Introduction

This article will investigate different probabilities in Texas Hold 'Em (7-card poker), starting with probabilities of the different hands and then reviewing some common conditional probabilities. My motivation for writing this paper is that resources regarding the combinatoric derivations of probabilities in Texas Hold 'Em is scarce. For those not familiar with the binomial coefficient $\binom{n}{k}$, its definition is as follows.

\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

This equation describes the number of ways to choose \( k \) objects from a set of \( n \) distinct objects without replacement and where order does not matter. To see why, consider first selecting \( k \) objects in order: there are \( n \) choices for the first object, \( n - 1 \) for the second, and so on, down to \( n - k + 1 \) choices for the \( k \)-th object. This gives a total of \( \frac{n!}{(n-k)!} \) ordered selections. However, since the order of selection does not matter, we must divide by the number of ways to order the \( k \) selected objects, which is \( k! \).

Hand Probabilities

I will order the hands in this article from strongest to weakest, as the derivations tend to increase in difficulty the more common the poker hand. Since there are 52 cards in a standard deck and a "hand" in Texas Hold 'Em is 7 random cards from a standard deck, that means there are a total of $\binom{52}{7}$ different possible hands in 7 card poker.

Royal Flush

A royal flush is the rarest hand in poker and is obtained when a player has the cards 10, J, Q, K, A (in no particular order) all of the same suit. There are 4 possible suits to choose from and $\binom{47}{2}$ ways to arrange the leftover two cards. This means there are

\[4 \binom{47}{2} = 4324\]

ways of obtaining a royal flush with a probability

\[P(\text{RF}) = \frac{4 \binom{47}{2}}{\binom{52}{7}} = 0.0032\%\]

or odds of 30,940:1.

Straight Flush

There are 10 different card sequences that are considered straights in poker, A-5 through 10-A, making 9 possible straight flush sequences when we exclude the royal flush. However, if we consider all $\binom{47}{2}$ hands that are possible for each straight flush, we will end up double counting six-card and seven-card straights. To circumvent this, we simply remove one card from each hand. Observe with an example (where n represents any possible non-repeating card).

\[\begin{pmatrix} 4_s & 5_s & 6_s & 7_s & 8_s & 9_s & n \end{pmatrix} \text{would be counted again in the next straight,} \begin{pmatrix} 5_s & 6_s & 7_s & 8_s & 9_s & 4_s & n \end{pmatrix}\]

By only removing the rank \emph{above} the given straight flush sequence, we ensure that all possible 6 and 7 card sequences are included only once. Now there are

\[\begin{pmatrix} 9 (4) \binom{46}{2} = 37,260 \end{pmatrix}\]

ways of obtaining a straight flush with a probability

\[\begin{pmatrix} P(\text{SF}) = \frac{9 4 \binom{46}{2}}{\binom{52}{7}} = 0.0309\% \end{pmatrix}\]

or odds of 3,590:1.

Four of a Kind

A four-of-a-kind is a hand that contains all four of the same card (i.e. four jacks). There are 13 ways to select a rank and $\binom{48}{3}$ ways of selecting the remaining cards. Therefore there are

\[\begin{pmatrix} 13 \binom{48}{3} = 224,848 \end{pmatrix}\]

ways of obtaining a four of a kind with a probability of

\[\begin{pmatrix} P(\text{FK}) = \frac{13 \binom{48}{3}}{\binom{52}{7}} = 0.1681 \% \end{pmatrix}\]

Full House

The calculation slightly increases in difficulty here. First, let's consider the number of full house hands that contain 2 three of a kind. There are $\binom{13}{2}$ ways to select the ranks, $\binom{4}{3}$ ways to arrange each three of a kind, and 44 cards left to choose from without creating a four of a kind.

\[\begin{pmatrix} \binom{13}{2} \binom{4}{3}^2 (44) \end{pmatrix}\]

Next, let's consider the number of full house hands that contain a three of a kind and two pair. There are 13 ways to select the triplet rank, $\binom{4}{3}$ ways to arrange the three of a kind, $\binom{12}{2}$ ways to select the pair ranks, and $\binom{4}{2}$ ways to arrange each pair.

\[\begin{pmatrix} 13 \binom{4}{3} \binom{12}{2} \binom{4}{2}^2 \end{pmatrix}\]

