What is the probability of a 5 card poker hand consisting of a king, the ace of hearts and the rest spades?

[steven1111]

Hi, I would like to ask if my solution to this problem is correct. I interpreted the question to imply that there can only be 1 king in this hand. So I calculated 2 separate cases.

Case 1: The king is not a spade.
So this part of the probability would be
(3C1)(1C1)(12C3)/(52C5).
Which is respectively, the number of ways to choose 1 king from 3 kings, which excludes the king of spades, the ace hearts from itself and 3 spades from 12 spades, which excludes the king of spades. All divided by 5 cards from 52 cards.

Case 2: The king is a spade. So this part of the probability would be (1C1)(1C1)(12C3)/(52C5). Which is respectively, the number of ways to choose the king of spades from itself, the ace hearts from itself and 3 spades from 12 spades, which excludes the king of spades. All divided by 5 cards from 52 cards.

So the total probability would be
((3C1)(1C1)(12C3)+(1C1)(1C1)(12C3))/(52C5)
which is approximately 0.00034.

Thanks

 

[Steven G]

I have learned from a few people that one should do a probability problem two different ways to see if you get the same answer. If you do get the same answer then you are probably correct. Having said that it seems that your work is fine. I agree with you that there should be exactly one king.

 

[pka]

What is the probability of a 5 card poker hand consisting of a king, the ace of hearts and the rest spades?

I agree with you that there should be exactly one king.

I too agree that the question says "a king", the article a means exactly one.
It does not change your answer by much.
So that means the answer would be \(\dfrac{\dbinom{4}{1}\cdot(1)\cdot\dbinom{12}{3}}{\dbinom{52}{5}} \approx 0.000338597\)

 

[steven1111]

Thank you to both Jomo and pka for their comments and insights. pka's comment was quite enlightening because although my solution is equivalent to pka's mathematically. (i.e. they give the same numerical result in the calculator). pka's solution is more succinct and elegant. It turns out that if you factor the numerator of my above formula it gives (3C1+1C1)(1C1)(12C3) where the first factor is equal to the (4C1) factor in pka's formula. What I have learned from pka's solution is that once you choose any one of the four kings. Regardless of its suit, you are forced to choose from only 12 spades! (i.e. the king is already a spade or you cannot choose the king of spades).