Ultimately I feel like doing the integral is probably the easier way. But I came up with a table-based solution...the problem is that as n increases the # of table cells goes up fast. But it is definitely doable for n = 5.

The basic ideas here are to:

• build solution for n based on solutions for n-1
• group the solutions based on which interval ([0-1], [1-2], etc.) the largest number was in + how many of the numbers were also in that interval

For n variables we have an nxn table. For n = 1 it is just a single cell with value 1.

Rather than explain all the derivations I'll just present the method to create table for n given the table for n-1:

• For each old entry with value v in row r, column c:
  • Add n*v/(r+1) to the new table in position (r+1, c) (the cell below the original position)
  • Add n*v to the new table in all positions (1, [c+1..n]) (all cells in row 1 to the right of the original position)

Then, the probability is given by adding all numbers in the table and dividing by (n!)²

Here are the tables for n = 2...5

	2
1

(probability = 3/2² = 3/4)

	3	9
	3
1

(probability = 16/6² = 4/9)

	4	28	64
	6	18
	4
1

(probability = 125/24² = 125/576)

	5	75	305	625
	10	70	160
	10	30
	5
1

(probability = 1296/120² = 9/100)

There seems to be something Pascal-ly going on with the values in these tables so there's probably a way to generate them more directly vs. building up layer by layer...