Probability of exactly 1 ace in each hand of 13 cards

Question

I am attempting to find the probability that each hand of 13 cards from a deck of 52 cards contains exactly one ace. My approach is to use conditional probability. Below is my approach, and I would appreciate guidance as to what is specifically causing a wrong final answer

Let's define A, B, C, D as the events that the 1st hand has 1 ace, the 2nd hand has 1 ace, the 3rd hand has 1 ace, the 4th hand has 1 ace.

Then, we need P(ABCD) = P(A) * P(B/A) * P(C/AB) * P(D/ABC)

P(A) = 4 * $\binom{48}{12}$ / $\binom{52}{13}$

P(B/A) = 3 * $\binom{48}{12}$ / $\binom{51}{13}$ --> My logic here is that I only know the first player's hand has 1 ace. So there are 51 cards that are equally likely to be in the 2nd person's hand, hence the $\binom{51}{13}$ denominator. For the numerator, we have 3 aces left and can have any of the 48 non-aces in the hand with equal probability, of which we choose 12.

P(C/AB) = 2 * $\binom{48}{12}$ / $\binom{50}{13}$

P(D/ABC) = 1 * $\binom{48}{12}$ / $\binom{49}{13}$

Multiplying these gives me 0.02, which is not the correct answer of 0.105. I would appreciate anyone's help in correcting the mistake with this approach, thanks!

Answer 1

The easiest approach is place the aces one-by-one.

• The Ace of Spades is in one of the hands with probability $\frac{52}{52}$
• Given the position of the Ace of Spades, the Ace of Hearts is in one of the other three hands with probability $\frac{39}{51}$
• Given the position of the Aces of Spades and Hearts in different hands, the Ace of Diamonds is in one of the other two hands with probability $\frac{26}{50}$
• Given the position of the Aces of Spades, Hearts and Diamonds in different hands, the Ace of Clubs is in the remaining hand with probability $\frac{13}{49}$

Multiply these together and you get about $0.105498$.

As a check, you would hope this would be a little bigger than $\frac44\times \frac34\times \frac24\times \frac14=0.09375$. It is.

---

Using your method, there is a clear problem with your final $1 \times \binom{48}{12} / \binom{49}{13}$ as, given the other three hands each have exactly one ace, the fourth hand also has exactly one ace with probability $1$; you should have had $1 \times \binom{12}{12} / \binom{13}{13}$ as you had already allocated $39$ cards including $3$ aces to the previous hands. There are similar issues with your second and third conditional probabilities.

Your multiplication of probabilities should have been

$\Big(4 \times \binom{48}{12} / \binom{52}{13}\Big) \times \Big(3 \times \binom{36}{12} / \binom{39}{13}\Big) \times \Big(2 \times \binom{24}{12} / \binom{26}{13}\Big) \times \Big(1 \times \binom{12}{12} / \binom{13}{13} \Big)$

and this too would have given you about $0.105498$.

Answer 2

An alternative approach, that is not as elegant as the answer of Henry, is to attack the problem with combinatorics.

The probability will be expressed as

$$\frac{N}{D} ~: ~D = \binom{52}{13} \times \binom{39}{13} \times \binom{26}{13} = \frac{52!}{(13!)^4}.$$

In computing $~D,~$ I am assuming that the 4 players are distinguishable from each other. So, I will have to make the same assumption when computing $~N.~$

When computing $~N,~$ it is as if there are two separate decks: one deck has 4 cards, which are the $~4~$ Aces, and the second deck has the other $~48~$ cards.

So, the number of ways of dealing $~13~$ cards to the first player is $~\displaystyle \binom{4}{1} \times \binom{48}{12}.$

Then, the number of ways of dealing $~13~$ cards to the second player is $~\displaystyle \binom{3}{1} \times \binom{36}{12}.$

Then, the number of ways of dealing $~13~$ cards to the third player is $~\displaystyle \binom{2}{1} \times \binom{24}{12}.$

Then, the fourth player automatically gets the remaining cards.

Therefore,

$$N = 4! \times \frac{48!}{(12!)^4}.$$

So, the final computation is

$$\frac{N}{D} = \frac{4! \times (48!) \times (13!)^4}{(52!) \times (12!)^4}$$

$$= \frac{4! \times (13)^4}{52 \times 51 \times 50 \times 49}$$

$$= \frac{52 \times 39 \times 26 \times 13}{52 \times 51 \times 50 \times 49}.$$