Probability that a 5-card poker hand will not contain any card less than 6, when it contains at least one card higher than 8

Question

The problem is:

Find the probability that a 5-card poker hand will not contain any card less than 6, when it contains at least one card higher than 8. (Assume aces have the highest-value cards)

This is my atempt:

Probability of containing 5 cards greather than or equal to 6 $P(A)=\frac{52-16\choose5}{52\choose5} = \frac{36\choose5}{52\choose5}$.

52 are all the cards from the deck and 16 are the cards which the values are less than 6.

And, the probability of at least one card being greather than 8 $P(B) = \frac{52-20\choose1}{52\choose1} = \frac{32\choose1}{52\choose1}$

Being 20 the cards greater than 8 (assuming the aces having the highest value).

So, the probability $P(A{\mid}B) = \frac{P(A^{\complement}B)}{P(B)}$

Solving this I couldn't achieve the answer. As my attempt is probably wrong I would like your help to solve the problem.

Answer 1

There are $\binom {52}5$ possible poker hands. There are $4\times 7=28$ cards $≤8$, hence there are $\binom {28}5$ possible poker hands that do not contain any card higher than $8$. Thus there are $$D= \binom {52}5-\binom {28}5=2,500,680$$ possible poker hands that contain at least one card $>8$. This will be our denominator.

To get the numerator: There are $4\times 4=16$ cards $<6$, hence there are $36$ cards $≥6$ and there are $\binom {36}5$ poker hands that contain only cards $≥6$. There are $12$ cards of value $6,7,8$, hence there are $\binom {12}5$ hands made up of cards that are all $≥6$ but $≤8$. Thus there are $$N=\binom {36}5-\binom {12}5=376,200 $$

possible poker hands that contain only cards $≥6$ which also contain at least one card $>8$. That is our numerator. Thus our answer is $$\frac ND = .15044$$

Sanity check: Ignoring the condition (which is very likely to be met) the probability that a random poker hand contains only cards $≥6$ is $$\binom {36}5 \Big / \binom {52}5=.145$$ Thus, as expected, the condition slightly increases the probability of the desired outcome.

Warning: while (I think) the method is sound, this sort of calculation is prone to arithmetic error. In particular it is prone to off-by-one errors (as I might have mangled the $<$ vs $≤$ choices which arise throughout). I suggest going through it carefully to check for blunders.

Answer 2

Guide:

Let $X$ denote the number of cards in the hand that have higher score than $5$ and let $Y$ denote the number of cards in the hand that have higher score than $8$.

Then to be found is:$$P\left(X=5\mid Y>0\right)$$

Make use of:

$$P\left(X=5\mid Y>0\right)=\frac{P\left(X=5,Y>0\right)}{P\left(Y>0\right)}=\frac{P\left(X=5\right)-P\left(X=5,Y=0\right)}{1-P\left(Y=0\right)}$$