# Probability of getting 4 aces in a stack of 52 cards

A deck of 52 cards is shuffled thoroughly. What is the probability that the four aces are all next to each other?

What I tried

To choose the first ace I took $$\left( \begin{array}{c} 13 \\ 4 \\ \end{array} \right)/\left( \begin{array}{c} 52 \\ 4 \\ \end{array} \right)$$

Is my working correct. Could anyone explain. Thanks

• Why did you take $\binom{13}{4}$? – Patrick Stevens Aug 24 '15 at 16:50
• Indeed. $\dbinom{13}4\Big/\dbinom{52}4$ is the probability that four cards drawn from a shuffled deck will all come from a particular suit. – Graham Kemp Aug 25 '15 at 1:23

We have $52!$ different shuffles. The aces have $4!$ and the others have $48!$ different permutations. Now, we insert the aces somewhere into the permutations of the others as 1 card, we have $49$ places to to that (before the first... after the last), thus the answer is $$\frac{49\cdot48!\cdot4!}{52!}=\frac{24}{50\cdot51\cdot52}$$
• I think this simplifies to $\frac{1}{5525}$. – user84413 Aug 24 '15 at 17:11
Possible positions for a stack of 4 aces in the {interstices + ends} of the 48 other cards = 49 against total possible positions for aces of ${52\choose 4}$
$$\text{Probability} = \frac{49}{52\choose 4} = \frac{1}{5525}$$
Since the aces are supposed to appear in a row, we need locate only the first ace as we check the cards in the order they appear in the deck. This may be the top card, the next, and so on, until the 49th. After fixing this, the four aces can still permute among themselves in 4! ways and so can the non-aces (in 48! ways). So, the probability is: $$\frac{(49\cdot4!)\cdot48!}{52!}=\frac{24}{50\cdot51\cdot52}$$