Hello,

I have spent the last few months working on the prospect of finding the distribution (or curve) of the odds of winning (pre-flop) for all heads up holdem hands and am looking for people interested in finding this curve.
For instance the most dominated heads up contest in holdem is KK vs K2o in which pocket kings are favored to win at 17.6 to 1, more common is JTs vs 22 in which JTs is only slightly favored to win at 1.166 to 1. But what if one could account for the odds to win for every heads-up hold'em contest? What would the distribution of this curve look like? This is what I am tabulating for and if you are curious enough about this please read on and consider engaging with this problem.

Reviewing the task at hand:
1.) Find the correct count of all heads up preflop holdem hands:
If the total number of heads up holdem hands is 52 C 2 = 1326, then the total number of heads up holdem contests should be 1326 C 2 = 878,475.

2.) Use PokerStoves Equity table for "all" preflop matchups as a base to find the odds of each heads up contest:

http://pokerstove.com/site/analysis/...atchups.txt.gz

PokerStove uses "canons" to display the categories of identical preflop heads up holdem contests. Given that there are 169 kinds of holdem hands,
PokerStove has tabulated for 169 C 2 = 14,196 canons.
For example AKs vs JTo would be a heads up match-up that accounts for only one canon of the 14,196 canons. But there are 48 ways in which AKs vs JTo can occur: 4 ways to make AKs times 12 ways to make JTo = 48.

3.) Address 2 problems with PokerStove's list of canons of preflop matchups for holdem hands:
3a.) PokerStove's list of preflop canons is missing all 169 identical heads up match-ups. This is because they have used combinations with a category rather than using combinations with an actual count of all possible hands. Their combination of 169 choose 2 = 14,196 does not allow for choosing a combination of hands that comes from the same canon. Admittedly all these identical match-ups will have 50/50 probability, and account for less than 1% of the total count but still, they are missing from the distribution and should be included for an accurate count.
For Example:
AA vs AA is not in PokerStove's list
J7o vs J7o is not in on the preflop list etc...
Then the canon count of possible heads-up match ups should be 169 C 2 + 169 = 14,196 + 169 = 14,365

3b.) Because PokerStove uses canons to group identical kinds of preflop heads up hands, these match-ups have not been weighted for frequency of occurrence and cannot be used as is to plot a distribution of odds until their frequencies have been accounted for and categorized.

Counting occurrences per canon in regards to suitedness and pairs:
suited versus suited hands: SvS = 4x4 = 16 occurrences per canon
suited versus suited with 1 shared card: SSM = 4x3x1 = 12 occurrences
suited versus suited identical: SSMM = 4x2x1 = 6 occurrences
suited versus offsuited hands: SvO = 4x12 = 48
suited versus offsuited with 1 shared card: SvOM = 4x3x3 = 36
suited versus offsuited identical: SvOMM = 4x3x2 = 24
offsuited versus offsuited hands: OvO = 12x12 = 144
offsuited versus offsuited with 1 shared card: OvOM = 12x3x3 = 108
offsuited versus offsuited identical: OvOM = 12x3x3 = 42
Pair vs Suited: PvS = 6x4 = 24
Pair vs Suited with one shared card: PvSM = 6x2 = 12
Pair vs Offsuited: PvO = 6x12 = 12
Pair vs Offsuited with one shared card: PvOM = 6x2x3 = 36
Pair vs Pair Distinct: 6x6 = 36
Pair vs Pair identical: = 3

4.) With the frequency of occurrence per canon accounted for in regards to suitedness and pairs, account for the relative ranks of the cards in these categories of match-ups. This will organize the data set for symmetry and make it easier to find any discrepancies in the total count.

A key will be useful in abbreviating classifications:
S = 2 Suited cards
O = 2 Off-suited cards
P = Pair
NP = Non-paired
F = Favored Hand
D = Dog hand
FFDD= Distinct rank of the 4 cards of the Favored and Dog Hands.
M = Shared rank between 2 hands

Two tables below shows what I have for an existing count:

NP vs NP 4 Distinct Ranks 1 Shared Rank 2 Shared Ranks Sums
F vs D: FFDD FDFD FDDF MFD FMD FDM MM
Examples:AK v J9 AJ v K9 A9 v KJ AJ v A9 AJ v J9 A9 v J9 AJ v AJ
SvS 16x715 16x715 16x715 12x286 12x286 12x286 6x78 45,084
SvO 48x715 48x715 48x715 36x286 36x286 36x286 24x78 135,720
OvS 48x715 48x715 48x715 36x286 36x286 36x286 0 133,848
OvO 144x715 144x715 144x715 108x286 108x286 108x286 42x78 404,820
Sums 183040 183040 183040 54912 54912 54912 5616 719,472

P vs NP NP distinct from P 1 Shared Card PvP PvP Sums
F vs D: PDD DPD DDP PMD DMP Distinct Same
Examples:KK v J9 JJ v K9 KJ v 99 AA v A9 AJ v JJ KK v JJ AA v AA
PvS 24x286 24x286 24x286 12x78 12x78 0 0 22,464
PvO 72x286 72x286 72x286 36x78 36x78 36x78 3x13 70,239
Sums 27456 27456 27456 3744 3744 2808 39 92,703


These tables did not transfer over as well as I would have liked, But suffice it to say there is a Lot of Symmetry here. The total count from the tables above is 719,472 + 92,703 = 812,175
This value is short by 66,300 of the true count which is 878,475.

