Probability of picking more balls of one color Suppose you flip a 3-sided coin $n$ times. The sides are denoted: $A$, $B$, and $C$. The probability of a coin flip turning one of the sides is given by $p_A$, $p_B$, and $p_C$, respectively. 
What is the probability that you end up flipping side $A$ more times than side $B$?

I know this looks like a homework problem, but it is not. It is a rewording of a problem I ran into in my research. It looks easy, but I am having trouble getting it started. Any hints would be very welcome. 

The probability of flipping side $A$ $j$ times and side $B$ $k$ times is
$$\frac{n!}{j!k!(n-j-k)!}p_A^jp_B^kp_C^{n-j-k}$$
and so the probability that you flip $A$ more than you flip $B$ is
$$\sum_{k=0}^{n-1}\sum_{j=k+1}^n\frac{n!}{j!k!(n-j-k)!}p_A^jp_B^kp_C^{n-j-k}.$$
This does not strike me as a very useful expression. Assuming $p_A<p_B$, you should be able to use Stirling's formula to prove an asymptotic of the form $e^{-n\alpha}$ for some $\alpha>0$.
