# Probability of cards out of a deck

What is the probability that a hand of five cards chosen randomly and without replacement from a standard deck of 52 cards contains the ace of hearts, exactly one other ace, and exactly two kings?

I have the following solution for this problem.

1*3*3*4/(52*51*50*49) = 3/541450

Is this correct, if no, how should I approach it?

• Two kings can be chosen in $\binom{4}{2}$ ways, one ace other than hearts can be chosen in $\binom{3}{1}$ ways, and the rest of the card can be neither an ace not a King and hence can be chosen in 44 ways. – user348749 Sep 3 '16 at 4:09

There are a couple mistakes:

• It seems like you are computing the probability of drawing those cards in that order, when instead you want to compute the probability of having a hand containing these cards in some order.
• You have not ensured that the fifth card is not an ace or a king.

There are various ways to get the right answer; here is one of them.

There are $\binom{52}{5} = 52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 / 5!$ possible hands of five cards, without accounting for order. This will be your denominator.

For the numerator we need to count how many valid hands are possible.

There are $3$ choices for the non-heart ace.

There are $\binom{4}{2}=6$ choices for the pair of kings.

There are $44$ cards that are not kings or aces.

So the numerator is $3 \cdot 6 \cdot 44$.

Another way is to count the number of valid ordered hands and divide by the total number of ordered hands $52 \cdot 51 \cdot 50 \cdot 49 \cdot 48$. What you will end up with is $5!$ times the numerator computed by the above method, so the final answer is the same.

You have assumed the ace of hearts is the first card drawn and (I suspect) the first $3$ is for the other ace, which you require to be the second card drawn. It could be you draw a king first, the stray card second, then the two aces and the other king. You need to account for all the orders you can draw the acceptable hands. Also the fifth card needs not to be an ace or king, so there are not $48$ choices for it.