# Talk:Poker probability (Texas hold 'em)

## Opening comment

You know a big problem with this page is I read "barring the miracle flush or straight" a lot. Why would we bar it? It's a part of the game it needs to be figured into the odds.....

## Values in "odds" columns

Is it just me, or are all the numbers in the "odds" column 1 lower then what they should be?

I will change them but if someone else points out where my math is going wrong, then my apoligies.

I can assure you, it was correct before. It's okay, it's a common mistake. The formula is: the odds are defined by (1/p) − 1 : 1, where p is the probability. So, if p = 1/2, the odds should be 1:1, not 2:1. What you're forgetting about is that its the relative frequency of winning to losing, not winning to total action. Revolver 07:34, 5 Jul 2004 (UTC)

## I'd like to see...

• If you flop a flush, the probabilities that another opponent flopped a higher flush. I'm picturing a table high cards down the side and number of opponents across the top.--Toms2866 03:01, 10 May 2006 (UTC)
OK, I have those...I'll add that, but it may be a little while as I'm going to be out of town. —Doug Bell talkcontrib 01:25, 11 May 2006 (UTC)
• Also, if you flop a (say) heart flush, odds someone has a single bigger heart (drawing to a bigger flush) by number of opponents. Brian Alspach has some interesting stats about losing flushes (assuming all opponents see the river).  I was thinking about putting his findings in, but sans the derivations (too complicated for article). Think it would be a good addition?--Toms2866 02:25, 11 May 2006 (UTC)
• What I miss a little bit is the probability to hit something on the flop. E.g. with JTs, how likely is it to have a straight draw (open-ended and/or gutshot), or a flush draw, a pair, two pair,... I found some useful tables at Mike Caros University of Poker Library (XVIII-XXVI for Hold 'em) with some numbers (e.g. 8.14:1 against having a flush draw). WhoCares01 21:43, 10 January 2007 (UTC)
And even some simpler odds have been left out. For instance, if you have any unpaired hand, the odds of hitting a pair or better (using at least one of your hole cards) is about 32.4%. I'd like to see other useful odds, such as the odds of hitting two pair or better on the flop, starting with an unpaired hand and using both of your hold cards. Deepfryer99 (talk) 17:29, 26 August 2008 (UTC)

## Starting hands

I think the formula {52 \choose 2} = 1326 is going to be pretty incomprehensible to 99% of readers. Would it not be clearer to say 52 times 51 divided by 2 = 1326? We non-mathematicians can understand that the first card may be any of 52, that for each first card the second may be any of 51, and that we divide by 2 because each combination may be produced by either Card A followed by Card B or Card B followed by Card A.

In any case, for holdem, 169 is the magic figure. Distinguishing hands such as 5C 3H from 5C 3S is irrelevant and misleading. Obviously those two examples play exactly the same and have the same chance of winning. I've changed this, but my derivation is pretty clunkily worded, and probably not necessary anyway. Stevage 17:23, 3 December 2005 (UTC)

I changed the discussion to include both ${52 \choose 2}=1,326$ and 52 × 51 ÷ 2 = 1,326. I think by discussing both means of representation, the combinatorial math can be introduced in a way that makes is comprehensible to at least most of the other 99% of readers. It gets really messy trying to show the calculations without binomial coefficients once you get beyond choosing two from a set. I also expanded and hopefully made less clunky the explanation of the 169 different strength starting hands. – Doug Bell talkcontrib 07:23, 6 February 2006 (UTC)

What does the "any specific (no/)pair" phrase mean in the starting hands table? Does that mean that any pair in your starting hand as the same odds as AA? That doesn't seem to make sense. Revise the numbers please and link this to combinatorial game theory. 70.111.251.203 23:28, 11 February 2006 (UTC)

That wording existed in the article before I began editing it. The word specific is the key to understand the meaning. AA and KK have the same odds. AK and T2 have the same odds. AKs and 78s have the same odds. Each of these is an example of a specific hand with the same characteristics (hand shape). – Doug Bell talkcontrib 11:47, 23 February 2006 (UTC)

## Calculations for probability of facing larger pocket pairs from multiple opponents

The equations I entered are wrong. I will fix them soon, but please leave them there for the moment unless you want to fix them. They are actually reasonably close approximations. The function cannot use (1 - psingle)players as the events are not independent. The calculation and explanation needs to use (players × psingle) - pmultiple where players is the number of opponents faced, psingle is the probability that a single opponent has a higher pair and pmultiple is the probability that multiple opponents have a higher pair. – Doug Bell talkcontrib 23:08, 6 February 2006 (UTC)

