The probability of getting heads in an biased coin is $p<\frac{1}{2}$ . We toss the coin $2n$ times.We win if the number of times heads appear is more than that of tails. We need to determine n (should be finite and greater than zero) before the game so that we win the game, what should be our best strategy to choose n?
According to my calculations probability of getting heads more than tails is $ P=\ \ ^{2n} C_{n+1} p^{n+1} (1-p)^{n-1} + ^{2n} C_{n+2} p^{n+2} (1-p)^{n-2} +.......+p^{2n}$
But how to simplify and interpret it or am I getting something wrong?