# Probability of drawing an ace. [closed]

I don't know how to solve this problem:

Five cards are dealt from a shuffled deck. What is the probability that the dealt hand contains...

• Exactly one ace?
• At least one ace?

## closed as off-topic by user296602, Watson, N. F. Taussig, Leucippus, ShaileshSep 4 '16 at 0:33

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• I just started and I have no idea how to do this. Im sorry im new – Zipcoder Sep 3 '16 at 21:06
• @Zipcoder For the first part, what is the probability of picking an ace first, followed by 4 non-ace cards? – gowrath Sep 3 '16 at 21:35
• Welcome to the site. You are getting downvoted and getting votes to close. Most of us expect you to say where you encountered the problem and to show some effort towards solution. – BruceET Sep 3 '16 at 22:28

## 3 Answers

Hint.

1. To get the chance that there is exactly one Ace in five draws, we count the ways to get one Ace. There are four Aces: spades, hearts, diamonds, and clubs. There are $\binom 41$ ways to choose one of these. Then we count the number of ways to choose the non-Aces. There are 48 of those, and we need 4 of them, so $\binom{48}{4}$. There are a total of $\binom{52}{5}$ ways to draw five cards. So the answer is $$\frac{\binom41\binom{48}{4}}{\binom{52}{5}}$$
2. Instead, consider the complement. The probability of getting at least one ace is equal to 1 minus the probability that you get no aces. Calculate the chance that you get no aces.

Note: $\binom nk = \frac{n!}{k!(n-k)!}$

There are $52$ possibilities for the first card of the hand, $51$ for the second and so on so there are $52*51*50*49*48$ or $\frac {52!}{47!}$ possible total hands.

But we aren't going to care about the exact order of the cards. If a hand consists of 5 cards $a,b,c,d,e$ there are $5!$ ways to arrange the cards in the hand. There are $5$ places to put the $a$; $4$ places to put the $b$, etc. So for any possible set of $5$ there are $5!$ possible hands that have exactly those cards.

So the number of different hands without regard to order is $\frac{52!}{47!*5!}$. (We call that number ${52 \choose 5}$: "52 choose 5. ${a \choose b} = \frac{a!}{(a-b)!b!}$. ${a \choose b }$ represents the number of ways one can choose $b$ objects from a set of $a$ without regard to the order.

To get exactly 1 ace there are $4$ choices for the ace. Then there are a $48$ choices for the the remaining $4$ cards. So there are $4*{48 \choose 4}$ sets of cards that have exactly 1 ace.

So the probability of exactly one ace is $4*{48 \choose 4}$ hands out of ${52 \choose 5}$ total hands. So:

$P = \frac{4*{48 \choose 4}}{{52 \choose 5}}= 4*\frac{\frac{48!}{4! 44!}}{\frac{52!}{47!5!}}=$

$4\frac{48!}{4! 44!}\frac{47!5!}{52!}=$

$4\frac{5!}{4!}\frac{48!}{44!}\frac{47!}{52!}=$

$\frac{4*5*48*47*46*45}{48*49*50*51*52}=\frac{4*5*47*46*45}{49*50*51*52}$

which we can simplify further if we want.

Perhaps coming later in your course is the hypergeometric distribution', often used in models where sampling is without replacement. Consider that you want the number $X$ of Aces in $n = 5$ draws from a population containing $4$ Aces and 48 non-Aces. [Maybe see if you can find 'hypergeomeetric distribution' in your text--or in the Wikipedia article.]

$$P(X = x) = \frac{{4 \choose x}{48 \choose 5-x}}{{52 \choose 5}},$$ for $x = 0, 1, 2, 3, 4.$

For practice using factorials, you should try to get the numerical values for $P(X = 0)$ and $P(X = 1).$ [See the Answer by @fleablood, which appeared while I was making the graph below.]

Here is a table of probabilities *rounded to 5 places) for all legal values of $x$ computed in R statistical software. ($X$ is a random variable and PDF stands for its 'probability distribution function'.)

 x = 0:4;  pdf = round(dhyper(x, 4, 48, 5), 5)
cbind(x, pdf)
x     pdf
## 0 0.65884
## 1 0.29947
## 2 0.03993
## 3 0.00174
## 4 0.00002
`

Your questions ask for $P(X = 1)$ and $P(X \ge 1) = 1 - P(X = 0)$. [The second equation illustrates the 'complement' method in the Answer of @Max.]

Here is a plot of the PDF. [Try to match the heights of the bars with the PDF values from the computer.] 