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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 $6180020$?

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  • $\begingroup$ 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). $\endgroup$ – Mantas Kandratavicius Jan 23 at 12:07
  • $\begingroup$ 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. $\endgroup$ – N. F. Taussig Jan 23 at 13:40
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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:

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

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