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?

This is not really an answer but purely involves your "naive first thought".
You base that on the fact that A and B have the same expected value of heads if only the first $10$ times are taken into account.
This is not correct.

Imagine a game where A throws $5$ coins and $B$ throws $4$ coins.


The first coin of A has probability $1$ on resulting in head, and the second, third and fourth have probability $\frac13$ on resulting in head and finally the fifth has probability $\frac12$ on heads.


The first coin of B has probability $0$ on resulting in head, and the second, third and fourth have probability $\frac23$ on resulting in head.


Then for both the expected number of heads gained in the first $4$ throws is $2$.


However the probability that A throws in total more heads than B in the whole game appears to be $\frac{706}{1458}<0.5$.

It would be correct to base it on the fact that A and B have the same probability of winning if only the first $10$ times are taken into account.
Think about it like this: A and B want to play a fair game ($10$ flips for each) but foreseeing that the game can end in a draw (which they want to prevent) they make the following agreement:

If the game ends in a draw then A throws a coin. If it is a head then A is declared to be the winner and otherwise B is declared to be the winner. Also if the game does not end in a draw A throws a coin but then evidently its result has no real impact on the question who wins.

Then evidently this game is fair and gives both players a chance of 50% to win.
Further it is true that A wins iff he throws more heads than B.
