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Given a shuffled deck of 52 cards, if you draw three cards (for example), how do you calculate the probability that at least one of those cards will be an Ace?

I've figured out that, for the first card drawn, the probability is $4/52$
And for the second card drawn, the probability is $4/51$ (unless the first card was an Ace, in which case it would be $3/51$)
(i.e. the drawn cards are not placed back into the deck.)

But I'm having trouble wrapping my head around the bigger picture.
I don't understand how you calculate the overall probability of any of the three cards being an Ace, accounting for the fact that multiple cards will be drawn.

What is the correct approach for this type of problem?
How do you break it down into smaller pieces, etc?

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    $\begingroup$ I presume this is without replacement. The event is complementary to that of picking no aces in three draws. $\endgroup$
    – Angina Seng
    Commented Aug 14, 2018 at 17:16
  • $\begingroup$ @LordSharktheUnknown I edited to clarify, but I think you understood correctly. $\endgroup$
    – Giffyguy
    Commented Aug 14, 2018 at 17:20
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    $\begingroup$ Take a look at math.stackexchange.com/questions/2254231/… -- does this help? $\endgroup$
    – BallBoy
    Commented Aug 14, 2018 at 17:21
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    $\begingroup$ $P_{\text {At least one card is an ace}}=1 - P_{\text {None of the cards is an ace}}$. $\endgroup$
    – For the love of maths
    Commented Aug 14, 2018 at 17:22

1 Answer 1

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Consider the hypergeometric distribution.

This is a problem in which it is easier to find the probability of the complementary event and then subtract from $1.$

Let $X$ be the number of Aces in three draws without replacement. You seek $P(X \ge 1) = 1 - P(X = 0),$ where $P(X = 0) = \frac{{4 \choose 0}{48 \choose 3}}{{52 \choose 3}} = 0.7826.$

In R statistical software dhyper is a hypergeometric PDF:

dhyper(0, 4, 48, 3)
[1] 0.7826244

The probability $P(X \ge 1)$ can also be simulated in R. With a million 3-draw games, one can expect two or three places of accuracy.

deck = 1:52     # Let the Aces be 1, 2, 3, & 4
x = replicate(10^6, sum(sample(deck, 3) <= 4))
mean(x >= 1)
[1] 0.21758

Note: Related problem. If cards were drawn with replacement, then the number of Aces drawn would be $Y \sim \mathsf{Binom}(n=3, p=12/13)$ and $P(Y = 0) = (12/13)^4 = 0.7260.$

Addendum per question in Comment:

Probability of exactly two Aces in three draws without replacement. Using 'binomial coefficients' such as ${4 \choose 2} = \frac{4!}{2!\cdot 2!}=6:$ $$P(X = 2) = \frac{{4 \choose 2}{48 \choose 1}}{{52\choose 3}}.$$

choose(4,2)*choose(48,1)/choose(52, 3)
[1] 0.01303167

Using dhyper in R:

dhyper(2, 4, 48, 3)
[1] 0.01303167

Probability of geting at least 2 Aces in three draws is the sum of two probabilities:

$$P(X \ge 2) = \frac{{4 \choose 2}{48 \choose 1}}{{52\choose 3}} + \frac{{4 \choose 3}{48 \choose 0}}{{52\choose 3}}.$$

sum(dhyper(2:3, 4, 48, 3))
[1] 0.01321267

Graph of entire distribution for the number of Aces obtained in three draws without replacement from a standard deck.

x = 0:3;  pdf = dhyper(x, 4, 48, 3)          # 4-vectors
plot(x, pdf, type="h", lwd=3, col="blue")
abline(h=-.005, col="green3")                # ref line just a bit below 0 ...
                                             #  ... to avoid hiding last bar

enter image description here

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  • $\begingroup$ Thanks so much for this. I had never seen the choose operator before, and I've been reading about it. Follow-up question: How would you modify your choose equation above, to find the probability of drawing two aces in three attempts? $\endgroup$
    – Giffyguy
    Commented Aug 17, 2018 at 14:56
  • $\begingroup$ In the "4 over 0", what does the 0 refers to ? $\endgroup$
    – thanos.a
    Commented Aug 31, 2018 at 9:22
  • $\begingroup$ In my example, I have 10000 coffee cups and only 1000 of them give a gift. If I drink 10 coffees then I interpret that the probability for me to win (once?) is given in R by: dhyper(0, 1000, 9000, 10). Correct ? $\endgroup$
    – thanos.a
    Commented Aug 31, 2018 at 9:29
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    $\begingroup$ In the hypergeometric formula the factor ${4 \choose 0}$ [read "4 choose 0"] would be the number of ways to choose no Aces out of the four Aces in the deck. Of course, ${4 \choose 0}=\frac{4!}{0!\cdot 4!} = 1,$ because $0!=1.$ In combinatorics "Do nothing" often counts as "one way." $\endgroup$
    – BruceET
    Commented Aug 31, 2018 at 9:31
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    $\begingroup$ @thanos.a This video explains it very well, and makes it much easier to understand how to adapt this to any given scenario: youtube.com/watch?v=BCeFgnh6A1U $\endgroup$
    – Giffyguy
    Commented Aug 31, 2018 at 13:16

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