I am off by 7.5% which is a rather large error.

Do you see anything in this count that would account for this error? The only hint I can see here is that the value of 66300 is cleanly divisible by 78, one of the common counts for the sub categories.
Does the value of 850 in the product of 78x850 = 66,300 reveal where this glitch is coming from?

This is your error. It should be

C(52,2)*C(50,2) / 2 = 812,175.

It's not C(1326,2) because the first player's cards reduces the number of hands for the second player to C(50,2) = 1225, not 1325.


.

Wow, Thank you BruceZ!

I had overlooked that entirely. Yes, there are 1326 unique holdem hands but we can not choose 2 from this group as the 2 cards that form the first hand will not be available for the second hand... However there is still a small correction here C(52,2)*C(50,2) = 1,624,340 which is 2x812175
Then the correct count would be accounted for by ½* C(52,2)*C(50,2) = 812,175

Great! thank you much, Now onward to the plotting of the distribution,

Yes, however, C(52,2)*C(50,2)/2 = 812,175 is total matchups without regard to which player has which hand, and C(52,2)*C(50,2) = 1,624,340 distinguishes our hand from our opponent's hand, so be sure to use the right one.

Right, of course.

I am plotting these curves always from the perspective of the favored hand. Using the odds to win for the favored hand. The odds will range from 1:1 to 17.6:1

I have the general curve finished but it is so compressed that I have chosen to expand the view. I am now working on the curves for individual classes of hands: Suited vs Suited, OffSuited vs Offsuited, etc....

Thanks for the help, I will post some images soon once I finish..

Okay, I have 28 distributions for those that are curious about the distributions of Heads-Up match-ups in Texas Holdem.

Check out the document at: http://users.humboldt.edu/tpayer/Holdem_Distributions/

That document is long gone; the Wayback Archive shows the link was dead over two years ago.

Your "canon count" undercounts the number of distinct matchups because it fails to address the various suit combinations that can occur. For example, just saying "KK vs. K2o" does not adequately describe the most dominated heads up contest. The reason is because the suit of the deuce matters: the K2o has a better chance of winning the hand if it matches the suit of the missing K, than it does if the deuce's suit matches the suit of one of the paired kings (the reason is obvious: the K2o has more winning flushes in the former case).

The true "canon count"---that is, the number of distinct hold'em matchups, where two matchups are considered equal if one can be gotten from the other by permuting the suits---happens to be 47,008. In April 2012 I put together a website that gives the high-hand equity and tie percentages for all of these matchups. It can be found here:

http://www.mathematrucker.com/poker/matchups.php

I used two distinct methods to calculate these percentages and came up with nothing but matches, so they are all almost certainly correct.

I should also mention that when I put together my page, again it's at

http://www.mathematrucker.com/poker/matchups.php,

everything went into a database that contains the "distribution (or curve)" that you were attempting to gather when you wrote your post. This distribution contains exactly 44,784 distinct values. Since there are 47,008 total, this means there are quite a few distinct matchups with identical high-hand winning percentages (again by "distinct" I mean you can't get one matchup from the other by simply permuting suits).

The biggest jump in the distribution is 0.34% (90.63% down to 90.29%). The second-biggest one happens to be the difference (0.26%) between the most dominated matchup (K2o vs. KK with the suit of the deuce matching the suit of one of the paired kings - KK wins 94.92% of the time) and the second-most dominated one (Q2o vs. QQ with the suit of the deuce matching the suit of one of the paired queens - QQ wins 94.66% of the time). What this sort of says is, the K2o vs. KK matchup is especially dominated!

There are zillions of questions one could ask about this data set. For example:

"Assuming that each of the 812,175 possible hold'em matchups are equally likely, what is the conditional probability given a pocket pair vs. two unpaired overcards (suited or not), that the overcards have a better than 50% chance of winning the hand preflop?"

Offhand I would guess the answer to this question is somewhere around 5%.

Lastly it's worth noting that page 719 of Stewart Ethier's book "The Doctrine of Chances: Probabilistic Aspects of Gambling" (http://www.amazon.com/Doctrine-Chanc...dp/3540787828/) gives a detailed hand-type vs. hand-type account of the 812,175 * 2 ordered matchups (each matchup is an ordered pair of two starting hold'em hands).