Petya and Vasya are playing the following game. Petya chooses a non-negative random value $\xi$ with expectation $\mathbb{E} [\xi ] = 1$, after which Vasya chooses his own value $\eta$ with expectation $\mathbb{E} [\eta ] = 1$ without reference to the value of $\xi$. For which maximal value $p$ can Petya choose a value $\xi$ in such a way that for any choice of Vasya's $\eta$, the inequality $\mathbb{P}[\eta \geq \xi ] \leq p$ holds? 
1. **Understanding the problem**: Petya and Vasya are playing a game where Petya chooses a non-negative random variable $\xi$ with expectation $\mathbb{E}[\xi] = 1$. Vasya then chooses his own random variable $\eta$ with expectation $\mathbb{E}[\eta] = 1$, without knowing $\xi$. We need to find the maximal value $p$ such that $\mathbb{P}[\eta \geq \xi] \leq p$ for any choice of $\eta$ by Vasya.

2. **Initial analysis**: Since $\mathbb{E}[\xi] = 1$ and $\mathbb{E}[\eta] = 1$, we need to consider the distribution of these variables. Vasya can choose $\eta$ to be distributed similarly to $\xi$. This implies that $\mathbb{P}[\eta \geq \xi]$ should be analyzed under the assumption that $\eta$ and $\xi$ have the same distribution.

3. **Distribution and probability**: If $\eta$ is distributed just as $\xi$, then $\mathbb{P}[\eta \geq \xi]$ is the probability that a random variable from the same distribution is greater than or equal to another random variable from the same distribution. For symmetric distributions, this probability is typically $1/2$.

4. **Formal proof**:
   - Let $F(x)$ be the cumulative distribution function (CDF) of $\eta$. Then $F(x) = \mathbb{P}(\eta \leq x)$.
   - The probability that $\eta \geq \xi$ can be written as:
     \[
     \mathbb{P}(\eta \geq \xi) = \int_0^\infty \mathbb{P}(\eta \geq x) dF_\xi(x)
     \]
     where $F_\xi(x)$ is the CDF of $\xi$.
   - Since $\eta$ and $\xi$ are identically distributed, $F_\xi(x) = F(x)$.
   - Therefore:
     \[
     \mathbb{P}(\eta \geq \xi) = \int_0^\infty (1 - F(x)) dF(x)
     \]
   - Using integration by parts, let $u = 1 - F(x)$ and $dv = dF(x)$, then $du = -f(x)dx$ and $v = F(x)$:
     \[
     \int_0^\infty (1 - F(x)) dF(x) = \left[ (1 - F(x)) F(x) \right]_0^\infty + \int_0^\infty F(x) f(x) dx
     \]
     where $f(x)$ is the probability density function (PDF) of $\eta$.
   - Since $F(x) \to 1$ as $x \to \infty$ and $F(0) = 0$:
     \[
     \left[ (1 - F(x)) F(x) \right]_0^\infty = 0
     \]
   - Thus:
     \[
     \int_0^\infty (1 - F(x)) dF(x) = \int_0^\infty F(x) f(x) dx = \mathbb{E}[F(\eta)]
     \]
   - Given that $\mathbb{E}[\eta] = 1$, and for symmetric distributions, $\mathbb{E}[F(\eta)] = \frac{1}{2}$:
     \[
     \mathbb{P}(\eta \geq \xi) = \frac{1}{2}
     \]

5. **Conclusion**: The maximal value $p$ such that $\mathbb{P}[\eta \geq \xi] \leq p$ for any choice of $\eta$ by Vasya is $\boxed{\frac{1}{2}}$.