# What is the probability of getting 3 aces, a king and a queen

Is the probability of this event: $$\frac{{4\choose 3}\cdot4\cdot4}{52\choose 5}$$

-
Yes, correct, and it is obvious where the numerator comes from. You might be expected to write a few words. –  André Nicolas Sep 26 '11 at 2:40

Yes, probability is the ratio of the number of desirable configurations over the number of total configurations, the number of total configurations is $52$ for the first card, times $51$ for the second, and so on, i.e. $\prod_{k=0}^4 (52-k)$.

The number of desired configurations $4 \times 3 \times 2$ for aces, $4$ for a king and $4$ for a queen, and then multinomial of 5 choose 3, 1, 1 which is $\frac{5!}{3! \cdot 1! \cdot 1!} =20$.

$$p = \frac{ (4 \cdot 3 \cdot 2) \cdot 4 \cdot 4 \cdot 20 }{ 52 \cdot \ldots \cdot 48 } = \frac{20}{812175} = \frac{4}{162435}.$$

Code follows ( takes a minute to run):

quintuples =
Subsets[Flatten[
Outer[List,
Range[2, 10]~Join~{"J", "Q", "K", "A"}, {"\[DiamondSuit]",

Count[quintuples,
x_List /;
Count[x, {"A", _}] == 3 && Count[x, {"K", _}] == 1 &&
Count[x, {"Q", _}] == 1]/Length[quintuples]

-
In your calculation the order matters. First choose aces, then king, last queen. Not sure if that is what lord12 wants. –  user14947 Sep 26 '11 at 1:15
@peanodo I realized that and corrected the answer. –  Sasha Sep 26 '11 at 1:19
You are counting the possible hands differently depending on the order in which they are dealt. This is not usually the case: you want to consider the "hands", not the order in which they are dealt... –  Arturo Magidin Sep 26 '11 at 1:37
Ah, I see; your denominator "should" be divided by $5!$; and your numerator by $3!$ if you want to count combinations. But that factor that you are "missing" in the count is compensated by using the multinomial coefficient. Fair enough. –  Arturo Magidin Sep 26 '11 at 2:13