1. Define the problem and variables:
We are given a positive integer $n$ and need to find the expected value $M_n$ of the largest of three nonnegative integers $a, b,$ and $c$ such that $a + b + c = n$. We are interested in the limit of $\frac{M_n}{n}$ as $n$ approaches infinity.

2. Count the number of ordered triples:
The number of ordered triples $(a, b, c)$ such that $a + b + c = n$ is given by the stars and bars theorem:
$$\binom{n+2}{2}$$

3. **Define $W_k(n)$:**
Let $W_k(n)$ denote the number of ordered triples $(a, b, c)$ such that $a + b + c = n$ and $\max(a, b, c) = k$.

4. **Symmetry and counting $W_k(n)$:**
By symmetry and the properties of the triangular array, we can determine $W_k(n)$ for different ranges of $k$:
- For $k \geq \lceil \frac{n}{3} \rceil$, the values of $W_k(n)$ depend on the residue of $n \mod 3$:
$$W_{\lceil \frac{n}{3} \rceil + k}(n) = \begin{cases} 3(3k + 2) & \text{if } n \equiv 1 \mod 3 \\ 3(3k + 1) & \text{if } n \equiv 2 \mod 3 \\ 9k & \text{if } 3 \mid n \end{cases}$$

5. Expected value calculation:
To compute $M_n$, we need to average the values of $k \cdot W_k(n)$ over all $k$ from $\lceil \frac{n}{3} \rceil$ to $n$:
$$M_n = \frac{\sum_{i=\lceil \frac{n}{3} \rceil}^{n} (i \cdot W_i(n))}{\binom{n+2}{2}}$$

6. Simplify the sum:
Using the properties of binomial coefficients and the Hockey Stick Identity, we can simplify the sum:
$$\sum_{i=1 + \lfloor \frac{n}{2} \rfloor}^{n} (i \cdot W_i(n)) = 3 \lfloor \frac{n}{2} \rfloor \binom{\lfloor \frac{n}{2} \rfloor + 2}{2} + 3 \binom{\lfloor \frac{n}{2} \rfloor + 3}{3}$$

7. Asymptotic behavior:
As $n$ approaches infinity, the floors and ceilings become negligible. We focus on the leading terms of the numerator and denominator:
$$\frac{\sum_{i=\lceil \frac{n}{3} \rceil}^{n} (i \cdot W_i(n))}{n \binom{n+2}{2}}$$

8. Leading coefficients:
The leading third-degree coefficients of the numerator and the denominator determine the limit. After simplification, we find:
$$\lim_{n \to \infty} \frac{M_n}{n} = \frac{11}{18}$$

The final answer is $\boxed{\frac{11}{18}}$