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?

An approximation. The coin flipping experiment is a binomial distribution, the binomial distribution can be approximated using a normal distribution (under certain conditions), the result then is the convolution of the two approximated normal distributions, which has a simple analytical solution.
If $X \sim \mathcal{B}(n,p)$ then $X \sim \mathcal{N}(np,\sqrt{(np(1-p)})$, if $np \geq 5$ and $n(1-p) \geq 5$ (rule of thumb).
For your specific case (R code)
ap=0.5
an=11

bp=0.5
bn=10

pnorm(
  0,
  (ap*an-bp*bn)/2,
  sqrt((an*ap*(1-ap)+bn*bp*(1-bp))/2),
  lower.tail=F
)

[1] 0.5613147

