In a variation of poker called Texas Hold' em players, each player receives two cards; the dealer also deals a total of five community cards in three phases: in the first phase the dealer deals the first three cards, in the remaining two phases the dealer deals one card. Many games will end before all five cards have been dealt, but regardless I'd like to calculate the total number of Texas Hold' em games that are played all the way to the end.
There are at least two and at most ten players, and the game is played with a standard deck of $52$ cards. If there are $p$ players, this means the total number of cards drawn from the deck by the end of a game is $2 \cdot p + 5$, which means the upper bound for the total number of possible poker games is $52 \choose{2 \cdot p + 5}$. For just two players, this yields around $3.679 \times 10^9$ games, while for ten players this number increases to a whopping $4.7755 \times 10^{14}$ games.
However, this is an overestimate, because the order of each pair of player cards does not matter. Neither does the order of the three cards in the first phase. So taking into account this duplication, I figured I should divide the total like this:
$$ \begin{align} T &= {52 \choose{2 \cdot p + 5}} \Leftrightarrow \\ T' &= \frac{T}{\left(2! \right)^p \cdot3!} \end{align} $$
But for some $p$ the division by $3!$ yields a non-integer number of games, which isn't possible. What am I doing wrong?
Your $T = {52 \choose{2 \cdot p + 5}}$ is incorrect (too small as it ignores order) and should be $T = \frac{52!}{(52- 2 \cdot p - 5)!}$
This is divisible by $(2!)^p \cdot 3!$ to give an integer