<s>[INST] Probability of a 7 card sequence (same suit) in a 32 card game Deck of four suits of 8 cards each (32 cards total). Two Players with 12 cards each. 
The probability of 8 sequential cards is supposably:
$$\binom{4}{1} \times \frac{\binom{8}{8}\binom{24}{4}}{\binom{32}{12}} $$
What is the probability of getting 7 sequential cards? Is it:
$$ 2 \times \binom{4}{1} \times \frac{\binom{8}{7}\binom{1}{0}\binom{23}{5}}{\binom{32}{12}} $$
 [/INST] There are four suits, with 8 cards each. 
If one player has 12 cards out of those 32, then the probability is indeed 
$\frac{\color{red}{4\cdot 1 } \cdot \color{green}{ \binom{24}{4}} }{\color{blue}{\binom{32}{12}}}$, where $\color{red}{4\cdot 1}$ is the number of ways to choose a sequence of 8 colors, $\color{green}{ \binom{24}{4}}$ the number of ways to choose the remaining cards and $\color{blue}{\binom{32}{12}}$ the total number of ways to draw 12 cards. 

However, if we take into account that there two players, the probability becomes:
$$\large{\frac{\color{red}{4\cdot 1 } \cdot \color{green}{ \binom{24}{4}} \cdot \color{green}{ \binom{20}{12}} \, + \color{red}{4\cdot 1 } \cdot \color{green}{ \binom{24}{4}} \cdot \color{green}{ \binom{20}{12}} \, - \color{red}{12\cdot 1 } \cdot \color{green}{ \binom{16}{4}} \cdot \color{green}{ \binom{12}{4}}  }{\color{blue}{\binom{32}{12}\binom{20}{12}}}}$$
where $\large{\color{red}{4\cdot 1}}$ is the number of ways to choose a sequence of 8 colors, which is $\large{\color{red}{12\cdot 1}}$ for the probability that both have a 8 card sequence (4&3 possibilities for the suits), $\color{green}{ \binom{24}{4}\binom{20}{12}} $ the number of ways to choose the remaining cards  for one of the players and all cards for the other player, and $\color{blue}{\binom{32}{12}\binom{20}{12}}$ the total number of ways to draw 12 cards for two players. We substract the probability that both players have a sequence of length 8.

As Doug M pointed out, you made a small mistake in the seven card case. He explains the one player case, so here is the two player case:
$$\large{\frac{\color{red}{4\cdot 2 } \cdot \color{green}{ \binom{24}{5}} \cdot \color{green}{ \binom{21}{12}} \, + \color{red}{4\cdot 2 } \cdot \color{green}{ \binom{24}{5}} \cdot \color{green}{ \binom{21}{12}} \, - \color{red}{12\cdot 4 } \cdot \color{green}{ \binom{16}{5}} \cdot \color{green}{ \binom{11}{5}}  }{\color{blue}{\binom{32}{12}\binom{20}{12}}}}$$
Here, the red, green an blue numbers play the same role. 
So $\large{\color{red}{4\cdot 2}}$ is the number of ways to choose a sequence of 7 colors, which is $\large{\color{red}{12\cdot 4}}$ for the probability that both have a 7 card sequence (4&3 possibilities for the suits times 2 times 2 because they are sequential), $\color{green}{ \binom{24}{4}\binom{20}{12}} $ the number of ways to choose the remaining cards  for one of the players and all cards for the other player, and $\color{blue}{\binom{32}{12}\binom{20}{12}}$ the total number of ways to draw 12 cards for two players. We substract the probability that both players have a sequence of length 7.
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