Conditional Probability with Poker Hands

Question

I am trying to answer this question:

Draw $5$ cards at random from a standard deck of $52$ cards. Find:

$\text{a) } P(\text{full house} \mid \text{full-house or three-of-a-kind}) $

$\text{b) }P(\text{full house} \mid \text{at least two aces})$.

$\text{c) }P(\text{at least two aces} \mid \text{full-house}) \\$.

These are my relevant equations I'm working with:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

I'm having difficulty calculating all of it, but specifically the intersection and union probabilities are hardest for me. Any help would be appreciated!

Edit: I fixed the equation mentioned, thanks. Also, would we say that $P(\text{Full house} \cup \text{Three-of-a-kind})$ = P(Full house) + P(Three-of-a-kind) - P(Full house $\cap$ Three-of-a-kind)? And then P(Full house $\cap$ Three-of-a-kind) = 0 since you can't have both these hands?

Answer 1

Find $\Pr(\text{full house} \mid \text{full house or three of a kind})$.

Our sample space consists of hand in which a full house or three of a kind is obtained.

Full house: We must choose a rank from which three cards are drawn, choose three cards of that rank, choose a rank from one of the remaining ranks from which two cards are drawn, and choose two cards of that rank. This can be done in $$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}$$ ways.

Three of a kind: We must choose a rank from which three cards are drawn, choose three cards of that rank, choose two of the remaining ranks for the singletons, and choose one card from each of those ranks, which can be done in $$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2$$

Another way of describing our sample space is that it contains hands in which there is one rank from which exactly three cards are drawn. Thus, we must choose that rank, choose three cards from that rank, and choose two cards from the remaining ranks, which can be done in $$\binom{13}{1}\binom{4}{3}\binom{48}{2}$$ ways.

As you can verify, $$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2} + \binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2 = \binom{13}{1}\binom{4}{3}\binom{48}{2}$$

Therefore, \begin{align*} \Pr(\text{full house} \mid \text{full house or three of a kind}) & = \frac{\Pr(\text{full house})}{\Pr(\text{full house or three of a kind})}\\ & = \frac{|\text{full house}|}{|\text{full house or three of a kind}|}\\ & = \frac{\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{1}\dbinom{4}{2}}{\dbinom{13}{1}\dbinom{4}{3}\dbinom{48}{2}} \end{align*} where $$\frac{\Pr(\text{full house})}{\Pr(\text{full house or three of a kind})} = \frac{|\text{full house}|}{|\text{full house or three of a kind}|}$$ because the denominator for both probabilities on the left-hand side of the expression is $\binom{52}{5}$, so it cancels out.

Since a full house and a three of a kind are mutually exclusive, we can simply add their probabilities, as you determined.

Find $\Pr(\text{full house} \mid \text{at least two aces})$.

The event at least two aces can occur in three mutually exclusive ways:

1. Exactly two aces and three other cards from the deck.
2. Exactly three aces and two other cards from the deck.
3. Exactly four aces and one other card from the deck.

Since our sample space consists of those hands with at least two aces, there are two favorable cases:

1. Two aces and three cards of another rank.
2. Three aces and two cards of another rank.

Find $\Pr(\text{at least two aces} \mid \text{full house})$.

Our sample space consists of those hands containing a full house.

Our favorable cases are those full houses containing:

1. Two aces and three cards of another rank.
2. Three aces and two cards of another rank.

I leave the calculations for the second and third questions to you.