Poker Heads Up Odds

In this lesson we’re going to run through a number of heads-up match-ups that will help give you an idea of where you stand in a variety of pre-flop situations when playing hold’em. Be aware that we’re only going to focus on individual hand match-ups. When playing hold’em it’s essential that you put your opponent on a range of hands, rather than specific holdings. However, knowing the odds of common pre-flop match-ups is a good starting point. Pick out and study what will help you. While it’s not essential that these statistics be committed to memory, it won’t hurt you if you do.

Let’s start by looking at hand match-ups when holding a pair:

Pair vs. Pair

The higher pair is an 80 percent favourite. We can get very technical and highlight the fact that if the underpair didn’t have any clean suits and/or the maximum number of straight outs then the high pair’s equity would increases by one or two percent.

Pair vs. Overcards

Doug Polk now installed as a -555 favorite in the upcoming heads-up challenge against Daniel Negreanu. A string of shaky results in warm-up matches hasn’t hurt Doug Polk’s odds against Daniel Negreanu in the eyes of PokerShares. The poker betting platform shifted the odds even more in. Using a Poker odds Calculator. Want to know how far ahead or behind you are in a Texas Hold’em hand against one, two or more opponents? Our poker calculator is the perfect medium for finding out the odds in any given situation. Simply plug in your hand, your opponents’ hands, and the board, and you’ll be on the way to figuring out your. The poker odds calculators on CardPlayer.com let you run any scenario that you see at the poker table, see your odds and outs, and cover the math of winning and losing poker hands. Texas Hold'em Omaha.

This is the classic coin flip hand that you’ll see many times late in tournaments with one player being all-in. The term coin flip indicates an even money situation which is really a 55 to 45 percent situation, as the pair is a slight favourite.

Pair vs. Undercards

In this situation the pair is normally about a 5-to-1 favourite and can vary depending on whether the two undercards are suited and/or connectors.

Pair vs. Overcard and an undercard

The pair is about a 70 percent favourite. Another example of this holding would be J-J against A-9. The underdog non-paired hand has three outs while the favourite has redraws.

Pair vs. Overcard and one of that pair

The classic example of this situation is the confrontation between a pair of cowboys and big slick. Casino gold coast restaurants. The A-K has three outs and it becomes a 70-30 percent situation or a 2.3-to-1 dog for the cowboys. This is a far cry from the next situation where even though one of the pair is matched the other card is lower.

Pair vs. Undercard and one of that pair

The non pair has to hit its undercard twice or make a straight or flush to prevail. The pair is better than a 90 percent favourite or slightly better than 10-to-1 odds. I’ll take those odds anytime.

Pair vs. Lower suited connectors

You see this match-up late in tournaments when a player is getting desperate and pushes all-in with middle suited connectors. A hand such as Q-Q against 7-6 suited would be a prime example. The pair is a strong favourite to win.

Poker Heads Up Odds Against

Pair vs. Higher suited connectors

Here is the real coin flip situation. A pair of eights heads-up against a suited Q-J is a fifty-fifty proposition. The higher suited cards would have an edge against a lower pair, such as 2’s or 3’s, since the board itself can sometimes destroy little pairs.

Common Pre-Flop Match-Ups (Non Pairs)

The following heads-up confrontations contain no pairs.

Two high cards vs. Two undercards

The two higher cards are usually a 65% favourite to win, but it can vary depending on whether any of the cards are suited and/or connectors.

High card, low card vs. Two middle cards

In this match-up the high card gives it the edge. But it’s only a marginal winner, approximately 57% to the hand containing the high card.

High card, middle card vs. Second highest, low card

Poker Heads Up Odds Nfl Week 7

The edge is increased by around 5% when the low card becomes the third highest card, as shown in this example, which gives approx 62% to 38% for high card/middle card combination.

High card, same card vs. Same card, low card

In this example the A-J is in a very strong position. If we discount any flush or straight possibilities, it only leaves the player holding J-8 with three outs (the three remaining 8’s).

Same high card, high kicker vs. Same card, low kicker

The high kicker gives this hand a fairly big edge. It’s very common for A-K run into A-Q, A-J, and lower, and it’s why Ace-King is such a powerful hand, particularly at the business end of no-limit hold’em tournaments when people move all-in with any sort of Ace.

Texas Holdem Heads Up Odds

Statistical Variations

Odds

For any math maniacs reading this who do not find these odds precise enough, I acknowledge that the math is rounded and for the most part does not take into account the possibilities of ties and back door straights and flushes. What players need to be equipped with is the general statistical match-up – not the fact that in the example of a pair of eights against a suited Q-J the percents are exactly 50.61 for the eights to 48.99 for the suited connectors with the balance going to potential ties. I call that a fifty-fifty proposition.

