In a 52 cards deck, what is the probability of getting exactly one ace and exactly one king in your openning hand (5 cards)?

At first I was calculating the odds like this:

5C2 (44/52) (43/51) (42/50) (4/49) (4/48) = 0.04

But then I realized I have 3 types of cards, aces, kings and non-aces/non-kings. So I'm not sure I can use the binomial factor in this exercise.


  • $\begingroup$ You are right, you can't use the binomial coefficient because there are 3 types. In alphabetic form, you are choosing AAABC, so you can use the multinomial coefficient, $\binom{5}{3,1,1}$ which is the same as the permutations of such a word, $\frac{5!}{3!1!1!}$ $\endgroup$ Dec 20, 2015 at 17:41

2 Answers 2



Exactly 1 Ace could be picked in ${4\choose1}$ ways, similarly exactly 1 King could be picked in ${4\choose1}$. Now you have 8 cards removed from the deck and the remainder 3 cards should come from 44 cards which are non - ace and non-king. That you can choose in ${44\choose3}$. Lastly, 5 cards could be chosen in ${52\choose5}$

Thus the probability is $$\approx \dfrac{{4\choose1}.{4\choose1}.{44\choose3}}{{52\choose5}}$$


Other way is :

The Ace could be chosen with a probability $= \frac{4}{52}$

The King could be chosen with a probability $= \frac{4}{51}$

Rest of the three cards could be chosen with a probability $= \frac{44}{50}.\frac{43}{49} .\frac{42}{48}$

These five cards could be picked in $=\frac{5!}{1!.1!.3!} $

Now multiply $$=\frac{4}{52}.\frac{4}{51}.\frac{44}{50}.\frac{43}{49} .\frac{42}{48}.\frac{5!}{1!.1!.3!}$$

The first method is an approximate where as the second one is the most accurate.

  • $\begingroup$ Thanks a lot. I get it now but, can this be expressed as my original post? I'm having a hard time trying to figure it out. I mean. I understand your reasoning but I can't express the procedure in a different way. $\endgroup$ Dec 20, 2015 at 16:06
  • $\begingroup$ I just edited it, I did not realize that up until now the first method is usually an approximate solution where is the the second one is the right method that takes into account without replacement which might playout to be diferent in other problems. In this problem it is exact $\endgroup$ Dec 20, 2015 at 17:46
  • $\begingroup$ I just edited before I looked at your comment just for the integrity sake. Of course I had wrongly typed it. $\endgroup$ Dec 20, 2015 at 17:48
  • $\begingroup$ Fine, I have deleted my earlier comment. However, It is not only in this problem that the first method is exact, you will find that it is so in all such problems. $\endgroup$ Dec 20, 2015 at 17:48
  • $\begingroup$ @true blue anil Take an example: A standard 52-card deck contains cards of 4 suits and 13 numbers, with exactly one card for each pairing of suit and number. If Maya draws two cards with replacement from this dec k, what is the probability that the two cards have the same suit or have the same number, but not both? The exact answer is 15/52. $\endgroup$ Dec 20, 2015 at 18:10

This is what is called a hypergeometric distribution (drawing w/o replacement)

I would prefer to solve it using combinations for choosing from the $3$ distinct groups

$Pr = \dfrac{\binom41\binom41\binom{48}3}{\binom{52}5}$

Your attempt will give the same answer if you use the appropriate multinomial coefficient as noted against your question.

  • $\begingroup$ Thanks a lot everyone. I understand the logic behind the procedure now. $\endgroup$ Dec 20, 2015 at 17:45

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