Problem 1268

1. Let $p$ be the probability that among the first $2023$ coins, Maria has more heads than Bilyana. By symmetry, this is also the probability that Bilyana has more heads than Maria. Let $q$ be the probability that Maria and Bilyana have the same number of heads among the first $2023$ coins. Since these are the only possible outcomes, we have:
$$2p + q = 1$$

2. Now consider the $2024$-th coin that Maria has. There are two cases to consider:
- The $2024$-th coin is a tail.
- The $2024$-th coin is a head.

3. If the $2024$-th coin is a tail, then Maria wins if she has more heads than Bilyana among the first $2023$ coins. The probability of this happening is $p$.

4. If the $2024$-th coin is a head, then Maria wins if she has at least as many heads as Bilyana among the first $2023$ coins. The probability of this happening is $p + q$.

5. Since the $2024$-th coin is fair, the probability of it being a head or a tail is $\frac{1}{2}$. Therefore, the total probability of Maria winning is:
$$\frac{1}{2} \cdot p + \frac{1}{2} \cdot (p + q)$$

6. Simplifying the expression, we get:
$$\frac{1}{2} \cdot p + \frac{1}{2} \cdot (p + q) = \frac{1}{2} \cdot p + \frac{1}{2} \cdot p + \frac{1}{2} \cdot q = \frac{2p + q}{2}$$

7. From step 1, we know that $2p + q = 1$. Substituting this into the expression, we get:
$$\frac{2p + q}{2} = \frac{1}{2}$$

Therefore, the probability of Maria winning the game is $\frac{1}{2}$.

The final answer is $\boxed{\frac{1}{2}}$.