Of greater importance than quibbling over tenths of a percent is the fact that in most heads-up confrontations you can never be a prohibitive underdog. That is one reason why poker is so challenging and fun. Of course, while true, I’m not attempting to embolden the reader to ignore the odds and become a maniac. Math is the underpinning of poker and if you regularly get your money into the middle with the worst of it you will go broke.

One statistic that hasn’t been mentioned, and it’s one that I particularly like is this – the odds of both players being dealt Aces when playing heads up (one on one) is 270,724-to-1. It’s my favourite statistic because it provides me with almost total confidence when I’m playing heads up and receive pocket Aces that I’m the boss! That confident feeling lasts right up to the river when my Aces get cracked by some rotten piece of cheese which my opponent elected to play. As mentioned already, rarely are you a prohibitive underdog – so remember that to keep those losing hands in perspective.

Related Lessons

By Tom 'TIME' Leonard

Tom has been writing about poker since 1994 and has played across the USA for over 40 years, playing every game in almost every card room in Atlantic City, California and Las Vegas.

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In this article we define and publish the exact pre-flop probabilities for each possible combination of two hands in Textas Hold’em poker. An online tool at tools.timodenk.com/poker-odds-pre-flop makes the data visually accessible.

Table of Contents

Introduction

A deck of French playing cards, as it is used in Texas Hold’em, contains 52 different cards; in a heads-up game two players are playing against each other. Both of them get two private cards dealt pre-flop face down. There are $binom{52}{2}=1326$ different possible pairs of cards that players can get. In this work we determine the odds of each starting hand to win against any other starting hand. There is no equation or easy way of calculating the winning probability of a given hand, since it would be required to contain all the rules and mechanics of the game.

Statistical approaches can determine the winning odds of one starting hand against another very quickly. This gives results that are statistically accurate but not guaranteed to be exact. However, for mathematical analysis of certain properties of pre-flop situations, the precise numbers are a requirement.

Win Function

Hold’em cards can have 13 different ranks and four different suits
$$begin{align}
mathcal{R}=&left{text{Ace}, text{2}, text{3}, text{4}, text{5}, text{6}, text{7}, text{8}, text{9}, text{10}, text{Jack}, text{Queen}, text{King}right},
mathcal{S}=&left{text{Club}, text{Heart}, text{Spade}, text{Diamond}right},.end{align}$$A card is an ordered pair of rank and suit; the set of all cards that exist in a the card deck is denoted as
$$begin{align}mathcal{C}=mathcal{R}timesmathcal{S},end{align}$$ with $leftlvertmathcal{C}rightrvert=52$. The set of possible starting hands, every possible combination of two distinct cards, is defined by
$$begin{align}mathcal{H}=left{left{c_1inmathcal{C},c_2inmathcal{C}right}mid c_1neq c_2right}end{align}$$ and has a cardinality of $lvertmathcal{H}rvert=binom{52}{2}=1326$. The set is containing unordered pairs because the two private cards are not ordered either.

Calculating the Cartesian product of the set of possible hands with itself gives a new set of ordered pairs, that is $mathcal{M}=mathcal{H}timesmathcal{H}$, with $leftlvertmathcal{M}rightrvert=1,758,276$. This set contains all combinations of two starting hands as ordered pairs. The order matters, because we define the meaning of each of the pairs as a pre-flop situation where the first starting hand plays against the second.

In this work we search for the winning odds function
$$begin{align}
o:mathcal{M}rightarrowleft{xinmathbb{Q}mid0le xle1right},
end{align}$$ that outputs for any pre-flop situation $left(h_1,h_2right)$, the odds of starting hand $h_1$ to win against $h_2$. Since every card exists only once in a deck, two players cannot play against each other if $h_1cap h_2neemptyset$. For these cases $o$ is undefined.

Post-flop, five community cards are dealt. Three on the flop, followed by the turn card, and finally the river card. $binom{52}{5}=2598960$ different combinations are possible. We define $mathcal{P}$ as the set of all possible community cards, where each $pinmathcal{P}$ is itself a set of five different cards. The community cards $p$ determine which player wins, that is the one whose starting hand builds the best showdown hand. If both showdown hands are of equal rank at showdown, the pot is split. Given a pre-flop situation $m=(h_1,h_2)$ where the starting hand $h_1$ plays against $h_2$, only a subset of $mathcal{P}$, namely $$begin{align}mathcal{P}_{(h_1,h_2)}=left{pinmathcal{P}mid pcap h_1=pcap h_2=emptysetright}end{align}$$ can be dealt, as some cards are already taken from the deck.

