Problem:
What is the probability of getting a flush (5 cards of the same suit)
in 7 card poker?
Note: For purposes of this question, a straight flush or a royal flush
counts as a flush.
Answer:
Let $p$ be the probability we seek.
\begin{align*}
p &= \dfrac{ 4 { {13} \choose {5}}{ {47} \choose {2} }}
{ {{52} \choose {7}} } \\
p &= \dfrac{ 99 } { 2380 } \\
p &\doteq 0.0415966
\end{align*}
I have good reason to believe the answer is: $0.058511$. Where did I go wrong?
Best Answer
The probability of 5.82% is the probability of getting a flush or better. https://en.wikipedia.org/wiki/Poker_probability#7-card_poker_hands
The probability of getting a flush, excluding royal flush and straight flush, is 3.03%
\begin{aligned}&\left[{4 \choose 1}\times \left[{13 \choose 7}-217\right]\right]\\+&\left[{4 \choose 1}\times \left[{13 \choose 6}-71\right]\times 39\right]\\+&\left[{4 \choose 1}\times \left[{13 \choose 5}-10\right]\times {39 \choose 2}\right]\end{aligned}
The above equals 4,047,644. The derivation is here: http://people.math.sfu.ca/~alspach/comp20/
The probability of a royal flush is the following, divided by ${52 \choose 7}$,
$$ {4 \choose 1}{47 \choose 2}$$
The above corresponds to picking a suit, and then 2 arbitrary cards after selecting the 5 sequential ones that correspond to the royal flush.
The probability of a straight flush is the following, divided by ${52 \choose 7}$,
$$ {9 \choose 1}{4 \choose 1}{46 \choose 2}$$
The above corresponds to picking a suit, and a reference card different from ace to start the straight flush, followed by 2 arbitrary cards after selecting the 5 sequential ones that correspond to the straight flush.