3 Corrected by considering difference distributions; deleted 2 characters in body; added 185 characters in body

The sum of independent identically distributed random variables follows the Irwin-Hall distribution. For $n$ variables that are uniformly distributed on (0,1), the probability density function is $$f_X(x)=\frac{1}{2\left(n-1\right)!}\sum_{k=0}^{n}\left(-1\right)^k{n \choose k}\left(x-k\right)^{n-1}\text{sgn}(x-k).$$

So for $n=11$ you will have to integrate an explicit piecewise polynomial function.

Edit: Tony is right, we need not quite the integral of the Irwin-Hall distribution. Rather, we need the difference distribution of two random variables that are distributed according to the above distribution (each of the RVs representing the total round-off error for each cyclist). So we need $$\int_0^3\left(f_X(x)\int_{x+8}^{11}f_X(y)dy\right)dx.$$ This is the probability that the difference between the round off errors is between 8 and 11. It is obtained by integrating over all permissible combinations (one error, other error).

The sum of independent identically distributed random variables follows the Irwin-Hall distribution. For $n$ variables that are uniformly distributed on (0,1), the probability density function is $$f_X(x)=\frac{1}{2\left(n-1\right)!}\sum_{k=0}^{n}\left(-1\right)^k{n \choose k}\left(x-k\right)^{n-1}\text{sgn}(x-k).$$
So for $n=11$ you will have to integrate an explicit piecewise polynomial function.
The sum of independent identically distributed random variables follows the Irwin-Hall distribution. For $n$ variables that are uniformly distributed on (0,1), the probability density function is $$f_X(x)=\frac{1}{2\left(n-1\right)!}\sum_{k=0}^{n}\left(-1\right)^k{n \choose k}\left(x-k\right)^{n-1}\text{sgn}(x-k).$$
So for $n=11$ you will have to integrate an explicit piecewise polynomial function.