In hybridizing, the flower count of the average offspring is not the arithmatic but the geometric mean of that of the parents. So, instead of adding the flower counts and diving by two, you multiply the flower counts and take the square root.
If I remember correctly, St. Swithin tops out around five blooms, and delenatii usually has one, but can have up to two. Thus, the root of (5 x 2) is 3 (rounded off). So a really well-grown Deli Saint with good parentage could have up to three, but two is much more likely (e.g., if we change the numbers above to a less exceptional 4 for St. Swithin and 1 for delenatii, we get (4 x 1) = 4, whose square root is of course 2).
Hope that made sense.