I know the easier version of this, which is when $A$ has $9$ (or $n$) coins and $B$ has $10$ (or $n+1$) coins; then we may use symmetry to conclude $1/2$ (since the probability that $B$ has more heads than $A$ is equal by symmetry to the probability that $B$ has more tails than $A$ and the two are mutually exclusive and cover all possibilities).
But now that there's a difference of $2$ between them, we can't use the same logic anymore. Notably, it is possible to have both more heads and more tails than $A$ at the same time.
Something I've tried:
First restrict yourself to $B$'s first $9$ coins. Using the logic above, the probability that $B$ scores more heads within the first $9$ tosses than $A$ is $1/2$.
Now consider the probability that $B$ scores (within the first $9$ tosses) equally many heads as $A$.
$P(\text{# of heads in first $9$ of $B$ = # of heads in $A$}) = P(\text{# of tails in first $9$ of B = # of tails in A + 1)}$
This can be calculated as: $\sum_{i = 0}^n P(\text{A has $i$ tails})P(\text{B has $i+1$ tails}) = \sum_{i=0}^n {n \choose i}0.5^n {n+1 \choose i}0.5^{n+1} = 0.5^{2n+1}\sum_{i=0}^n {n \choose i}{n+1 \choose i}$
Then the total probability would be:
$P(\text{$B$ scores more heads in the first $9$ coin tosses}) + P(\text{$B$ scores equally many heads in the first $9$ tosses}) P(\text{$B$'s final toss is a head})$
$ = 0.5 + 0.5^{2n+2} \sum_{i = 0}^n {n \choose i} {n+1 \choose i}$
Oh, and $n = 8$ above.
I put this in a Wolphram alpha and got $0.592735$.
But given the structure of the question I feel like there should be some easier, cleverer way of doing this similar to the $n, n+1$ question.
Is there a way to leverage symmetry or some other neat property here?
There is a clever argument.
Let $X$ be the number of heads that $A$ throws, and let $Y$ be the number of heads that $B$ throws. Since $Y$ is equidistributed with $10-Y$, the probability of the event $$ Y>X $$ is the same as the probability of the event $$ 10-Y>X. $$ But this is just the probability that $X+Y<10$, or in other words, the probability that there are at most $9$ heads among the flips of $18$ coins. By symmetry, the probability there are at most $9$ heads when $18$ coins are flipped is $$ \frac12\left[1 + \frac1{2^{18}}\binom{18}9\right] $$
This is essentially the same as the case where $A$ has $n$ coins and $B$ has $n+1$ coins. When you do the same trick, the result is the probability that $X+Y<n+1$, which by symmetric is $1/2$.