Is the probability of this event: $$\frac{{4\choose 3}\cdot4\cdot4}{52\choose 5}$$
What is the probability of getting 3 aces, a king and a queen
0
$\begingroup$
probability
combinatorics
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0Yes, correct, and it is obvious where the numerator comes from. You might be expected to write a few words. – 2011-09-26
1 Answers
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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 $\frac{5!}{3! \cdot 1! \cdot 1!} =20$.
$$ p = \frac{ (4 \cdot 3 \cdot 2) \cdot 4 \cdot 4 \cdot 20 }{ 52 \cdot \ldots \cdot 48 } = \frac{20}{812175} = \frac{4}{162435}. $$
This matches your answer as well.
Added: Simulation illustrating the answer:
Code follows ( takes a minute to run):
quintuples =
Subsets[Flatten[
Outer[List,
Range[2, 10]~Join~{"J", "Q", "K", "A"}, {"\[DiamondSuit]",
"\[ClubSuit]", "\[HeartSuit]", "\[SpadeSuit]"}], 1], {5}];
Count[quintuples,
x_List /;
Count[x, {"A", _}] == 3 && Count[x, {"K", _}] == 1 &&
Count[x, {"Q", _}] == 1]/Length[quintuples]
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0In your calculation the order matters. First choose aces, then king, last queen. Not sure if that is what lord12 wants. – 2011-09-26
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1@peanodo I realized that and corrected the answer. – 2011-09-26
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1You 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... – 2011-09-26
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0Ah, 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. – 2011-09-26