Bulgarian Winter Tournament

Let $p$ be the probability that Maria has more heads than Bilyana after tossing the first 2023 of Maria's coins. Then, for reasons of symmetry, the probability that Maria has fewer heads than Bilyana is also $p$, and therefore the probability that Maria and Bilyana have thrown an equal number of heads is $1-2p$. If Maria has thrown less heads than Bilyana to this moment, the probability of her winning is 0 (regardless of the last coin), if she flipped more heads, the probability of her winning is 1 (again, regardless of the last coin), and if they flipped an even number, the probability of her winning is $\frac{1}{2}$ (here the last coin must be heads).

Thus we get that the probability that Maria wins is $p+\frac{1-2p}{2}=\frac{1}{2}$.