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?
 A: 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.
A: 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.
A: 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.
A: 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 $1/2.$
A: 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

