One extra coin

Idea. Alice has one more coin than Bob, so she must have either strictly more heads or strictly more tails than him, never both. Heads and tails play symmetric roles, so each case has probability $\frac{1}{2}$.

Step 1: exactly one of two events. Let $H_A,T_A$ be Alice's heads and tails and $H_B,T_B$ Bob's. Since $H_A+T_A=11$ and $H_B+T_B=10$, $$(H_A-H_B)+(T_A-T_B)=1.$$ Two integers adding up to $1$: exactly one of them is at least $1$. So exactly one of the events $\{H_A>H_B\}$ and $\{T_A>T_B\}$ happens.

Step 2: symmetry. Swapping heads and tails on every coin does not change the joint law (the coins are fair), and it turns one event into the other. So both have the same probability, and that probability is $\frac{1}{2}$.

The other route. Compare Alice's first $10$ coins with Bob's. By symmetry, “Alice ahead” and “Bob ahead” have the same probability $q$, and “level” has probability $1-2q$. Alice wins if she was ahead, or if they were level and her last coin is heads: $$q+\frac{1}{2}(1-2q)=\frac{1}{2}.$$