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?

Your "naive first thought" is the clever (standard) solution.
To make it rigorous, let $\mathscr E_0$ be the event "A and B are tied after each has tossed 10 times;" let $\mathscr E_A$ and $\mathscr E_B$ be the events "A has more heads than B after 10 tosses each" and "B has more heads than A after 10 tosses each," respectively.  Let $\mathscr F$ designate the event "A has more heads than B after all tosses are made."
Notice:

*

*$\mathscr E_0,$ $\mathscr E_A,$ and $\mathscr E_B$ are mutually exclusive: no two have any outcomes in common and collectively they include all the possibilities.  Therefore $$\Pr(\mathscr E_0) + \Pr(\mathscr E_A) + \Pr(\mathscr E_B)=1.$$


*$\Pr(\mathscr F\mid \mathscr E_A) = 1$ (A has won by the first 10 tosses); $\Pr(\mathscr F\mid \mathscr E_B) = 0$ (A is behind after 10 tosses and therefore cannot win with the last toss); and $\Pr(\mathscr F\mid \mathscr E_0) = 1/2$ (if both are tied after 10 tosses, A's 11th toss is the tiebreaker).


*$\Pr(\mathscr E_A) = \Pr(\mathscr{E_B})$ (after 10 tosses the game is symmetric -- both players are equally situated -- and therefore they have equal chances of being ahead at that point).
By the law of total probability,
$$\begin{aligned}
\Pr(\mathscr F) &= \Pr(\mathscr F\mid \mathscr E_0)\Pr(\mathscr E_0) + \Pr(\mathscr F\mid \mathscr E_A)\Pr(\mathscr E_A) + \Pr(\mathscr F\mid \mathscr E_B)\Pr(\mathscr E_B)\\
& = \Pr(\mathscr E_0)\left(\frac{1}{2}\right) + \Pr(\mathscr E_A)(1) + \Pr(\mathscr E_B)(0)\\
&= \frac{1}{2}\left(\Pr(\mathscr E_0) + \Pr(\mathscr E_A) + \Pr(\mathscr E_A)\right)\\
&= \frac{1}{2}\left(\Pr(\mathscr E_0) + \Pr(\mathscr E_A) + \Pr(\mathscr E_B)\right)\\
& = \frac{1}{2}\left(1\right) = \frac{1}{2}.
\end{aligned} $$

As an alternative approach, you wish to evaluate the double sum
$$\sum_{a \gt b} \binom{11}{a}\binom{10}{b} = \sum_{a \gt b} \binom{11}{11-a}\binom{10}{10-b} = \sum_{a^\prime \le b^\prime} \binom{11}{a^\prime}\binom{10}{b^\prime}.$$
(In case the algebra isn't obvious, the first equality exploits the Binomial coefficient symmetry and the second is the change of variable $a^\prime = 11-a,$ $b^\prime = 10-b.$  We can be a vague about the endpoints of the summations because whenever $a$ or $a^\prime$ is not in the range from $0$ through $11$ or $b$ or $b^\prime$ is not in the range from $0$ through $10$ the Binomial coefficients are zero.)
Because the indexes in the two sums on the left and right sides (1) never overlap and (2) cover all the possibilities (since either $a\gt b$ or $a\le b$ but never both), together they give the total probability, which is $1.$  Consequently, since those sums are equal, each is $1/2,$ QED.

This figure shows the rotational symmetry of the distribution under the mapping $(a,b)\to(11-a,10-b).$  The blue circles are rotated around the yellow dot into red triangles of exactly the same probability.  The desired sum is the total of the blue circles, which therefore must be $1/2.$
