why does this simple shuffle algorithm produce biased results? what is a simple reason?

Question

it seems that this simple shuffle algorithm will produce biased results:

# suppose $arr is filled with 1 to 52

for ($i < 0; $i < 52; $i++) { 
  $j = r         


        

Answer 1

The clearest answer to show the first algorithm fails is to view the algorithm in question as a Markov chain of n steps on the graph of n! vertices of all the permutation of n natural numbers. The algorithm hops from one vertex to another with a transition probability. The first algorithm gives the transition probability of 1/n for each hop. There are n^n paths the probability of each of which is 1/n^n. Suppose the final probability of landing on each vertex is 1/n! which is a reduced fraction. To achieve that there must be m paths with the same final vertex such that m/n^n=1/n! or n^n = mn! for some natural number m, or that n^n is divisible by n!. But that is impossible. Otherwise, n has to be divisible by n-1 which is only possible when n=2. We have contradiction.