Here's my attempt at an intuitive explanation: After the first swap, the distribution of numbers at the first position is uniform. What happens in the second s

Here's my attempt at an intuitive explanation: After the first swap, the distribution of numbers at the first position is uniform. What happens in the second swap? Well, the second position contains 1 with probability 1/n, and 2 with probability (n-1)/n. With probability 1/n, this position that is highly biased towards 2 (for n > 2) will get swapped with the first position. So 2 is more likely to be found in the first position. Of course, it gets messy to carry this out for the n swaps and I still haven't really answered your question yet. To do that let's make the following assumption: after the ith swap, the distribution of numbers at position i is uniform. (I think you can prove this inductively but I haven't tried working it out - I was studying for a midterm before I got nerd-sniped.) Before the ith swap, position i is biased towards number i. By our assumption, position i-1 is uniform. By making position i uniform we are essentially "smoothing out" the bias at position i to the other positions. Hence, as in the example I started with, it is more likely after swap i that position i-1 will contain number i, at the expense of the other numbers. This then reduces the likelihood of finding number k in positions 1, 2, etc. as opposed to position k-1: 2 is more likely in 1, 3 is more likely in 2, and so on.