Matrix of all 169 x 169 hold 'em starting hand matchups?

[Phat Mack]

Are you saying that if SB went all in, and you had AA, you would only call if the bet was less than 449.16?

[allenciox]

whoops again, it should be subtract 1, not 2.

[allenciox]

I forgot to edit that from my program output. All amounts above 82 are estimated. Obviously AA should be infinite.

[Phat Mack]

OK. I'm just trying to make sure I understand the numbers. Let's say that sb calls 1 and raises 62. In other words, he puts in 21 times the dead money in the pot. You'd call with ATs/66 or better. Correct? With KK, you could call a bet of 118 x's dead money...

Nice work, allenciox.

[George Rice]

This matrix has AKs as a 5% favorite over AKo. These numbers include ties which is not very helpful. This is the same mistake Andy Glazer made. AKs is a 7:2 favorite over AKo.

[rockoon]

[ QUOTE ]
Does anyone have a matrix giving the EV of all possible Texas hold 'em starting hand matchups --- this would assume that both hands are playing all the way to the end, and would average across suit combinations --- for example, the matchup between AA and KK would do a weighted average of the EVs of all different suits (1 combination), one suit overlapping (4 combinations), and both suits the same (2 combinations).

It seems intuitive that SOMEONE has done this at some point... if they could share it with me, or point me to it, I will be happy to share my analyses --- I want to try to build on the Sklansky-Karlson results.



[/ QUOTE ]

The problem with using a 169x169 matrix is that it doesnt address a real problem. You need a 1326x1326 matrix to get the math 100% right.

Some steps can be taken to reduce the problem somewhat to a smaller matrix and retain 100% accuracy but its still very large compared to the 169x169 matrix.

My lookup table has 428415 elements. This is not as small as it can be but its close enough for my system ram [img]/images/graemlins/laugh.gif[/img]. Some redundant information still exists.

To point people in the right direction for this type of reduction, note that there are only 15 *meaningfully unique* ways to arrange the suits of 4 cards. (as a side note, 15x169x169 = 428415)

[Bozeman]

I get only 11:
ssss
shdc
sshh
shhs
shsh
sshd
shsd
shds
hssd
hsds
hdss

Which ones am I missing?


And one could get to 1326/4*1326=439569 very easily, just assume (without loss of generality) that the first card is a spade.

Craig

[rockoon]

[ QUOTE ]
I get only 11:
ssss
shdc
sshh
shhs
shsh
sshd
shsd
shds
hssd
hsds
hdss

Which ones am I missing?


And one could get to 1326/4*1326=439569 very easily, just assume (without loss of generality) that the first card is a spade.

Craig

[/ QUOTE ]

You are missing hand matchups that have 3 cards of the same suit.

How do you get a division by 4 exactly? If you are assuming the first card is a spade (this is similar to my indexing technique, BTW) then you have 13*25*1326 = 430950

My indexing technique is partly based on the speed at which I can produce the index for access the table.

[Bozeman]

Oops, knew I shouldn't do this late at night.

Obviously one could do better indexing, I just wanted to make a stupid, quick estimate. The factor of 4 is not exact because pairs have no definitive suit blind order.

Craig

[JonCooke]

So if there is a tie do they redeal the flop in your game?