Maria and Bilyana play the following game. Maria has $2024$ fair coins and Bilyana has $2023$ fair coins. They toss every coin they have. Maria wins if she has strictly more heads than Bilyana, otherwise Bilyana wins. What is the probability of Maria winning this game?
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 \( \boxed{\frac{1}{2}} \).