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?

I like your first approach, it's neat.
Here's how I often approach this kind of 'probability that player A scores more than player B' problem (which I first came up with when dealing with similar questions involving dice, many years ago). It's not clever, but it generalizes in various ways and does reduce the calculations involved. I'll try to explain in fair bit of detail but in practice the whole process takes a few moments of thought before a simplified calculation in this case involving no summing or combinatorics at all.
Let's score each head '1' and each tail '0' and sum the scores (this is just counting but emphasizes the fact that we're summing outcomes). Let $X_A$ be the score for $A$ and similarly for $B$ ($X_B$).
Now consider a different game where player B counted tails instead of heads, and call that variable $Y_B$. The probabilities of interest with fair coins are unchanged, so we would get the same answer if we responded to this question (i.e. if we computed $P(X_A>Y_B)$). Now for this new game, player B's score $Y_B=10-X_B$. Consequently:
$P(X_A>X_B) = P(X_A>Y_B) = P(X_A>10-X_B) = P(X_A+X_B>10)$
Now $X_A+X_B$ is just the number of heads in 21 tosses of a fair coin, which has expectation 10.5.
By symmetry the probability that this total exceeds 10 is $\frac12$. $\qquad\square$

Note that this strategy works for other questions like "Player A tosses 12 times and player B tosses 9 times, what's the probability player A beats player B by more than 3?".
In that case $P(X_A>X_B+3) = P(X_A>(9-X_B)+3) = P(X_A+X_B>12)$. This is a straight binomial problem, though we can use symmetry to simplify it further to calculating just two adjacent binomial probabilities.
With sums of dice, you take one more than the number of faces to flip it around. e.g. for a six-sided die, it's equivalent to subtracting a six-sided die from 7. This lets you put B's rolls on A's side of the probability again, reducing comparing two sums of dice to comparing a single sum (with more dice) to a constant number, reducing it to a straight convolution, which is easy to automate calculations for.
