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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.

Jamie Carr
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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 $1$.

Ross Millikan
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    Agreed. We need only seek the probability for obtaining 2 of 13 cards in 1 from 4 suits, when selecting 2 from 52 cards. If you did want to include the remaining three cards in the numerator, you would also need the denominator to include selecting which 2 among all 5 cards would be the first. (Hence why they are off by a factor of 10.)$$\left.\binom 41\binom{13}2\binom{50}3 \middle/ \binom {52}5\binom 52\right. = \left.\binom 41\binom{13}2\middle/\binom{52}2\right.$$ – Graham Kemp Oct 14 '21 at 22:53
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    Alternatively, @JamieCarr : $\mathbb P(F\mid T)$ is the probability for obtaining 3 more cards from 11 remaining in the same suit as the first two (given that they are of the same suit), when selecting 3 cards from the 50 remaining in the deck. Thus you can check that your answer simplifies down to that.$$\left.\binom{11}3\middle/\binom{50}3\right.= 33/3920$$ – Graham Kemp Oct 14 '21 at 23:10
  • Thanks for your detailed answer! I'm writing an e-mail to the provider of original solution. – Jamie Carr Oct 14 '21 at 23:28
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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).

Gwendolyn Anderson
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  • I think the reason that this reasoning is wrong is because that for a given hand (5 cards) it must contain two cards of the same suit. So, it doesn't make sense to calculate the probability of it because it's automatically 1. – Jamie Carr Oct 14 '21 at 23:34
  • What about the first two cards? Rather than any two cards? – Gwendolyn Anderson Oct 15 '21 at 00:12