How many possible poker hands with missing cards and too many of other cards?

Question

The problem:

You got a deck with 52 cards. Two cards are missing: heart 10 and 9. But you also got some extra cards: you got 2 diamond 7 and 6 to make up for the missing cards. This is what the deck looks like:

Hearts: $\hspace{2em}$ A K Q J $\hspace{1.5em}$ 8 7 6 5 4 3 2

Diamonds: $\hspace{0.7em}$ A K Q J 10 9 8 7 7 6 6 5 4 3 2

Spades: $\hspace{1.9em}$ A K Q J 10 9 8 7 6 5 4 3 2

Clubs: $\hspace{2.4em}$ A K Q J 10 9 8 7 6 5 4 3 2

What are the probability that you get a full house with this deck?

What I am thinking: There are a total of $\binom{52}{5}$ different hands, then I should subtract for the missing cards but also add the number of combinations due to the extra cards. I have watched some examples of full house combinations with a regular deck but this modified deck is throwing me off.

How do I face this kind of problem?

Answer 1

Strategy

There are six cases:

1. A full house drawn from two ranks containing four cards.
2. A full house drawn from one rank containing four cards and another rank containing five cards.
3. A full house drawn from one rank containing four cards and another rank containing three cards.
4. A full house drawn from the two ranks containing five cards.
5. A full house drawn from one rank containing five cards and another rank containing three cards.
6. A full house drawn from the two ranks containing three cards.

A full house drawn from two ranks containing four cards: There are $13 - 2 - 2 = 9$ such ranks. Choose the rank from which three cards are drawn, choose three of the four cards of that rank, choose the rank from which two cards are drawn, and choose two of the four cards of that rank.

$$\binom{9}{1}\binom{4}{3}\binom{8}{1}\binom{4}{2}$$

A full house drawn from one rank containing four cards and another rank containing five cards: This can occur in two ways: Three cards are drawn from a rank with four cards and two cards are drawn from a rank with five cards or two cards are drawn from a rank with four cards and three cards (two of which may be identical) are drawn from a rank with five cards.

For the case in which three cards are drawn from a rank with four cards and two cards are drawn from a rank with five cards, choose which of the ranks with four cards from which three cards will be drawn, choose three of the four cards of that rank, choose which of the two ranks with five cards from which two cards will be drawn, and choose two of the five cards of that rank.

For the case in which two cards are drawn from a rank with four cards and three cards are drawn from a rank with five cards, choose which of the ranks with four cards from which three cards will be drawn, choose two of the four cards of that rank, choose which of the two ranks with five cards from which two cards will be drawn, and choose three of the five cards of that rank.

$$\binom{9}{1}\binom{4}{3}\binom{2}{1}\binom{5}{2} + \binom{9}{1}\binom{4}{2}\binom{2}{1}\binom{5}{3}$$

A full house drawn from one rank containing four cards and another rank containing three cards: This can occur in two ways: Three cards are drawn from a rank with four cards and two cards are drawn from a rank with five cards or two cards are drawn from a rank with four cards and all three cards are drawn from a rank with three cards.

$$\binom{9}{1}\binom{4}{3}\binom{2}{1}\binom{3}{2} + \binom{9}{1}\binom{4}{2}\binom{2}{1}\binom{3}{3}$$

A full house is drawn from the two ranks containing five cards: Choose which of the two ranks containing five cards from which three cards will be drawn, choose three of the five cards of that rank, choose two of the five cards of the remaining rank.

$$\binom{2}{1}\binom{5}{3}\binom{5}{2}$$

A full house is drawn from one rank containing five cards and another rank containing three cards: There are two cases: Three cards are drawn from a rank with five cards and two cards are drawn from a rank with two cards or two cards are drawn from a rank with five cards and all three cards are drawn from a rank with three cards.

$$\binom{2}{1}\binom{5}{3}\binom{2}{1}\binom{3}{2} + \binom{2}{1}\binom{5}{2}\binom{2}{1}\binom{3}{3}$$

A full house is drawn from the two ranks containing three cards: Choose which of the two ranks containing three cards from which three cards will be drawn, choose all three cards of that rank, choose two of the three cards of the remaining rank.

$$\binom{2}{1}\binom{3}{3}\binom{3}{2}$$

Since the six cases are mutually exclusive and exhaustive, the number of favorable cases is found by adding the above amounts. Dividing by the $\binom{52}{5}$ ways of drawing five cards from the modified deck yields the probability of obtaining a full house.

Answer 2

Let a group of cards be composed from all cards that have the same rank.

Then there are $9$ groups of $4$, $2$ groups of $3$ and $2$ groups of $5$ cards.

Full house will appear if from a specific groups $3$ cards are drawn and form another specific group $2$ cards are drawn.

We discern the following possibilities:

• $3$ are drawn from a $4$-group and $2$ are drawn from another $4$-group.

• $3$ are drawn from a $4$-group and $2$ are drawn from a $3$-group.

• $3$ are drawn from a $4$-group and $2$ are drawn from a $5$-group.

• $3$ are drawn from a $3$-group and $2$ are drawn from a $4$-group.

• $3$ are drawn from a $3$-group and $2$ are drawn from the other $3$-group.

• $3$ are drawn from a $3$-group and $2$ are drawn from a $5$-group.

• $3$ are drawn from a $5$-group and $2$ are drawn from a $4$-group.

• $3$ are drawn from a $5$-group and $2$ are drawn from a $3$-group.

• $3$ are drawn from a $5$-group and $2$ are drawn from the other $5$-group.

For these cases we find in the same order the following number of possibilities:

• $\binom{9}{1}\binom{4}{3}\binom{8}{1}\binom{4}{2}$

• $\binom{9}{1}\binom{4}{3}\binom{2}{1}\binom{3}{2}$

• $\binom{9}{1}\binom{4}{3}\binom{2}{1}\binom{5}{2}$

• $\binom{2}{1}\binom{3}{3}\binom{9}{1}\binom{4}{2}$

• $\binom{2}{1}\binom{3}{3}\binom{1}{1}\binom{3}{2}$

• $\binom{2}{1}\binom{3}{3}\binom{2}{1}\binom{5}{2}$

• $\binom{2}{1}\binom{5}{3}\binom{9}{1}\binom{4}{2}$

• $\binom{2}{1}\binom{5}{3}\binom{2}{1}\binom{3}{2}$

• $\binom{2}{1}\binom{5}{3}\binom{1}{1}\binom{5}{2}$

Where the first factor corresponds with the number of ways a group can be chosen from which $3$ cards are drawn.

Second factor corresponds with the number of ways that $3$ cards can be drawn from this group.

Third factor corresponds with the number of ways a group can be chosen from which $2$ cards are drawn under condition that the group from which $3$ cards are drawn has been chosen already.

Fourth factor corresponds with the number of ways that $2$ cards can be drawn from that group.

The summation of these numbers divided by $\binom{52}5$ equals the probability of drawing a full house in this situation.