If two different integers between 1 and 100 inclusive are chosen at random, what is the probability that the difference of the two numbers is 15?

Question

My work:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 (1-20)

20-5


19-4

18-3


17-2

16-1
 so 5


5^5=3125 Am I missing a step? Or am I correct?

Answer 1

Hint: How many pairs have a difference of $15?$ How many pairs are there? Divide to get the probability.

Added: if the first number is $1-15$ or $86-100$, a $0.3$ probability, you have $\frac 1{99}$, while if the first number is $16-85$ you have $\frac 2{99}$, so the overall chance is $$0.3 \cdot \frac 1{99}+0.7 \cdot \frac 2{99}=\frac {17}{990}$$
Another way to see it is that there are $85$ pairs $15$ apart out of ${100 \choose 2}=4950$ total pairs, so the chance is $$\frac {85}{4950}=\frac {17}{990}$$