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

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

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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]",

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