For the remaining standard full house hands, there are $13 \binom{4}{3}$ ways to choose the triplet, $12 \binom{4}{2}$ ways to choose the pair, and $\binom{11}{2} 4^2$ ways to choose the other two cards (we don't want to include another pair in this hand, so we choose a unique combination of the remaining 11 ranks, with 4 suits for each rank). Putting this all together and the number of ways to create a full house is

\[\begin{pmatrix} \binom{13}{2}\binom{4}{3}^2 (44) + 13 \binom{4}{3} \binom{12}{2} \binom{4}{2}^2 + 13 \binom{4}{3} (12) \binom{4}{2} \binom{11}{2} (4^2) = 3,473,184 \end{pmatrix}\]

With a probability of

\[\begin{pmatrix} P(\text{FH}) = \frac{3,473,184}{\binom{52}{7}} = 2.596 \% \end{pmatrix}\]

Flush

A flush is a hand that contains five or more cards of the same suit. First we will consider the number of flushes with 7 cards of the same suit. There are $\binom{13}{7}$ ways to choose the ranks and 4 ways to choose the suit. For the number of flushes with 6 cards of the same suit, there are $\binom{13}{6}$ ways to choose the ranks, again 4 ways to choose the suit, and 39 ways to choose the remaining card (it must be of a different suit than the other 6 cards). Finally, for the number of flushes with 5 cards of the same suit there are $\binom{13}{5}$ ways to choose the ranks, 4 ways to choose the suit, and $\binom{39}{2}$ ways to choose the remaining two cards. Adding this together and subtracting the number of straight flushes we get

\[\begin{pmatrix} 4 \left( \binom{13}{5}\binom{39}{2} + \binom{13}{6}(39) + \binom{13}{7} - 9\binom{46}{2} - \binom{47}{2}\right) = 4,047,644 \end{pmatrix}\]

With a probability of

\[\begin{pmatrix} P(\text{F}) = \frac{4,047,644}{\binom{52}{7}} = 3.025\% \end{pmatrix}\]

Straight

In my opinion, this is the most difficult 7-card poker hand combinatorics problem, as the simplest approach to the problem isn't quite obvious. Excluding the intersection of straights and flushes also requires some careful thinking. Let's start by naively setting up the problem. There are 10 possible straight-sequences, each with 5 cards, each card with 4 suits. There are 47 leftover cards to choose from for each hand. Therefore we have $10( 4^5) \binom{47}{2}$ possible straight hands according to our naive calculation.

What did we double count here? If you read the straight flush section, you might realize that we are double counting the possible 6-straights and 7-straights.

\[\begin{pmatrix} 6_{h} & 7_{s} & 8_{c} & 9_{h} & 10_{d} & \text{J}_s & n \end{pmatrix} \text{\&} \begin{pmatrix} 7_{s} & 8_{c} & 9_{h} & 10_{d} & \text{J}_s & 6_h & n \end{pmatrix}\]

Another sneaky hand form that we need to avoid counting multiple times are those that contain a pair, two pair, and three of a kind of a rank contained in the straight.

\[\begin{pmatrix} 6_{h} & 7_{s} & 8_{c} & 9_{h} & 10_{d} & \text{6}_s & n \end{pmatrix} \text{\&} \begin{pmatrix} 6_{s} & 7_{s} & 8_{c} & 9_{h} & 10_{d} & 6_h & n \end{pmatrix}\]

Therefore we must be careful when setting up the problem. Let's start with the number of ways that there is to form a straight with no pairs. There are 9 sequences of which we can choose $\binom{7}{2}$ remaining ranks, and one sequence from which we can choose any of the $\binom{8}{2}$ remaining ranks. Since all cards are of a different rank, there are $4^7$ possible ways to arrange the suits.

\[\begin{pmatrix} 4^7(9\binom{7}{2} + \binom{8}{2}) \end{pmatrix}\]

The above equation is the total number of straights with no coupled ranks. But how do we subtract the number of flushes? Well, there are 4 ways to select all 7 ranks from the same suit, 7(4)(3) ways for only one rank to be of a different suit (7 ways to choose the different card, 4 ways to choose the suit of the same cards, and 3 ways to choose the suit of the different card), and $\binom{7}{2}(4)(3^2)$ ways to choose strictly 5 ranks of the same suit. Since these equations describe choosing suits and not ranks, we subtract this from the $4^7$ factor to get

\[\begin{pmatrix} \left(4^7 -4 - 7(4)(3) - \binom{7}{2}(4)(3^2)\right)\left(9\binom{7}{2}+\binom{8}{2}\right) \end{pmatrix}\]

number of straights with all distinct ranks.

