We consider the following game for one person. The aim of the player is to reach a fixed capital $C>2$. The player begins with capital $0<x_0<C$. In each turn let $x$ be the player’s current capital. Define $s(x)$ as follows:
$$s(x):=\begin{cases}x&\text{if }x<1\\C-x&\text{if }C-x<1\\1&\text{otherwise.}\end{cases}$$Then a fair coin is tossed and the player’s capital either increases or decreases by $s(x)$, each with probability $\frac12$. Find the probability that in a finite number of turns the player wins by reaching the capital $C$.
1. **Define the problem and the goal:**
   - The player starts with an initial capital \(0 < x_0 < C\).
   - The goal is to reach a fixed capital \(C\).
   - The player's capital changes according to the function \(s(x)\) defined as:
     \[
     s(x) := \begin{cases}
     x & \text{if } x < 1 \\
     C - x & \text{if } C - x < 1 \\
     1 & \text{otherwise.}
     \end{cases}
     \]
   - A fair coin is tossed, and the player's capital either increases or decreases by \(s(x)\) with equal probability \(\frac{1}{2}\).

2. **Expected value and martingales:**
   - At each step, the player's expected capital remains the same. This is a key property of martingales.
   - Let \(X_n\) be the player's capital after \(n\) turns. The sequence \(\{X_n\}\) is a martingale because:
     \[
     \mathbb{E}[X_{n+1} \mid X_n] = X_n
     \]
   - This property implies that the expected value of the player's capital does not change over time.

3. **Stopping time and probability of reaching \(C\):**
   - Define the stopping time \(T\) as the first time the player's capital reaches either 0 or \(C\).
   - By the Optional Stopping Theorem for martingales, we have:
     \[
     \mathbb{E}[X_T] = X_0
     \]
   - Since \(X_T\) can only be 0 or \(C\), we can write:
     \[
     \mathbb{E}[X_T] = 0 \cdot \mathbb{P}(X_T = 0) + C \cdot \mathbb{P}(X_T = C)
     \]
     \[
     \mathbb{E}[X_T] = C \cdot \mathbb{P}(X_T = C)
     \]

4. **Solving for the probability:**
   - Given \(\mathbb{E}[X_T] = X_0\), we have:
     \[
     X_0 = C \cdot \mathbb{P}(X_T = C)
     \]
   - Solving for \(\mathbb{P}(X_T = C)\), we get:
     \[
     \mathbb{P}(X_T = C) = \frac{X_0}{C}
     \]

5. **Conclusion:**
   - The probability that the player reaches the capital \(C\) in a finite number of turns is \(\frac{X_0}{C}\).

The final answer is \( \boxed{ \frac{X_0}{C} } \).