A triangle $ABC$ with area $1$ is given. Anja and Bernd are playing the following game: Anja chooses a point $X$ on side $BC$. Then Bernd chooses a point $Y$ on side $CA$ und at last Anja chooses a point $Z$ on side $AB$. Also, $X,Y$ and $Z$ cannot be a vertex of triangle $ABC$. Anja wants to maximize the area of triangle $XYZ$ and Bernd wants to minimize that area.
What is the area of triangle $XYZ$ at the end of the game, if both play optimally?
1. **Initial Setup and Strategy**:
   - Given a triangle \(ABC\) with area 1.
   - Anja and Bernd play a game where Anja chooses a point \(X\) on side \(BC\), Bernd chooses a point \(Y\) on side \(CA\), and Anja chooses a point \(Z\) on side \(AB\).
   - Anja aims to maximize the area of triangle \(XYZ\), while Bernd aims to minimize it.

2. **Analyzing Bernd's Strategy**:
   - Suppose Bernd chooses \(Y\) such that \(XY\) is parallel to \(AB\).
   - Let \(\frac{BX}{BC} = k\). By similar triangles, the ratio of the height from \(Z\) to \(XY\) to the height from \(C\) to \(AB\) is also \(k\), and the ratio \(\frac{XY}{AB} = 1 - k\).
   - Therefore, the ratio \(\frac{[XYZ]}{[ABC]} = k(1 - k)\), which means \([XYZ] = k(1 - k)\).

3. **Maximizing the Area**:
   - The function \(k(1 - k)\) is a quadratic function that reaches its maximum value when \(k = \frac{1}{2}\).
   - Substituting \(k = \frac{1}{2}\), we get \([XYZ] = \frac{1}{2} \cdot \left(1 - \frac{1}{2}\right) = \frac{1}{4}\).
   - Thus, the maximum area Anja can achieve is \(\frac{1}{4}\) if Bernd plays optimally.

4. **Optimal Play by Anja**:
   - Suppose Anja chooses \(X\) to be the midpoint of \(BC\), i.e., \(k = \frac{1}{2}\).
   - If Bernd chooses \(Y\) closer to \(C\) than \(A\), Anja should choose \(Z = A\). This creates a triangle with area greater than \(\frac{1}{4}\) since the height from \(X\) to \(YZ\) is exactly half of the height from \(B\) to \(AC\) and also \(YZ > \frac{AC}{2}\).
   - If Bernd chooses \(Y\) closer to \(A\) than \(C\), Anja should choose \(Z = B\). This also creates a triangle with area greater than \(\frac{1}{4}\) since \(XZ\) is exactly half of \(BC\) but the height from \(Y\) to \(BC\) is more than half of the height from \(A\) to \(BC\).

5. **Bernd's Optimal Response**:
   - To minimize the area, Bernd should choose \(Y\) to be the midpoint of \(AC\), which forces the triangle to have area \(\frac{1}{4}\) regardless of \(Z\), since \(XY\) is parallel to \(AB\).

Conclusion:
\[
\boxed{\frac{1}{4}}
\]