You are not taking the index of the element that you are currently sampling into consideration. Suppose that the sample size is 1 and you are getting the secon

You are not taking the index of the element that you are currently sampling into consideration. Suppose that the sample size is 1 and you are getting the second item (index 1). You will call rand(1), which has 0 as the only possible outcome. So, you will always replace the first item (index 0). Whereas if you would call rand(2) (possible outcomes: 0 and 1), you replace the item in the sample with probability .5 (assuming that the random number generator is uniform).