re-using random number in reservoir sampling

It was asked in relation to another question recently: Given an unknown length list, return a random item in it by scanning it only 1 time

I know you shouldn't, I just can't put my finger on a canonical explanation of why not.

Look at the example code:

import random, sys

def rnd(): # a function that returns a random number each call
    return int(random.getrandbits(32))

class fixed: # a functor that returns the same random number each call
    def __init__(self):
        self._ret = rnd()
    def __call__(self):
        return self._ret

def sample(rnd,seq_size):
    choice = 0
    for item in xrange(1,seq_size):
        if (rnd() % (item+1)) == 0:
            choice = item
    return choice

dist = [0 for i in xrange(500)]
for i in xrange(1000):
    dist[sample(rnd,len(dist))] += 1
print "real",dist
print

dist = [0 for i in xrange(500)]
for i in xrange(1000):
    dist[sample(fixed(),len(dist))] += 1
print "reuse",dist

The choices for the proper reservoir sampling that generates a new random number per item is nicely evenly distributed as it should be:

real [1, 3, 0, 1, 2, 3, 2, 3, 1, 2, 2, 2, 2, 0, 0, 1, 3, 3, 4, 0, 2, 1, 2, 1, 1, 4, 0, 3, 1, 1, 2, 0, 0, 0, 1, 4, 6, 2, 3, 1, 1, 3, 2, 1, 3, 3, 1, 4, 1, 1, 2, 2, 5, 1, 2, 1, 0, 3, 1, 0, 2, 6, 1, 2, 2, 1, 1, 1, 1, 3, 2, 1, 5, 4, 0, 3, 3, 4, 0, 0, 2, 1, 3, 2, 3, 0, 2, 4, 6, 3, 0, 1, 3, 0, 2, 2, 4, 3, 2, 1, 2, 1, 2, 2, 1, 4, 2, 0, 0, 1, 1, 0, 1, 4, 2, 2, 2, 1, 0, 3, 1, 2, 1, 0, 2, 2, 1, 5, 1, 5, 3, 3, 1, 0, 2, 2, 0, 3, 2, 3, 0, 1, 1, 3, 0, 1, 2, 2, 0, 1, 2, 2, 3, 2, 3, 1, 1, 0, 1, 2, 2, 2, 2, 2, 3, 2, 1, 2, 2, 2, 1, 3, 3, 1, 0, 1, 1, 0, 1, 3, 2, 1, 4, 3, 4, 1, 1, 1, 2, 1,