Poker Heads Up Odds
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.
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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.
Pot odds is actually the least important in heads-up, because rarely will you call because you are 'getting odds'. If anything, if you have odds to call - BET! Play those flushies and straight draws aggressively, because whatever odds you are getting, aggression gives those odds a skyrocket value with folding equity tossed 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.
- You will improve it on the turn: 9.2=18% and real poker odds are around 19%; You have a straight draw on the flop with 8 outs. You will improve it on the turn: 8.2=16% and real poker odds are around 17%; You have two over cards on the turn with 6 outs. You will improve it on the turn: 6.2=12% and real poker odds are around 13%.
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.
- 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 $o(m)lvertmathcal{P}_mrvert$.
- 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.
Stripper Stacks
Strippers are cards that are shaped in such a way as to allow them to be easily controlled by pulling them out of the deck, no matter where they are located.
A stripper stack uses this concept, along with carefully chosen cards, to enable a cheat to easily stack the deck in their favour.
A well known example of a stripper stack is the ten-card poker deal, which is a concept often used by magicians.
The Ten-Card Poker Deal
The ten-card poker deal came from the card table; it originally used nine cards and was specifically designed for a 5-card straight poker heads-up scenario.
The cheat would take nine cards from the deck, and cut them in such way that they could be stripped out when required.
In play, after the mark had shuffled, the deck would be passed to the cheat to be cut. The cheat would use the motion of cutting the cards to strip out the nine target cards and place them on top.
The result of this would be that the nine cards are dealt between the two players, with the mark’s final card being a random card from the deck.
The nine cards consisted of three sets of three-of-a-kind, which we will denote A, B, and C.
So the nine cards are: AAABBBCCC. As such there are a relatively small number of possible outcomes:
The table shows the probabilities that either party will win, given the way the nine cards come out, and whether the random card is helpful to the mark or not.
In most cases the random card is not useful, and the cheat will win. But occasionally the random card will improve the marks hand, and in some cases that may be enough to beat the cheat.
Overall, the cheat will win 96.01% of the time, whereas the mark will only win 3.99% of the time.
The advantages of this particular method of cheating, are that
- It only requires a single “cut” type action, which is built into the game’s procedure,
- It’s done on the opponent’s deal,
- It has a very high success rate.
The main disadvantage is that it requires stripped cards, which need to be put into the game somehow.
Getting Strippers into the Game
Cutting cards to make strippers is not generally something that can be done at the table, which leaves three options:
- Making sure the deck in play has already had the cards doctored
- Switching the deck or the required cards during play
- Cutting the cards from the deck in-play
The first method could be achieved by stocking the local shop with a large number of doctored decks and then insisting that a new deck is purchased before the game begins. Another method would be to have an accomplice who pretends to purchase a deck, but in fact just provides a prepared deck.
Alternatively, the deck could be switched before the game; again with an accomplice that has access to where the decks are stored.
A deck switch could also be done by the cheat during the game, either when they handle the cards to deal, or when given the cards for a cut. In some cases an accomplice may be used to provide a distraction whilst the decks are switched.
Depending on the style of strippers used, it may actually be only the target cards that need to be gimmicked. In this case they could be slowly switched out, one at a time, throughout the night as game play continues.
Finally, the cards in play could be cut, which would require stealing them from the deck, taking them somewhere private to put the work in, and then reintroducing them back into the pack.
A Stripper Stack for Heads-Up Texas Hold’em
Currently, the most popular form of poker is Texas Hold’em, which has quite a different format to straight poker.
Two face down hole cards are dealt to each player, then between rounds of betting, five communal cards are dealt face up on to the table. The players must make the best 5-card hand using their hole cards and the communal cards.
The aim would be to find a set of cards that can be stripped to the top in a single cutting action, and provide a high win rate, when playing heads-up Texas Hold’em.
The introduction of communal cards means that in many cases the winning hand is not decided until the final card comes down. This means that, selecting a set of 11 cards, such that the river card is random, makes the final result highly variable. So this is not a particularly effective strategy.
If the rules were loosened slightly to allow two ‘moves’, then 7 cards of one suit could be stripped to the top in the first action, and the ace of that suit stripped to the top in a second action. So, instead of ensuring that the mark is dealt a bad card, it’s ensured that the cheat is dealt a good card. In this case, both players would make a flush, and the cheat would win with the ace. This scenario produces a 100% probability of the cheat winning, as the ace high flush can’t be beaten, regardless of which cards are dealt on the turn or the river. However, there is the risk that the cards the mark ends up with are not good enough for them to want to bet on; so whilst the cheat will still win the hand, it may not be the biggest pot.
