An variant of Knuth shuffle

Question

This is a very hard but interesting probability question related to Knuth shuffle.

When looping for each element, the swap is performed for the current element with

Answer 1

This page compares the algorithm you mention with Knuth's, showing why Knuth's is better.

Note: the below might be wrong, see the comments.

I'm not aware of an easy way to compute the probability you ask about and I can't seem to find any simple explanation either, but the idea is that your algorithm (usually referred to as the naive shuffle algorithm) considers n^n permutations of the array instead of n!. This is because element i can end up at each of n positions at step i. Since you have n possibilities at each step and n steps, that adds up to n^n. Since n^n is not always divisible by n! it follows that not all permutations have the same probability, which is why the algorithm is considered a bad shuffle.

Since not all permutations have the same probability, it follows that the probability you ask about is different for different values of i and j, but I'm not aware of a formula that computes it.