Select two distinct diagonals at random from a regular octagon. What is the probability that the two diagonals intersect at a point strictly within the octagon? Express your answer as $a + b$, where the probability is $\tfrac{a}{b}$ and $a$ and $b$ are relatively prime positive integers.
1. **Identify the total number of diagonals in a regular octagon:**
   - A regular octagon has 8 vertices. The number of diagonals in an \( n \)-sided polygon is given by the formula:
     \[
     \frac{n(n-3)}{2}
     \]
     For an octagon (\( n = 8 \)):
     \[
     \frac{8(8-3)}{2} = \frac{8 \cdot 5}{2} = 20
     \]
     So, there are 20 diagonals in a regular octagon.

2. **Calculate the total number of ways to select two diagonals:**
   - The number of ways to choose 2 diagonals out of 20 is given by the binomial coefficient:
     \[
     \binom{20}{2} = \frac{20 \cdot 19}{2} = 190
     \]

3. **Determine the number of ways two diagonals can intersect inside the octagon:**
   - For two diagonals to intersect inside the octagon, they must be formed by choosing 4 vertices out of the 8 vertices. Each set of 4 vertices forms a quadrilateral, and the two diagonals of this quadrilateral will intersect inside the octagon.
   - The number of ways to choose 4 vertices out of 8 is given by:
     \[
     \binom{8}{4} = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} = 70
     \]

4. **Calculate the probability:**
   - The probability that two randomly selected diagonals intersect inside the octagon is the ratio of the number of favorable outcomes to the total number of outcomes:
     \[
     \frac{\binom{8}{4}}{\binom{20}{2}} = \frac{70}{190} = \frac{7}{19}
     \]

The final answer is \(\boxed{\frac{7}{19}}\)