OK, I fixed the equation and the results table. – Doug Bell talkcontrib 01:28, 8 February 2006 (UTC)
There are very significant errors in the tables for 1 or more opponents having a higher pair, and for 2 or more opponents having a higher pair. The equations given are correct, but they were not followed correctly to obtain the table numbers. Also, computing the P's and summing them is not necessary. It requires extra work, and the equations for computing them are not given. These P's could be computed using a generalized form of inclusion-exclusion, but this is not necessary as the numbers in each table can be computed directly without the P's using inclusion-exclusion.
Instead of P's, define pk as the probability that k SPECIFIC opponents have higher pairs. Then we compute the probability of 1 or more of n opponents having a higher pair as
P = C(n,1)*p1 - C(n,2)*p2 + C(n,3)*p3 - C(n,4)*p4 + ... C(n,n)*pn
by inclusion-exclusion, where the final term is added for odd n or subtracted for even n. Note that C(n,k) is combinations of n things taken k at a time. This is the number of ways to choose the k opponents with higher pairs from the n opponents. Of course, while this gives the exact answer, only enough terms need be computed to obtain the answer to the required precision. Note also that
p1 = (84-6r)/1225.
For the probability of 2 or more of n opponents having a larger pair, we compute
P = C(n,2)*p2 - 2C(n,3)*p3 + 3C(n,4)*p4 - 4C(n,5)p5 + ... (n-1)C(n,n)*pn
by a generalization of inclusion-exclusion, where the final term is subtracted for odd n, and added for even n.
Below is shown the correct way to compute the table entries for a pair of deuces vs. 9 opponents. These methods are simpler than the article's methods, and produce greater accuracy for less work.
Here is the probability of 1 or more opponents having a higher pair with 9 opponents when you hold a pair of deuces:
9*72/C(50,2) -
C(9,2)*72*67/C(50,2)/C(48,2) +
C(9,3)*72*(66*62+1*66)/C(50,2)/C(48,2)/C(46,2) -
C(9,4)*72*(66*(60*57+2*61)+1*66*61)/C(50,2)/C(48,2)/C(46,2)/C(44,2) +
C(9,5)*72*(66*(60*(54*52+3*56)+2*(60*56+1*60))+1*6 6*(60*56+1*60))/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2) -
...
=~ 41.9%
Not 36.33% as the table shows. The first 2 terms above are already above 40%, and this is a lower bound. Actually 36.33% may have been erroneously computed from
(72/1225 * 9) - 2*P2
but this is wrong. We would want to subtract P2 + 2*P3 if we were using the article's method.
The above may look complicated, but the terms follow a pattern. To get each next term, replace each final *n in the previous term with
*(6m(n-5) + (n-6m)(n-1))
where 6m is the multiple of 6 just less than n. When n is a multiple of 6, just multiply by another factor of (n-5). Then multiply by the next combinatoric factor, divide by an extra combinatoric factor, and alternate + and -.
Here is the probability of 2 or more opponents having a higher pair with 9 opponents when you hold a pair of deuces. This requires generalized inclusion-exclusion, which here means the factors of 1,2,3... starting each line.
C(9,2)*72*67/C(50,2)/C(48,2) -
2*C(9,3)*72*(66*62+1*66)/C(50,2)/C(48,2)/C(46,2) +
3*C(9,4)*72*(66*(60*57+2*61)+1*66*61)/C(50,2)/C(48,2)/C(46,2)/C(44,2) -
4*C(9,5)*72*(66*(60*(54*52+3*56)+2*(60*56+1*60))+1 *66*(60*56+1*60))/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)
...
=~ 9.5%
Not 14.484% as the table shows. That is larger than just the first term above which is an upper bound.
These calculations have been confirmed by simulation.
Brucezas (talk) 19:05, 17 October 2011 (UTC)

## Latest changes

Just dropping by to say great work on the latest changes. This article could become one of the best within the WikiProject when completed. Look forward to seeing it progress! Essexmutant 00:04, 10 February 2006 (UTC)

## Combinatorial game theory and complexity

Added them to the related links, since they are part of it. 128.6.175.60 20:24, 20 February 2006 (UTC)

## Pictures

Though I love what's going on with this article, there seems to be an overdose of pictures. I'm on a DSL connection and all the pictures don't load within a short time. Is there a way we can reduce the amount, while still keeping all the good information? Perhaps just a single picture file that has the whole chart, instead of a chart with a lot of pictures? 128.6.175.60 20:35, 20 February 2006 (UTC)

First, some of it depends on the general responsiveness of Wiki. The "pictures" are all the math equations. If you set your preferences under the "Math" tab to "HTML if possible or else PNG", many of the equations will be rendered as HTML instead of images. There will still be a lot of images, but probably less than half. Try this and let me know how it works for you. – Doug Bell talkcontrib 21:22, 20 February 2006 (UTC)

## 3 mistakes so far (please check)

The 4th formula in the chapter "starting hands against multiple opponents" seems incorrect: "... and against n opponents is H =..." The passage is "50-2k" , it should be "52-2k" or it does not work out.

I think you are forgetting that 2 cards are already in the player's hand, leaving only 50 cards remaining in the deck to be distributed. – Doug Bell talkcontrib 18:42, 8 March 2006 (UTC)

Chapter "Pocket Pairs": The Link "Probabilites during play" does not work!

Fixed. I renamed the section and forgot to change the link, thanks for pointing that out. – Doug Bell talkcontrib 18:42, 8 March 2006 (UTC)

The chapter "Hands with one ace": The formula contains a "*2" in the second half ("3/50 * (13-x)*4*2/49"). Damn, where does this *2 come from? Without it, it should be in the end: 3/1225 + [6*(13-x)/1225].

Added an explanation for the *2. – Doug Bell talkcontrib 18:42, 8 March 2006 (UTC)

Correct me, if I am wrong. I'd be glad if (in case these are mistakes) get corrected soon.

Thanks.

Sam

Germany

Thank you for your comments, please feel free to either provide additional feedback or simply edit any problems you find in the article. – Doug Bell talkcontrib 18:42, 8 March 2006 (UTC)