As for hands that contain exactly one pair, there is only room for one rank outside of the given straight. For 9 of the straight sequences, there are 7 remaining ranks to choose from, and 8 ranks to choose from for the other sequence. For each hand arrangement, there are 6 distinct ranks that we can choose from to make the pair. As for suit combinations, there are $4^5$ ways to select suits for the non-coupled ranks and $\binom{4}{2}$ ways to choose suits for the pair.

\[\begin{pmatrix} 4^5\binom{4}{2}(6)(9(7)+8) \end{pmatrix}\]

To exclude flushes, note that there are 4 ways for the suits of the non-coupled ranks to be the same, with any $\binom{4}{2}$ combination of pair suits being a flush. There are 5(4)(3) for one suit to be different in the non-coupled ranks, with three possible arrangements of suits in the pair to produce a flush. Now we can update the pair equation to exclude flushes:

\[\begin{pmatrix} \left(4^5\binom{4}{2} - 4\binom{4}{2} - 5(4)(3^2)\right)\left(9(6)(7) + 6(8)\right) \end{pmatrix}\]

For hands that contain two pair, there are $\binom{5}{2}$ ways to choose the 2 ranks, $\binom{4}{2}^2$ ways to create the two pair and $4^3$ ways to choose the remaining suits. All 10 possible straight sequences obey the same rule. To exclude flushes, all three non-coupled ranks must be the same suit (4 choices) and for each pair, there are 3 possible arrangements that allow for a flush.

\[\begin{pmatrix} \left( 4^3 \binom{4}{2}^2 - 4(3^2) \right)\binom{5}{2}(10) \end{pmatrix}\]

Finally, for three of a kind, there are 5 ranks to choose from, $\binom{4}{3} = 4$ ways to create the three of a kind, and $4^4$ ways to choose the remaining suits. For each $4^4$ combination of non-coupled suits, there are only 3 ways to create a flush.

\[\begin{pmatrix} (4^4 -3)\binom{4}{3}(5)(10) \end{pmatrix}\]

Summing all of this together and applying some algebra we get the following equation for the number of possible straights in 7-card poker.

\[\begin{pmatrix} (4^7 - 844)(217) + (4^5-34)(6^2)(71) + (4^4-4)(3^2)(10^2) + (4^4-3)(4)(50) = 6,180,020 \end{pmatrix}\]

The probability is thus

\[\begin{pmatrix} P(\text{S}) = \frac{6,180,020}{\binom{52}{7}} = 4.619 \% \end{pmatrix}\]

Three of a Kind

The only tricky part of this combinatorics problem is removing the 3 of a kind hands that are flushes and straights--and we have already done some of this work in the previous section. For a general 3 of a kind hand with no pairs (as this would make it a full house) and no four of a kind, there are 13 ways to choose the rank, $\binom{4}{3}$ ways to arrange the suits, $\binom{12}{4}$ ways to choose the remaining 4 ranks and $4^4$ ways to choose the suits of those ranks.

\[\begin{pmatrix} 13\binom{4}{3} \binom{12}{4} (4^4) \end{pmatrix}\]

Looking at a 3 of a kind example, we can see that for every combination of suits for the 4 non-coupled ranks, there are exactly 3 that produce a flush.

\[\begin{pmatrix} 5_{h} & 5_{s} & 5_{c} & 6_{h} & 9_{h} & \text{10}_h & 3_h \end{pmatrix}\]

Therefore we adjust the $4^4$ term to be $(4^4-3)$. As for straights, we know from the straight section that the number of straight and three of a kind intersections is $5\binom{4}{3}(10)(4^4)$. Therefore we subtract a term of $10(5)(4)(4^4-3)$ so as not to subtract the number of straight $\cap$ flush $\cap$ three of a kind hands more than once. Our equation for the number of three of a kind hands is thus

\[\begin{pmatrix} (4^4-3)(13\binom{12}{4}\binom{4}{3} - 10(5)(4)) = 6,461,620 \end{pmatrix}\]

With a probability of

\[\begin{pmatrix} P(\text{TK}) = \frac{6,461,620}{\binom{52}{7}} = 4.830 \% \end{pmatrix}\]

Two Pair

For hands that contain exactly two pairs, there are $\binom{13}{2}$ ways to select the ranks for the pairs, $\binom{11}{3}$ ways to choose the ranks for the remaining cards, $\binom{4}{2}^2$ ways to arrange the suits for the pairs, and $4^3$ ways to choose the suits for the remaining three cards.

