For those interested, here is the reasoning from a combinatorics perspective. There are 52! different arrangements of a deck of cards. To see this, note that in any arrangement there are 52 locations for the Ace of Spades, once that choice is made there are 51 locations for the next card to end up in and so on until the last card must go in the only open location. Hence there are 52 factorial arrangements of cards in to a deck.<p>Shuffling, by swapping each card with any of the 52 other spots produces 52^52 (i.e. 52 * 52 * 52 * ...) arrangements.  How can this be when there are only 52! (i.e. 52 * 51 * 50 * ...) possible arrangements.  The answer is than many of the arrangements generated by this shuffling technique end up with the cards in the same order. This follows from the pigeon-hole theorem since 52^52 &gt; 52!<p>Furthermore, we know that 52^52 is not a multiple of 52! (to see this realize that, for example, 11 divides 52! but not 52^52).  Therefore, some bad shuffle generated arrangements will occur more than others and, consequently, that bad shuffle algorithm does not produce a &quot;fair&quot; shuffle.