A flips a fair coin 11 times, B 10 times: what is the probability A gets more heads than B? Question

A flips a fair coin 11 times, B 10 times, what is the probability A gets more head than B?

Naive first thought
For the first 10 times of A, he has the same expected number of heads as B.
So if the 11th flip of A results in H, he get more head than B, so the answer is $50\%$.
More careful thought
instead of making arguments, I like to find systematic solutions:
$$
\Pr[H_A = a] = C(11, a) ({1\over 2})^{a}  ({1\over 2})^{11-a} =  C(11, a) ({1\over 2})^{11}\\
\Pr[H_B = b] = C(10, b) ({1\over 2})^{b}  ({1\over 2})^{10-b} =  C(10, b) ({1\over 2})^{10}\\
\Pr[H_A > H_B] = \sum_{a=0}^{11} \Pr[H_A = a] \sum_{b=0}^{a-1} \Pr[H_B = b] = ({1\over 2})^{21} \sum_{a=0}^{11} C(11, a)  \sum_{b=0}^{a-1} C(10, b) 
$$
I don't know how to find the sum from the last line, can someone help?

After 10 throws, A and B are in the exact same position. The chance that A has more heads is p, the chance that B has more heads is also p (we don't know p), and the chance for same number of heads is the remainder, 1-2p.
After the 11th throw for A only, A has more heads if he had more heads after 10 throws, or the number of heads was the same after 10 throws, and he throws head. We add the probabilities: p + (1 - 2p) / 2 = p + (0.5 - p) = 0.5.
