Playing Bridge

The probability $A$ of receiving more Kings than Queens is the same as the probability of receiving more Queens than Kings. The only other option is to have the same number of Kings and Queens; let this option occur with probability $B$ .

We then have that $2A + B = 1 \Longrightarrow A = \dfrac{1 - B}{2}$ .

So we now need to calculate $B$ . (The reason I'm taking this approach is that there are $10$ cases to deal with to calculate $A$ and only $5$ to deal with to calculate $B$ .)

Now we can have $n$ of both Kings and Queens in $\dbinom{4}{n} \dbinom{4}{n} \dbinom{44}{13 - 2n}$ ways, where $0 \le n \le 4$ . We thus have that

$B = \dfrac{\displaystyle\sum_{n=0}^{4} \binom{4}{n}^{2} \binom{44}{13 - 2n}}{\dbinom{52}{13}} = 0.316151778......$ ,

which in turn gives us $A = \dfrac{1 - B}{2} = \boxed{0.3419}$ to $4$ decimal places.