The question asks to randomly draw 13 cards from a standard deck (52 cards). What is the probability that there are 3 aces in those 13 cards you drew.

Let $n(s)$ be the number of $13$ card combinations that can be created with a deck of $52$ cards. Let $n(a)$ be the number of ways I can obtain $3$ aces. Let $P(aces)$ be the probability that there are 3 aces in my hand.

$$n(s) = \binom{52} {13}$$ Since I want to pick $13$ cards from $52$ $$n(a) = \binom{13} {1} \times \binom{4} {3} \times \binom{48} {10}$$ Since there are $13$ different types of cards in a deck namely ${ace,king,.....one}$. So I need to pick an ace from the $13$. After I have chosen the ace I need to pick $3$ of the $4$ aces. Finally I need to pick $10$ cards from $48$ since I can only have $3$ aces.

Now $P(aces) =$ $n(a) \over n(s)$ but I know $n(a)$ was calculated wrong so can someone tell me where I went wrong in my thought process.

  • $\begingroup$ Your random draw is equivalent to two independent operations: (1) Pick three aces. (2) Pick ten non-aces. Since these operations are independent, their probabilities will multiply. That accounts for two of the terms in your product. Where does the third one, $\binom{13}{1}$, come from? To what operation does it correspond? $\endgroup$
    – whuber
    Aug 12, 2015 at 16:23
  • $\begingroup$ I thought that we had to first pick which rank of the card we needed hence the $\binom{13} {1}$ $\endgroup$
    – Daniel
    Aug 12, 2015 at 16:44

1 Answer 1


$n(a) = \binom{4}{3}\binom{48}{10}$, since you have to pick $3$ of $4$ aces and other $10$ of $48$ cards; and $n(s) = \binom{52}{13}$ since you have to pick $13$ random cards of a $52$ cards deck.

$$P(aces) = \frac{n(a)}{n(s)} = \frac{\binom{4}{3}\binom{48}{10}}{\binom{52}{13}} = 0.04120048$$

  • $\begingroup$ Why is it not necessary to pick $\binom{13} {1}$ though. Wouldn't we need to first determine that the card is an ace before we can pick $3$ aces? $\endgroup$
    – Daniel
    Aug 12, 2015 at 16:38
  • $\begingroup$ No, I mean: you just pick 3 aces out of 4. There are no further things to specify. No reasons to add $\binom{13}{1}$! $\endgroup$ Aug 12, 2015 at 16:51
  • 1
    $\begingroup$ Okay I think I know where I went wrong but just to clarify I don't need to specify the rank because picking $3$ from $4$ $aces$ already specified that it was going to be an ace right. $\endgroup$
    – Daniel
    Aug 12, 2015 at 17:03

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