Solving for n in binomial distribution

Hi,

I just came over a little problem that has me puzzled.

If we have two teams, where one of the teams has a 55% chance of winning a match, how many times must the two teams meet such that we can be 95% certain that the best team wins?

What pops into my mind is the binomial distribution, but then I would have to solve the following equation for n,

$\displaystyle \sum^{(n/2+1)}_{k=0}\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k} > 0.95$,

which looks daunting. Perhaps I could use Markov chains in some way?

Any hints are appreciated!