A poker deck has $52$ cards, $13$ ranks ${A,2,...,10,J,Q,K}$ and $4$ suits {diamond, club, heart, spade}.

Question

A poker deck has $52$ cards, $13$ ranks $\{A,2,...,10,J,Q,K\}$ and $4$ suits {diamond, club, heart, spade}. If $5$ cards are dealt, what is the probability that there is an ace of diamonds?

Here is my thinking: event over the sample space, which is event of getting (ace of diamonds)/ $\begin{pmatrix} 52 \\ 5 \end{pmatrix}$.

then for the event: _ _ _ _ _. We got 5 chances which is $\begin{pmatrix} 5 \\ 1 \end{pmatrix}$, then for the ace of diamonds $\begin{pmatrix} 52 \\ 1 \end{pmatrix} $since there is only one ace of a diamond from the $52$ cards. so $\frac{ \begin{pmatrix} 5 \\ 1 \end{pmatrix} \begin{pmatrix} 52 \\ 1 \end{pmatrix}} {\begin{pmatrix} 52 \\ 5 \end{pmatrix}}$? is this right?

What about if we just count of the p(ace) and P(a diamond)? Thank you in advance.

Answer 1

You are on the right track. For the event, there's the ace of diamonds and some other $4$ cards from the remaining $51$, which can be chosen in $\binom{51}4$ ways. Hence, the answer is

$$\frac{\binom{51}4}{\binom{52}5}.$$

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For just aces, the idea is similar, but a bit more complicated. We must consider cases. A hand of $5$ cards that contains an ace comes in four flavors:

• It contains exactly one ace. Then we must choose one of four aces in $\binom 41$ ways and the remaining four cards from among the $48$ non-ace cards in $\binom{48}4$ ways.

• It contains exactly two aces. Then we must choose two of four aces in $\binom 42$ ways and the remaining three cards from among the $48$ non-ace cards in $\binom{48}3$ ways.

• It contains exactly three aces. Then we must choose three of four aces in $\binom 42$ ways and the remaining two cards from among the $48$ non-ace cards in $\binom{48}2$ ways.

• It contains exactly four aces. Then we must choose four of four aces in $\binom 44$ ways and the remaining one card from among the $48$ non-ace cards in $\binom{48}1$ ways.

This gives us

$$\frac{\binom 41\binom{48}4 + \binom 42\binom{48}3 + \binom 43\binom{48}2 + \binom 44\binom{48}1}{\binom{52}5} = \frac1{\binom{52}5}\left(\sum_{k=1}^4\binom 4k \binom{48}{5-k}\right).$$

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With this, do you think you can handle the case of just diamonds? It's similar.

Answer 2

An easier reasoning is: I am dealt $5$ cards out of $52$.

So possibilities I am satisfied are $5$ (cards I am dealt) out of $52$: $P=\frac{5}{52}$.

Answer 3

Another way of looking at the problem, which might be even easier to digest, is that the probability that you do get an ace of diamonds is very simply $1$ minus the probability that you don't get the ace of diamonds.

The number of five card hands total is $\binom{52}{5}$, each of which are equally likely to have occurred (which allows us to calculate probabilities by taking ratios). The number of five card hands which don't contain the ace of diamonds is $\binom{51}{5}$.

The probability is then:

$$1 - \dfrac{\binom{51}{5}}{\binom{52}{5}}$$

This generalizes easily to the other problems you seem interested in, the probability that within your five card hand you have at least one ace. The probability that you get at least one ace is $1$ minus the probability you got no aces. The number of hands with no aces would be $\binom{48}{5}$, so the probability of at least one ace is:

$$1 - \dfrac{\binom{48}{5}}{\binom{52}{5}}$$

Similarly, the probability of getting at least one diamond is $1 - \dfrac{\binom{39}{5}}{\binom{52}{5}}$