Probability of a flush given that the first two cards are of the same suit.

Question

there. I'm looking to write probability of a flush (suppose it contains royal flush, straight flush, etc.) given that the first two cards are of the same suit.

Here is the solution, let $F$ be the event that a flush appears, $T$ be the event that the first two are of the same suit. $$ \begin{aligned} \mathbb{P}(F|T) &=\frac{\mathbb{P}(T|F)\mathbb{P}(F)}{\mathbb{P}(T)} \\ &= \frac{\mathbb{P}(F)}{\mathbb{P}(T)} \\ &= \frac{{4\choose 1} {13\choose 5}/{52 \choose 5}}{{4 \choose1} {13 \choose 2}{50 \choose 3}/{52 \choose 5}} \end{aligned} $$

I agree with all the above calculation except the last probability $\mathbb{P} (T)$. I think the correct $\mathbb{P}(T)$ should be $$\frac{{4 \choose 1}{13\choose 2}}{{52 \choose 2}}$$ because the probability that there are two cards of the same suit in a hand is not equal to the probability that the first two cards are of the same suit.

Am I right or wrong? Any help is greatly appreciated.

Answer 1

You are correct. Your probability is $\frac 4{17}$, which makes sense in being a little less than $\frac 14$. The second card has to match the first in suit, which would be a chance of $\frac 14$ but there is one fewer of that suit available. Their calculation is $\frac {40}{17}$, which cannot be correct as it is greater than $1$.

Answer 2

All of the possible combinations of five-card hands is given by $52 \choose 5$. All of the possible combinations of the first two cards drawn is $52 \choose 2$. Note that $52 \choose 5$ = $2,598,960$ while $52 \choose 2$ $= 1,326$. Those are very different magnitudes. Also, there are $13 \choose 5$ $ = 1,287$ ways to construct a flush five-card hand from $13$ cards of the same suit, and only $13 \choose 2$ $ = 78$ combinations of two cards from $13$ of a particular suit. Notice that four possible suits - or $4 \choose 1$ suit selections - times $13 \choose 2$ gives 312 possible combinations of two matching suits in a two-card draw, encompassing all suits. If we want to say that the remaining three cards in the five card hand can come from any of the 50 remaining cards, $50 \choose 3 $$ = 19,600$. This result multiplied by $312$ would suggest $6,115,200$ card combinations which is a greater number than all possible five-card hands $52 \choose 5$ = $2,598,960$ so there must be something wrong with that reasoning. Is that what you saw? For one thing, I can't see why $50 \choose 3 $ would ever be multiplied by four (suits).