We know there are 100_C_3 ways to draw 3 numbers from 1 to 100 without replacement.
Let's look at counting the number of ways to draw three numbers that add up to 100 or less.
First, let's solve for drawing two numbers:
L = The limit
F = the first number
What is the number of ways to draw a second number S given the limit L and the first number F?
We know that F + 1 <= S <= L - F. If F + 1 > L - F, then there are 0 ways. If F + 1 <= L - F, then there are (L - F) - (F + 1) + 1 numbers in the range (F + 1) to (L - F). That reduces to L - 2F.
W_2(L,F) = 0, If 2F+1 > L
W_2(L,F) = L - 2F, If 2F+1 <= L
We can compute W_2(L), for all valid F, by summing W_2(L,F) for F ranging from 1 to (L-1)/2:
...skipping the algebra...
If L is odd, W_2(L) = (L^2 - 2L + 1) / 4
If L is even, W_2(L) = (L^2 - 2L) / 4
Now, we can use this result in calculating the three-number problem:
If we have a first number F, then our second number, S, and third number, T, must satisfy:
F < S < T
F + S + T <= L
Assuming we have chosen an F small enough that there are S and T that can satisfy F + S + T <= L,
W_3(L,F) = W_2(L - 3F)
W_3(L,F) = ((L-3F)^2-2*(L-3F)+1)/4, if 3F-L is odd
W_3(L,F) = ((L-3F)^2-2*(L-3F))/4, if 3F-L is even
Our upper limit for F is F + (F + 1) + (F + 2) <= L or F <= (L - 3) / 3
Finally, we can compute W_3(L) for all valid F, by summing W_3(L,F) for F ranging from 1 to I = Int((L-3)/3), where INT() rounds down to the nearest integer.
...skipping even more algebra...
If (L-3)/3 is even, W_3(L) = (6I^3-3(2L-5)I^2+2*(L^2-5L+5)I)/8
If (L-3)/3 is odd and L is even, W_3(L) = (6I^3-3(2L-5)I^2+2*(L^2-5L+5)I+1)/8
If (L-3)/3 is odd and L is odd, W_3(L) = (6I^3-3(2L-5)I^2+2*(L^2-5L+5)I-1)/8
For very large numbers, you can just use the first formula, since the adjustments only add or subtract 1/8.
To answer your question then,
When L = 100,
Int((L - 3) / 3) = 32
W_3(100) = 25,736
100_C_3 = 161,700
so your probability of drawing 3 balls whose sum exceeds 100 is:
(161,700 - 25,736) / 161,700 = 0.840841