\[\begin{pmatrix} \binom{13}{2}\binom{11}{3}\binom{4}{2}^2(4^3) \end{pmatrix}\]

For a two pair to contain a flush, all 3 suits of the non-coupled ranks must be the same (there are 4 options). For each of the pairs, one of the suits must match the suit of the non-coupled ranks, with 3 options to choose from for the other card. This means we adjust our $\binom{4}{2}^2(4^3)$ factor to be $\binom{4}{2}^2(4^3) - 4(3^2)$. For a two pair to contain a straight, we can refer to the straight subsection and note that there are $\left( 4^3 \binom{4}{2}^2 - 4(3^2) \right)\binom{5}{2}(10)$ possible hands that contain both a straight and a two pair.

\[\begin{pmatrix} \left( 4^3 \binom{4}{2}^2 - 4(3^2) \right)\left(\binom{13}{2}\binom{11}{3}-\binom{5}{2}(10)\right) \end{pmatrix}\]

However, we are not done. An important consideration is that hands which contain 3 pairs are counted as a two pair hand in the game of Texas Hold 'Em. There are $\binom{13}{3}$ ways to choose the ranks for the three pairs, $\binom{4}{2}^3$ ways to arrange the suits, and $10(4)$ remaining cards to choose from. We do not have to worry about counting flushes or straights here, as those configurations are not possible in a 3 pair hand. Adding this to our previous equation and we have

\[\begin{pmatrix} (4^3(6^2) - 36)\left(\binom{13}{2}\binom{11}{3} - (10^2)\right) + \binom{13}{3}(6^3)(40) = 31,433,400 \end{pmatrix}\]

ways to obtain a two pair with a probability of

\[\begin{pmatrix} P(\text{TP}) = \frac{31,433,400}{\binom{52}{7}} \end{pmatrix}\]

Pair

There are 13 ways to choose the rank for the pair, $\binom{12}{5}$ ways to choose the remaining ranks, $\binom{4}{2}$ ways to select suits for the pair, and $4^5$ ways to select suits for the remaining cards.

\[\begin{pmatrix} 13\binom{12}{5}\binom{4}{2}(4^5) \end{pmatrix}\]

Referring to the straight section, we can exclude hands that are flushes by correcting the factor of $\binom{4}{2}(4^5)$ to be $\left(4^5\binom{4}{2} - 4\binom{4}{2} - 5(4)(3^2)\right)$. To exclude straights, we simply subtract the number of straights that are also pairs $\left(4^5\binom{4}{2} - 4\binom{4}{2} - 5(4)(3^2)\right)\left(9(6)(7) + 6(8)\right)$. The number of total hands that contain at best a pair is

\[\begin{pmatrix} \left(13 \binom{12}{5} - 9(6)(7) - 6(8) \right) \left( 4^5(6) -4(6) - 5(4)(3^2) \right) = 58,627,800 \end{pmatrix}\]

with a probability of

\[\begin{pmatrix} P(\text{P}) = \frac{58,627,800}{\binom{52}{7}} = \end{pmatrix}\]

High Card

There are $\binom{13}{7}$ ways to choose 7 distinct ranks and $4^7$ ways to select suits.

\[\begin{pmatrix} \binom{13}{7}(4^7) \end{pmatrix}\]

For each selection of ranks, there are $4 + 7(4)(3) + \binom{7}{2}(4)(3^2)$ ways to create a flush (number of 7 card flushes + 6 card flushes + 5 card flushes). Therefore, we adjust our factor of $4^7$ to be $4^7 - 4-7(4)(3)-\binom{7}{2}(4)(3^2) = 4^7 - 844$. From the straight section, we can see that there are $9\binom{7}{2} + \binom{8}{2} = 217$ ways to select ranks for a straight with no pairs. Now we have our equation for the number of hands with at best a high card.

\[\begin{pmatrix} \left( \binom{13}{7} - 217\right) (4^7-844) = 23,294,460 \end{pmatrix}\]

Interestingly enough, you are more likely to obtain a hand that contains a pair or two pair than you are to have at best a high card. The probability of obtaining a high card hand is

\[\begin{pmatrix} P(\text{HC}) = \frac{23,294,460 }{\binom{52}{7}} = 17.412 \% \end{pmatrix}\]