Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

1 Answer

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:

enter image description here

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]
share|improve this answer
    
In your calculation the order matters. First choose aces, then king, last queen. Not sure if that is what lord12 wants. –  testito Sep 26 '11 at 1:15
1  
@peanodo I realized that and corrected the answer. –  Sasha Sep 26 '11 at 1:19
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.