Among all pairs of real numbers $(x, y)$ such that $\sin \sin x = \sin \sin y$ with $-10 \pi \le x, y \le 10 \pi$, Oleg randomly selected a pair $(X, Y)$.  Compute the probability that $X = Y$. 
1. We start with the given equation:
   \[
   \sin(\sin(x)) = \sin(\sin(y))
   \]
   This equation holds if and only if one of the following conditions is satisfied:
   - \( \sin(x) = \sin(y) \)
   - \( \sin(x) = \pi - \sin(y) \)

2. The first condition, \( \sin(x) = \sin(y) \), implies:
   \[
   x = y + 2k\pi \quad \text{or} \quad x = \pi - y + 2k\pi
   \]
   for some integer \( k \).

3. The second condition, \( \sin(x) = \pi - \sin(y) \), is not possible because the range of \(\sin(x)\) and \(\sin(y)\) is \([-1, 1]\), and \(\pi - \sin(y)\) would be outside this range.

4. Therefore, we only need to consider the first condition:
   \[
   x = y + 2k\pi \quad \text{or} \quad x = \pi - y + 2k\pi
   \]

5. We are given the interval \([-10\pi, 10\pi]\). We divide this interval into 10 subintervals of length \(2\pi\):
   \[
   [-10\pi, -8\pi), [-8\pi, -6\pi), \ldots, [8\pi, 10\pi]
   \]

6. For each fixed \( y \) in \([-10\pi, 10\pi]\), there are exactly 10 values of \( x \) that satisfy \( x = y + 2k\pi \) and 10 values of \( x \) that satisfy \( x = \pi - y + 2k\pi \). This gives a total of 20 possible values of \( x \) for each \( y \).

7. Among these 20 possible values, exactly one of them is \( x = y \).

8. Therefore, the probability that \( X = Y \) is:
   \[
   \frac{1}{20}
   \]

The final answer is \(\boxed{\frac{1}{20}}\).