Why Random Swaps Do Not Uniformly Shuffle

Count the total number of distinct sequences of swap choices versus the number of permutations.

Compare this to the Fisher-Yates shuffle, which restricts the range of the random index at each step.

1. 1. Count the outcome space

   This procedure makes $n$ independent choices each from $n$ options, producing $n^n$ equally likely sequences of choices, mapped onto only $n!$ permutations.

2. 2. Show the mapping is uneven

   Since $n^n$ is generally not a multiple of $n!$, the $n^n$ equally likely sequences cannot map onto the $n!$ permutations with equal counts, so some permutations occur more often than others; correct uniform shuffling (Fisher-Yates) instead restricts the $i$th swap to the remaining unplaced elements.

Answer: It does not give all permutations equal probability.