Your answer is very complicated and at this time I have my usual headaches (also English does not help).In that calculation you have nothing about possible matching.
You just calculated average distance between 2 neighbor numbers
from sorted list of 100,000 random generated intigers from interval [0, 2,147,483,648]
Lets say person X1 got one of 2,147,483,648 numbers.
Person X2 will get the same number with 1 / 2,147,483,648 chance = 0.0000000004656...
Person X2 will get different number with 2,147,483,647 / 2,147,483,648 chance = 0.9999999995343...
but we did't test all possible matching.
we tested only person X1 with person X2
we must test:
(X1 with X2) AND (X1 with X3) AND (X1 with X4) AND ... AND (X1 with X100,000)
(X2 with X3) AND (X2 with X34 AND (X2 with X5) AND ... AND (X2 with X100,000)
(X99,999 with X100,000)
There is 4,999,950,000 possible matching (Binomial coefficient "100,000 choose 2")
Probability that one pair have different numbers is:
2,147,483,647 / 2,147,483,648 chance = 0.9999999995343
Probability that all 4,999,950,000 pairs have different numbers is:
If we take the complement of that:
Probability that there is at least one pair that have same numbers is:
1 - 0.0974 = 0.9026 = 90.26 %
Maybe contra intuitive but, like this,
similar "paradox" is that in group of just 23 people
greater chance is that there exist pair of two people that celebrates birthday
on same day than there doesn't exist such pair.
And here is 365 / 23 = 15.87 average days distance between 2 neighbor birthdays
Hope this helps!
If you're right, a test is sufficient.
100 random pairs (or more, 100,000) and calculate how many of them are equal.