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10-21-2012, 05:56 PM   #2
BruceZ
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Join Date: Sep 2002
Posts: 11,877
Re: Plotting the distribution of odds for all preflop heads-up hands in Texas Holdem

Quote:
 Originally Posted by Corvus Corax 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.
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.

Quote:
 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
.

Last edited by BruceZ; 10-22-2012 at 07:47 PM. Reason: divide by 2

 10-22-2012, 04:17 PM #3 Corvus Corax stranger   Join Date: Feb 2012 Posts: 6 Re: Plotting the distribution of odds for all preflop heads-up hands in Texas Holdem 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, Last edited by BruceZ; 10-22-2012 at 07:42 PM. Reason: inadvertent
10-22-2012, 07:45 PM   #4
BruceZ
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Join Date: Sep 2002
Posts: 11,877
Re: Plotting the distribution of odds for all preflop heads-up hands in Texas Holdem

Quote:
 Originally Posted by Corvus Corax 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
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.

 10-23-2012, 08:56 PM #5 Corvus Corax stranger   Join Date: Feb 2012 Posts: 6 Re: Plotting the distribution of odds for all preflop heads-up hands in Texas Holdem 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..
 10-31-2012, 07:08 PM #6 Corvus Corax stranger   Join Date: Feb 2012 Posts: 6 Re: Plotting the distribution of odds for all preflop heads-up hands in Texas Holdem 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/
12-16-2017, 12:33 AM   #7
mathematrucker
stranger

Join Date: Feb 2012
Location: Las Vegas, USA
Posts: 6
Re: Plotting the distribution of odds for all preflop heads-up hands in Texas Holdem

Quote:
 Originally Posted by Corvus Corax 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.

12-18-2017, 12:08 AM   #8
mathematrucker
stranger

Join Date: Feb 2012
Location: Las Vegas, USA
Posts: 6
Re: Plotting the distribution of odds for all preflop heads-up hands in Texas Holdem

Quote:
 Originally Posted by Corvus Corax 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
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!

"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).

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