Furthermore a win function
$$begin{align}w:{(minmathcal{M},pinmathcal{P}_m)}rightarrow{0,1}end{align}$$is required, that assesses a situation at showdown and returns $1$, if the first starting hand in $m$ wins against the second hand, given the public cards $p$. In case of loss or split it returns $0$. $w$ is defined by the rules of poker. With these definitions at hand we can define the odds of a hand to win against another hand as
$$begin{align}o(m)=frac{displaystylesum_{pinmathcal{P}_m}w(m,p)}{lvert mathcal{P}_mrvert},.end{align}$$In words, $o$ is dividing the number of possible community cards where $h_1$ wins against $h_2$ by the number of community cards that can be dealt altogether.

Data and Results

With the odds function $o$ at hand, we can define the winning odds matrix $mathbf{M}inmathbb{Q}^{lvertmathcal{H}rverttimeslvertmathcal{H}rvert}$ as
$$M_{i,j}=o(mathcal{H}_i,mathcal{H}_j),.$$ The matrix contains the winning odds for every heads-up pre-flop situation and is undefined at places, where $mathcal{H}_i$ and $mathcal{H}_j$ cannot play against each other (because of shared cards). From its row vectors, i.e. $mathbf{M}_{i,:}$, we can compute the average odds of a hand $mathcal{H}_i$ to win against a random hand, by taking the average of all entries with values. The probabilities for a split between two hands are given by $1-M_{i,j}-M_{j,i}$.

We release $mathbf{M}$ in two ways.

  1. As an online tool with user interface at tools.timodenk.com where the user can pick their pre-flop situation and get the exact odds, i.e. $o(m)$ and $o(m)lvertmathcal{P}_mrvert$.
  2. As a serialized Java object holding the $o(m)lvertmathcal{P}_mrvert$ values for every $m$. The object can be imported into a Java program and processed from there.

We have already conducted some experiments related to the non-transitivity of the win function. For example, we found the three hands
$$begin{align}
h_1=&left{left(text{Ace},text{Club}right),left(text{2},text{Club}right)right}
h_2=&left{left(text{10},text{Spade}right),left(text{9},text{Spade}right)right}
h_3=&left{left(text{2},text{Heart}right),left(text{2},text{Diamond}right)right},
end{align}$$to satisfy$$begin{align}
oleft(h_1,h_2right)approx&0.54gt0.5
oleft(h_2,h_3right)approx&0.53gt0.5
oleft(h_3,h_1right)approx&0.61gt0.5,.
end{align}$$Expressed in words, this means for a hand $h_1$, which statistically beats $h_2$, which in turn beats a third hand $h_3$, it cannot be concluded that $h_1$ beats $h_3$ as well. The win function is not transitive.

The most uneven pre-flop situation exists, if the hands$$begin{align}
h_1=&left{left(text{King},text{Club}right),left(text{King},text{Diamond}right)right}
h_2=&left{left(text{King},text{Heart}right),left(text{2},text{Club}right)right}
end{align}$$play against each other (or suit permutations). The pair of kings has a $94.16%$ chance of winning. The pot is chopped in $1.53%$ of all possible outcomes.

Surprisingly, Aces do not appear in this constellation. The reason is that the hand A2 could hit a straight with just three cards (3, 4, 5). On the other hand, K2 needs four cards (A, 3, 4, 5 or 3, 4, 5, 6) in order to build a straight that uses the 2. Also noteworthy is the fact, that the King of Clubs of $h_1$ blocks flushes, which $h_2$ could have otherwise gotten using the Deuce of Clubs.

The lowest winning probability and highest probability for a split pot exists if $$begin{align}
h_1=&left{left(text{3},text{Club}right),left(text{2},text{Diamond}right)right}
h_2=&left{left(text{3},text{Diamond}right),left(text{2},text{Club}right)right}
end{align}$$battle each other. Both hands have an equal chance of winning, namely $0.71%$. Consequently, a split occurs with a likelihood of $98.57%$.

Interestingly, the get-together of 43 vs. 43 (same suits as above) comes with a higher winning probability ($0.73%$). That’s (at least partly) because there are a few more flushes, for which one of the hands does not only play the board.

The lowest split probability of just $0.19%$ is there, if the following hands play:$$begin{align}
h_1=&left{left(text{9},text{Club}right),left(text{8},text{Diamond}right)right}
h_2=&left{left(text{Ace},text{Spade}right),left(text{Ace},text{Heart}right)right},.
end{align}$$In which cases would the pot be chopped? For instance if a straight flush of Clubs from 2 to 6 occurs.

Still unsolved is the search for the longest non-transitive chain of mutually disjoint pre-flop hands, such that $$o(h_1, h_2)>0.5land o(h_2,h_3)>0.5landdotsland o(h_n,h_1)>0.5,.$$

Special thanks goes to Dominik Müller for fruitful discussions and many algorithm optimization ideas.