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8 pokers hands are dealt from a shuffled deck without replacement.

a. Find the probability that at least one of the 8 hands is a heart flush(all five cards are hearts).

Pr(at least one of 8 hands is heart flush) $= 1 -$ Pr(none of eight hands is heart flush) $=$

$ \left( 8\frac{\binom{13}{5}}{\binom{52}{5}}-28\frac{\binom{13}{5}}{\binom{52}{5}}\frac{\binom{8}{5}}{\binom{47}{5}}\right)$

Is this answer correct?

b. Find the expected value and variance of the total number of eight hands which are heart flushes.

Expected value $= 8\dfrac{13 \choose 5}{52 \choose 5}$

Variance $= 8\dfrac{13 \choose 5}{52 \choose 5}\left(1- \dfrac{13 \choose 5}{52 \choose 5}\right)$

Are these answers correct?

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    Using the simple formula for variance, I obtained: 8($1$3 choose 5)/(52 choose 5) - [8(13 choose 5)/(52 choose 5)]^2. Is this correct?2011-09-25

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Let $A$ be the probability the first player has a heart flush. Then $A=\frac{13 \choose 5}{52 \choose 5}.$

Let $B$ be the probability the first and the second player both have heart flushes. Then $B=\frac{13 \choose 10}{52 \choose 10}.$

The probability at least one player has a heart flush is then ${8 \choose 1}A-{8 \choose 2}B=8A-28B$ which is in effect what you have written for question (a).

The probability of zero heart flushes is $1-8A+28B$, of one $8A-56B$ and of two $28B$.

Your expected value of $8A$ in (b) is correct. You can see this directly, or as $0\times(1-8A+28B) +1\times(8A-56B)+2\times 28B$.

Your variance of $8A(1-A)$ or $8A(1-8A)$ is slightly wrong. You could work out the variance to be $0^2\times(1-8A+28B) +1^2\times(8A-56B)+2^2\times 28B - (8A)^2 = 8A(1-8A)+56B$.

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    @sasha: Would you have been happier with $\dfrac{{13 \choose 5}{8\choose 5}}{{52 \choose 5}{47 \choose 5}}$? It is the same as $\dfrac{{13 \choose 10}}{{52 \choose 10}}$ and they are both $\dfrac{1}{252}$. The reason they are the same is that they both correspond to $\displaystyle\prod_{n=0}^{9}\dfrac{13-n }{52-n}$2011-09-27