Probability: a specific hand of cards

1. Sep 1, 2008

mbrmbrg

1. The problem statement, all variables and given/known data

Find the probability of a bridge hand (13 cards) with the AKQJ of spades and no other spades.

2. Relevant equations

$$P(event)=\frac{\# \ \\favorable \ \\outcomes}{total \ \# \\outcomes}$$

$$nCr=\left(\begin{array}{c}n\\r\end{array}\right)=\frac{n!}{r!(n-r)!}$$

There are 52 cards in a deck, 13 of which are spades.

3. The attempt at a solution

• The total number of possible bridge hands is $$\left(\begin{array}{c}52\\13\end{array}\right)$$
• To get a hands with AKQJ of spades, pick them and then fill your hand with 9 other cards from the remaining 4 cards. So the committees containing AKQJ are $$\left(\begin{array}{c}48\\9\end{array}\right)$$
• But how many of those $$\left(\begin{array}{c}48\\9\end{array}\right)$$ hands don't have spades? Well, let's remove the 9 remaining spades from the deck, and see how many ways we now have to fill our hand: $$\left(\begin{array}{c}48-9\\9\end{array}\right)=\left(\begin{array}{c}39\\9\end{array}\right)$$
• Here's where I'm stuck. The result in the previous bullet tells me how many hands will have four spades. But I want 4 particular spades; and intuitively, I know that I'm less likely to pull a run of four spades than any group of 4 spades. I don't know where to go from here. Should I compute the probability of drawing 9 non-spades and then compute (separately) the probability of drawing a run of spades from the remaining cards? I don't know how to compute the probability for a run, nor do I know how to combine those two probabilities to get the answer to the problem.

Thanks in advance for any help!

Last edited: Sep 1, 2008
2. Sep 1, 2008