Hold'em %s Pre-flop => Flop

[abanger]

I'm trying to determine the %s for various hold'em heads-up matchups pre-flop => flop, given that you know exactly what your opponent has. Here's what I came up w/:

CC vs. AB (Overcards) = (6/48*45/47*44/46 + 46/48*6/47*44/46 + 46/48*45/47*6/46) = 35.12%

AC vs. BD (Livecards) = (6/48*44/47*43/46 + 45/48*6/47*43/46 + 45/48*44/47*6/46) = 33.57%

AB vs. CD (Undercards) = (6/48*41/47*40/46 + 42/48*6/47*40/46 + 42/48*41/47*6/46) = 29.15%

BB vs. AC (Overcard) = (3/48*45/47*44/46 + 46/48*3/47*44/46 + 46/48*45/47*3/46) = 17.56%

AB vs. AC (Dominated) = (3/48*44/47*43/46 + 45/48*3/47*43/46 + 45/48*44/47*3/46) = 16.79%

AA vs. BB (Underpair) = (2/48*45/47*44/46 + 46/48*2/47*44/46 + 46/48*45/47*2/46) = 11.71%

Is the logic behind my calculations correct, and are the percentages roughly correct (I'm not including the possibility of making two pair, straights, flushes, etc.)?

[AaronBrown]

Your logic is not quite right, and you'll have trouble extending these calculations without making errors. I recommend the following approach.

For CC vs. AB, let X represent any card that is not A, B or C. There are 3 A's, 3 B's, 2 C's and 40 X's in the 48 card deck. There are C(48,3) = 17,296 flops.

(1) XXX, there are C(40,3) = 9,880 of these. CC is ahead.
(2) CXX, there are 2*C(40,2) = 1,560 of these. CC is ahead.
(3) CCX, there are 40 of these, CC is ahead.
(4) AXX or BXX, there are 6*C(40,2) = 4,680 of these. AB is ahead.
(5) AAX or BBX, there are 6*40 = 240 of these. AB is ahead.
(6) AAA or BBB, there are 2 of these. AB is ahead.
(7) ABX, there are 9*40 = 360 of these. AB is ahead.
(8) AAB or ABB, there are 18 of these. AB is ahead
(9) ACX or BCX, there are 12*40 = 480 of these. CC is ahead.
(10) ACC or BCC, there are 6 of these. CC is ahead.
(11) AAC or BBC, there are 12 of these. AB is ahead.
(12) ABC, there are 18 of these. CC is ahead.

The nice thing about this is you can add up all the cases to make sure you haven't left anything out or made any mistakes. You can keep the same 12 combinations, just change who is ahead for your other situations (for example, if C is the highest card, CC is ahead instead of behind for AXX; I know you used letters to denote rank, but it's easier to let them stand for the same combinations).

11,984 of the combinations put CC ahead, 11,984/17,296 = 0.6929.

[abanger]

"There are C(48,3) = 17,296 flops."
How did you calculate this?

For case 11, isn't CC ahead (boat vs. trips)?

Thanks for your help.

[AaronBrown]

There are 48 cards not in the two hands. They can come down 48*47*46 ways. But there are 3*2*1 ways those cards can be arranged, and we don't care about the order. In Excel, you can type =COMBIN(48,3) to get the answer.

You're right about (11), I was sloppy.

[abanger]

Anyone want to verify if these are correct?

CC VS. AB = 2.3 : 1
AD VS. BC = 2.4 : 1
AC VS. BD = 2.6 : 1
AB VS. CD = 2.9 : 1
BB VS. AC = 4.8 : 1
AB vs. AC = 5.3 : 1
AA vs. BB = 7.9 : 1

[drbst]

[ QUOTE ]
Your logic is not quite right, [...]

[/ QUOTE ]

How can logic not be right? Logic always leads to correct results, right?

[AaronBrown]

[ QUOTE ]
How can logic not be right? Logic always leads to correct results, right?

[/ QUOTE ]
We need the interruption of the night
To ease attention off when overtight,
To break our logic in too long a flight,
And ask us if our premises are right.

--Robert Frost

[drbst]

When the premises differ, then the results of logical reasoning may differ. (I was sloppy in saying the results are correct.) But that is not the fault of the logic. Logic in itself can not be wrong.

Never mind, I liked your approach a lot, Aaron, I just tried to pull your leg a little. [img]/images/graemlins/wink.gif[/img]