Probability of getting full house in five-card poker with cards missing

Question

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 $\frac{1}{13}$.

Answer 1

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.

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Answer 2

NOTE: This is just the answer for a), not for b).

a) If I draw 5 cards randomly what are the chances of getting a full house?

You took out $5$ cards, so there would be $47$ cards leftover in the deck.

There are $\binom{47}{5}=1533939$ ways to choose any $5$ cards out of a $47$-card deck.

Then I used casework to find the ways to get a full house:

Case 1: 2 is chosen three times

There are $3$ $2$'s in the modified deck, so there is only $1$ way to choose $3$ out of $3$.

Subcase 1.a: 3 is chosen 2 times

There are $3$ $3$'s in the modified deck, so there would be $\binom{3}{2}=3$ ways to get a full house this way.

Subcase 1.b: Another rank(minus the king) is chosen 2 times

There would be $4$ cards to choose from, and there are $13-3=10$ other ranks, so that means that there would be $\binom{4}{2}*10=60$ ways to get a full house this way.

In total that would be $60+3=63$ ways for Case 1.

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Case 2: 3 is chosen three times

This is essentially this same thing as Case 1 so the number of ways would stay the same. $63$ ways.

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Case 3: Another rank(other than 2, 3, and king) is chosen 3 times

There are $4$ cards to choose from, so that would make it $\binom{4}{3}=4$ ways to choose $3$ cards out of $4$.

Subcase 3.a: 2 is chosen 2 times

There are $3$ cards with rank $2$, so that means the number of ways to choose $2$ cards out of $3$ is $\binom{3}{2}=3$ ways for this subcase.

Subcase 3.b: 3 is chosen 2 times

Essentially the same as subcase 3.a so $3$ ways.

Subcase 3.c: Another rank(not 2, 3, king, or the rank chosen for this case) is chosen 2 times

There are $9$ ranks leftover. Each of these ranks have $4$ choices to choose from, and you are choosing $2$ out of the $4$, so the number of ways for this subcase is $\binom{4}{2}*9=54$.

In total, that would be $10*4*(3+3+54)=2400$ ways for this case.

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So, the final answer is: $$\frac{63+63+2400}{1533939}=\frac{2526}{1533939}\approx 0.0016467$$