# Calculating combinations of a straight in Texas Hold 'Em

Similarly to a previous question that I've asked, I have now attempted to calculate the amount of all possible seven card combinations that would result in a straight (and nothing better).

The Wikipedia article about poker probabilities says, that the frequency of a straight in a 7-card poker game is $$6180020$$.

I've tried replicating the steps of calculating combinations of three of a kind except this time for a straight, but was unable to get $$6180020$$ as my final answer. This is what I've tried so far:

$$10 \cdot 4^{5}-10 \cdot 4=10200$$

$$10 \cdot 4^{5}$$ gets me all possible straights, minus $$10 \cdot 4$$ which gets rid of all straight flushes.

I then proceed to split $$10200$$ into two parts - $$1020$$ straights with the highest card being ace and $$9180$$ other combinations. Where ace is the highest card I would assume it doesn't matter what two cards I pick - resulting in:

$$1020 \cdot \binom{52-5}{2}=1102620$$

For other combinations I don't want to pick a cards with rank that could follow my straight meaning the formula will be:

$$9180 \cdot \binom{52-5-4}{2}=8289540$$

Adding them up: $$1102620+8289540=9392160$$

I would assume the next step is to subtract all the flushes out of $$9392160$$ however I realize that some straights will be counted multiple times if one of two cards after the straight have the same rank like a card in the straight (for example straight 2♥,3♥,4♥,5♦,6♦ + 3♦,A♣ and straight 2♥,3♦,4♥,5♦,6♦ + 3♥,A♣ will be both counted as two different combinations even though the cards are identical). Even after subtracting all the flushes the amount of combinations will be too large.

How can I proceed from here to arrive at the correct amounut of combinations, namely $$6180020$$?

• The word frequency was taken directly from the wikipedia article I linked. In 5-Card combinations, you would have 4 possible royal flushes. That $4$ appears in the Frequency column (which I assume is the same thing as number of occurrences). – Mantas Kandratavicius Jan 23 at 12:07
• Another issue is the possibility that there are six or seven consecutive ranks, as a six-card straight consists of two five-card straights and a seven-card straight counts as three five-card straights. – N. F. Taussig Jan 23 at 13:40

## 1 Answer

First, ignoring all flushes (we do not need to distinguish between flushes and straight flushes), we will consider each of the other hand ranks that can be concurrent with a straight:

• No Pair
• One Pair
• Two Pair
• Three-of-a-Kind

With no pair, we can have a $$7$$-card straight ($$8$$ possible), a $$6$$-card straight with disjoint rank for the $$7$$th card ($$47$$ possible), or a $$5$$-card straight with distinct disjoint ranks for the $$6$$th and $$7$$th card ($$162$$ possible). In total we have $$8+47+162=\color{blue}{217}$$ possible $$7$$-card hands that contain a straight and no pair. For each of these, there are $$4^7=16384$$ ways to choose the suits. Counting the number of flushes, we find $$1$$ way to have $$7$$ cards in suit, $$\binom76\cdot3=21$$ ways to have $$6$$ cards in suit, and $$\binom75\cdot3^2=189$$ ways to have $$5$$ cards in suit, for a total of $$211\cdot4=844$$ flushes. This leaves $$16384-844=\color{blue}{15540}$$ non-flushes.

$$15540\times217=3372180\text{ straights with no pair}$$

With one pair, we can have a $$6$$-card straight (9 possible) or a $$5$$-card straight with a disjoint rank for the $$6$$th card (62 possible). For each, there are $$6$$ choices for the paired card, for a total of $$71\cdot6=\color{blue}{426}$$ possible $$7$$-card hands that contain a straight and one pair. For each of these, there are $$\binom42\cdot4^3=6144$$ ways to choose the suits. Counting the number of flushes, we find $$3$$ ways to have $$6$$ cards in suit and $$3+\binom54\cdot3^2=48$$ ways to have $$5$$ cards in suit, for a total of $$51\cdot4=204$$ flushes. This leaves $$6144-204=\color{blue}{5940}$$ non-flushes.

$$5940\times426=2530440\text{ straights with one pair}$$

With two pairs, we can only have a $$5$$-card straight (10 possible) with $$2$$ of the cards paired. There are a total of $$10\cdot\binom52=\color{blue}{100}$$ possible $$7$$-card hands that contain a straight and two pairs. For each of these, there are $$\binom42^2\cdot4^3=2304$$ ways to choose the suits. Counting the flushes, there are $$3^2=9$$ ways to choose the second suits for the paired cards, for a total of $$36$$ flushes. This leaves $$2304-36=\color{blue}{2268}$$ non-flushes.

$$2268\times100=226800\text{ straights with two pairs}$$

With three-of-a-kind, we can only have a $$5$$-card straight (10 possible) with one of the ranks repeated on the $$6$$th and $$7$$th card. There are a total of $$10\cdot5=\color{blue}{50}$$ possible $$7$$-card hands that contain a straight and two pairs. For each of these, there are $$4^5=1024$$ ways to choose the suits. Counting the flushes, for each suit there are $$3$$ ways to choose the missing suit for the set, for a total of $$12$$ flushes. This leaves $$1024-12=\color{blue}{1012}$$ non-flushes.

$$1012\times50=50600\text{ straights with three-of-a-kind}$$

Finally, adding these together give the total:

$$\begin{array}{r}3372180\\2530440\\226800\\+\quad50600\\\hline6180020\end{array}$$