# poker hand probability question

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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Let $\mathcal{A}_i$ be the event that the $i$th hand is heart flush. In your answer to question a. you assume that the events $\mathcal{A}^c_i$ are independent. Are they? – Ben Derrett Sep 24 '11 at 17:36
If it was $8$ poker hands from $8$ shuffled decks, (a) would be right. But single deck, no. Expected value is right because of linearity of mean. Variance is wrong, if it is a single deck. I suggest inclusion/exclusion for (a). – André Nicolas Sep 24 '11 at 18:02
inclusion/exclusion is quite tricky since you have to consider all possible subsets. – lord12 Sep 24 '11 at 20:04
Not too bad, because there are at most $2$ heart flushes. Let $H_i$ be event there is heart flush on the $i$-th hand. You know $P(H_i)$. To find probability of at least one heart flush, add up the $P(H_i)$ (you did) and subtract sum of $P(H_i\cap H_j)$ where ($i\ne j$). There are $28$ pairs. For any one, say $i=1$, $j=6$, the work is not hard: $5$ hearts in hand $1$ (you know how to do that) followed by $5$ hearts in hand $6$ (easy, dealing from $47$ card deck that has $8$ hearts). The total correction term is a lot smaller than the $\sum P(H_i)$ you have already calculated. – André Nicolas Sep 24 '11 at 20:34
I edit the above answer for part a and was wondering if it was correct. – lord12 Sep 25 '11 at 0:10

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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This might not be the place for my comment, but I'll leave it here anyway. I can't convince myself that your $13 \choose 10$ is the number of ways of getting two heart flushes. It seems that in every one of those 10-card combinations, there are $\frac{1}{2}\cdot {10 \choose 2}$ ways of making two flushes. – sasha Sep 26 '11 at 19:06
I'm still not clear how you got the variance. – lord12 Sep 26 '11 at 23:00
@lord12: I used $var(X) = E[X^2] - (E[X])^2$ and $E[X^2]=\sum_i x_i^2 \Pr(X=x_i)$. – Henry Sep 27 '11 at 0:04
@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 $\displaystyle\prod_{n=0}^{9}\dfrac{13-n }{52-n}$ – Henry Sep 27 '11 at 0:12