An alternative would be only controlling the hole cards, risking the fact that the 5 communal cards could completely change the winning hand.
By stripping three aces to the top on the mark’s deal, the cheat will ensure a pair of aces for themselves and the mark will be dealt at least one ace.
There are two advantages to this; firstly, giving the mark an ace may entice them to stay in the hand longer, giving the cheat the opportunity to win more. Secondly, the cheat holds the other two aces, which makes it unlikely that the mark will pair their ace. Furthermore, if the mark makes a hand with their ace, then in many cases the cheat will at least be able to equal the hand, and in some cases beat it.
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To illustrate this advantage, the cheat’s pair of aces against any other two random cards, gives the cheat a win rate of 84.93%, the chances of drawing are 0.54%, and losing is 14.52% [results from wizardofodds.com]. Whereas, when the cheat’s aces are up against an ace and any other card, the cheat’s win rate is increased to 88.67%, the chances of drawing is increased to 3.17%, and crucially the probability of losing is now only 8.15% [results calculated with the help of cardplayer.com]. So by giving the mark an ace, the cheat’s chances of winning are increased, as is the chance of the mark making a bet.
Whilst this 88.67% probability of winning is not as high as 96.01% for the straight poker game, or the 100% for the flush scenario, it is still gives a reasonably high guaranteed edge.
To take this further, in The Mathematics of Poker, by Bill Chen and Jerrod Ankenman, a near optimal strategy is given for heads-up play when both player’s stacks are less than 50 big blinds. This strategy suggests that the only two moves in this scenario should be to jam or fold. The details of this strategy are described by a table, which give all possible hole card combinations, and assigns each a value. For the attacker, this value is the stack size in big blinds at which your play should switch from fold to jam. A second table is given for the defender, which dictates at what stack size their strategy should switch from fold to call.
When heads-up it’s common for the order of play to change. In standard multiway tables the player to the left of the dealer posts the small blind, the next player puts up the big blind, and the next player is the first to act. In this manner when heads-up, the dealer would put in the big blind and their opponent would act first. However, the change in play for heads-up means that the dealer acts first and their opponent puts in the big blind.
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As the cheat is stripping the cards on the mark’s deal, then if the special heads-up order of play is enforced, the cheat is considered the defender. Whereas, if this change of play is not introduced, then the cheat would be considered the attacker.
Whilst the jam-fold strategy was developed for the alternate heads-up order of play, with the dealer acting first, results were calculated considering the cheat as both the attacker and the defender.
The following graph shows the cheat’s probability of success for a given opponent’s stack size. Where success would mean that the mark did not fold, and the pocket aces won the hand.
The cheat’s actions are clear, with their aces, they should jam if they are the attacker or call if they are the defender. However, the mark’s actions will depend on their stack size and what random card they get with their ace.
The graph shows that the cheat is more likely to succeed if they are the defender, which suggests that the alternate heads-up order of play is beneficial for this method of cheating.
As an example, if the mark has a stack of 40 big blinds, then there is a 60.64% chance of the mark going all-in and the cheat winning the pot.
For a 50% or greater probability of success, the cheat should use this technique when their opponent has less than 42.6 big blinds in their stack. Or 29.2 big blinds if the alternate heads-up order of play is not used and the cheat is the attacker.
Conclusion
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The nine-card poker deal was designed for a game of straight poker, and gave the cheat a 96.01% chance of winning the pot. The required cards could be stripped out in a single cutting action, meaning that the play would take place on the mark’s deal.
Applying the same methodology to the game of Texas Hold’em is made more difficult with the introduction of communal cards. But by stripping only three aces, a similar outcome can be achieved; although the win rate drops to 88.67%.
Furthermore, using the near-optimal jam-fold strategy when the players are down to less than 50 big blinds, the cheats probability of success can be determined. These results suggest that the cheat has a greater than 50% success rate when their opponent has a stack of 42.6 big blinds or less.
As a final thought, after the mark calls all-in and the cards are flipped over, they will likely realise that they are a severe underdog. This opens up the opportunity for making a deal and potentially dividing up the pot in a cash game, or deciding finishing positions and prize money splits in a tournament. Making a deal negates the risk of running the board and losing, for both players.
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