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:
$$ 2 \times \binom{4}{1} \times \frac{\binom{8}{7}\binom{1}{0}\binom{23}{5}}{\binom{32}{12}} $$
 A: Not sure what you rationale for the ${1\choose 0}$ in the equation.  But since ${1\choose 0} = 1$ it has no influence.
the ${8\choose 7}$ is creating possibilities that don't apply.
If you have 7 cards in sequence (with 8 to choose from) there are only two possible sequence you might have.  (which is the 2 in the front of your equation)
but the ${8\choose 7} = 8$ is a suggestion that you have any 7 of the 8 cards in that color, which is not the case.
$\dfrac {{4\choose1}(2){24\choose 5}}{{32\choose 12}}$
A: 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.
A: 
The probability of 8 sequential cards is supposably:
  $$\binom{4}{1} \times \frac{\binom{8}{8}\binom{24}{4}}{\binom{32}{12}} $$

This is misleading you.   It is correct for the wrong reasons and that is throwing you off.
You have a deck of four suits of 8 cards each (32 cards total).   (PS: You should mention this at the start; you really should.)
Two players are dealt twelve cards.   The probability that one particular player draws a straight sequence of 8 cards of one suit is: $$\dfrac{\dbinom 41\dbinom {24} 4}{\dbinom{32}{12}}$$
Where we are counting: $\tbinom 4 1$ ways to select one of four suits for the eight cards and $\tbinom{24}{4}$ ways to select any four cards of any of the other suits, divided by the $\tbinom{32}{12}$ ways to select any 12 cards in the deck.
We don't actually count $\tbinom 8 8$ ways to select eight cards from eight cards in a suit.   Rather we count the $1$ way to select a straight of eight from a suit.   This has the same numerical value, but the logic behind the representation is in error.

So the probability that one particular player draws a straight of 7 cards from one suit is:
$$\dfrac{\dbinom 21~\dbinom 41 \dbinom {24} 5 - \dbinom{4}{1}\dbinom{24}{4}}{\dbinom{32}{12}}$$
As there are only two ways to draw a straight of seven cards from a suit of eight (by selecting the lowest card of the straight from a possibility of two).   However we must avoid over-counting the intersection — of drawing both possibilities by drawing a straight of eight — hence the subtracted term.   (The Principle of Inclusion and Exclusion should be familiar to you; if not, make it so.)

If you move on to calculating the probability that one particular player draws a straight of six, you also have to consider the possibility of drawing them in two suits.   (Because: twelve card hand.)

Also, these are the probabilities for draws by one particular player.   The probability that either one of the two players draws a straight of eight is:
$$\dfrac{\dbinom {2}{1}~\dbinom{4}{1}\dbinom{24}{4}\dbinom{20}{12}-\dbinom 4 1\dbinom{24}{4}\dbinom 3 1\dbinom{12}{4}}{\dbinom{32}{12}\dbinom{20}{12}}$$
Can you see what is being counted, and why?   Thence can you calculate the probability that either player draws a straight of seven cards?
