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/ 2,147,483,648chance = 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)

AND

(X2 with X3) AND (X2 with X34 AND (X2 with X5) AND ... AND (X2 with X100,000)

AND

.

.

.

AND

(X99,999 with X100,000)

There is 4,999,950,000 possible matching (Binomial coefficient "100,000 choose2")

So,

Probability thatone pairhave different numbers is:

2,147,483,647/ 2,147,483,648chance = 0.9999999995343

Probability thatall 4,999,950,000pairs havedifferent numbersis:

0.9999999995343^4,999,950,000=0.0974

If we take the complement of that:

Probability thatthere is at least one pairthat havesame numbersis:

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.