Missing cards from the deck: $\heartsuit3$, $\spadesuit$King, $\heartsuit$King, $\clubsuit$king and $\spadesuit2$
a) If I draw $5$ cards randomly what are the chances of getting a full house ?
b) Which other card can you remove to maximize your chances of getting full house?
I don't really understand how to solve it, for my understanding there are $13\diamondsuit$, $11\spadesuit$, $11\heartsuit$ and $12\spadesuit$ left in the deck. If randomly you pick a card out you have $\frac{1}{4}$ to draw $\frac{1}{11}, \frac{1}{11}, \frac{1}{13} \text{ or }\frac{1}{12}$. So depending on what suit you draw it goes from $\frac{1}{11}$ to $\frac{1}{13}$.
Best Answer
Clearly no full house can have a King. Other than the King, the cards fall into two types:
$A$: Cards with $4$ suits: there are ten of these.
$B$: Cards with $3$ suits: there are two of these (namely $2,3$).
We consider the various types of full houses (here, for instance type $(A,B)$ means that the three of a kind is of type $A$ and the pair is of type $B$).
Type $(A,A)$. There are $10$ ways to choose the rank for the triple, then $\binom 43$ ways to choose the triple. Then there are $9$ ways to choose the rank for the pair and $\binom 42$ ways to choose the pair. Thus $$10\times \binom 43\times 9 \times \binom 42=2160$$
Type $(A,B)$. There are $10$ ways to choose the rank for the triple, then $\binom 43$ ways to choose the triple. Then there are $2$ ways to choose the rank for the pair and $\binom 32$ ways to choose the pair. Thus $$10\times \binom 43\times 2 \times \binom 32=240$$
Type $(B,B)$. There are $2$ ways to choose the rank for the triple, then $\binom 33$ ways to choose the triple. Then there is $1$ way to choose the rank for the pair and $\binom 32$ ways to choose the pair. Thus $$2\times \binom 33\times 1 \times \binom 32=6$$
Type $(B,A)$.There are $2$ ways to choose the rank for the triple, then $\binom33$ ways to choose the triple. Then there are $10$ ways to choose the rank for the pair and $\binom 42$ ways to chpose the pair. Thus $$2\times \binom 33\times 10 \times \binom 42=120$$
Finally we sum to see that there are $2526$ possible full houses. As there are $\binom {47}5$ possible hands the answer is $$\boxed {\frac {2526}{\binom {47}5}=.001647}$$
To complete the problem, note that removing any card shrinks the denominator in the the same way (doesn't matter which you remove). However, if you remove the $\diamondsuit K$ then the numerator does not change at all, since no full house can have a King using this deck. As any other removal shrinks the numerator, we see that this is the best choice.
'