Probability of picking 6 cards of the same suit and colour I have recently been studying probability by working through a textbook however some of the steps outlined dont entirely make sense to me.
The question: "Six cards are dealt from a well shuffled deck of cards, if all six cards dealt are black what is the probability of all six being from the same suit".
So far I understand that in the deck there are 26 cards black and 26 red and ask such I would expect:
$$nCr = 2 \cdot 13C6 $$
This part I understand as it should result in 6 combinations from 26 cards. The next 2 steps confuses me:
$$2 \cdot \frac{13C6}{52C6}$$
1. Whilst I understand that it is 2*13 cards with six combinations, i do not understand why you must divide by the number of combinations possible in the entire deck....
The next step i also cannot follow:
$$2 \cdot \frac{13C6 }{52C6} \cdot  \frac{52C6 }{26C6}  $$ 
2. Why must the result of finding the number of combinations of black cards / number of combinations in the entire deck be multiplied by the number of cominbations possible again divided by all combinations of black cards?
Whilst I can understand the pattern and re-use it Im struggling with the logic of why it is done.
 A: If $A=\{\text{all six being from the same suit}\}$ and $B=\{\text{all six cards dealt are black}\}$, then you need to calculate conditional probability
$$\tag{1}\label{1}
\mathbb P(A\mid B)=\dfrac{\mathbb P(A\cap B)}{\mathbb P(B)}.
$$
The right hand side is the definition of conditional probability. 
To calculate $\mathbb P(B)$, one should divide the number of combinations of six black cards $\binom{26}{6}$ to the number of all possible conbinations of six cards from the whole deck $\binom{52}{6}$: 
$$\mathbb P(B)=\dfrac{\binom{26}{6}}{\binom{52}{6}}.$$
The event $A\cap B$ means that six cards are black and have the same suit. There are two possibilities to have the same suit: six cards from clubs or spades. Both cases have $\binom{13}{6}$ combinations. Probability of $A\cap B$ is then the ratio
$$
\mathbb P(A\cap B)=\dfrac{\binom{13}{6}+\binom{13}{6}}{\binom{52}{6}}.
$$
And substitute both $\mathbb P(A\cap B)$ and $\mathbb P(B)$ into (\ref{1}), get the solution that you found in a textbook.
You see from this solution that the desired probability can be obtained simply as a ratio of total number of combinations of six cards from clubs or spades and the number of combinations of six black cards:
$$
\mathbb P(A\mid B) = \dfrac{2\binom{13}{6}}{\binom{26}{6}}.
$$
A: The question can be rephrased as: what is the probability that all six cards are the same suit, given that they're all black. 
Let $S$ be the event that all the cards are the same suit, and $B$ be the event that they're all black. Provided that $P(B) > 0$, we have
$$P(S \mid B) = \frac{P(B \mid S)P(S)}{P(B)}.$$
If all of the cards are the same suit, they're either all red or all black (each case equally likely), so $P(B \mid S)$ is just $1/2$.
The probability that they're all the same suit is (the number of same-suited hands) divided by (the number of total hands possible), or
$$\frac{4\binom{13}{6}}{\binom{52}{6}}.$$
The probability that all six cards are black is (the number of all-black hands) divided by (the number of total hands possible), or
$$\frac{\binom{26}{6}}{\binom{52}{6}}.$$
Putting these facts together, we have
$$P(S \mid B) = \left(\frac{1}{2}\right) \cdot \left(\frac{4\binom{13}{6}}{\binom{52}{6}}\right) \cdot \left(\frac{\binom{52}{6}}{\binom{26}{6}}\right),$$
which simplifies to the answer you mentioned.
