024 ways to make any unpaired five card hand. 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
2. A 5 card hand containing 3 cards with specific value (ex. 2,3,7, x, x)

Basically, I'm trying to get a feel for where certain hands lie on the overall distribution of hands. With holdem we can look at tools like pokerstove and get a rough idea of what the top 10% of hands look like.

Thanks!
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Old 12-03-2010, 09:21 AM   #2
BruceZ
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Re: 2-7 Triple Draw Hand Combinations

Quote:
Originally Posted by OnceAgain View Post
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 (45). 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)
43*C(40,2) +
3*C(4,2)*42*40 +
C(3,2)*C(4,2)*C(4,2)*4 +
3*C(4,3)*42

= 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
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Old 12-03-2010, 03:59 PM   #3
OnceAgain
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Re: 2-7 Triple Draw Hand Combinations

Thank you very much!
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Old 12-03-2010, 05:08 PM   #4
BruceZ
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Re: 2-7 Triple Draw Hand Combinations

Quote:
Originally Posted by BruceZ View Post
Note that the 3 cards 2-7 don't have to be different ranks. If they need to be different ranks, multiply the answer to the next part by C(6,3) ways to choose the 3 ranks.
Actually, that doesn't quite work due to overlap, but if you multiply by C(6,3) and change 40 to 28, that gives you exactly 3 different ranks 2-7 allowing pairs or trips:

C(6,3)*
[43*C(28,2) +
3*C(4,2)*42*28 +
C(3,2)*C(4,2)*C(4,2)*4 +
3*C(4,3)*42]

= 657,600

Now if you want 3 or more ranks 2-7 allowing pairs and trips, that would be

C(6,3)*
[43*C(28,2) +
3*C(4,2)*42*28 +
C(3,2)*C(4,2)*C(4,2)*4 +
3*C(4,3)*42] +

C(6,4)*
[44*28 +
4*C(4,2)*43] +

C(6,5)*45

= 794,304

where I just added the cases for 4 and 5 ranks from 2-7, where only 1 pair is possible with 4 ranks.

These numbers are correct as I did an independent calculation to make sure everything adds up.

Last edited by BruceZ; 12-03-2010 at 06:45 PM.
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