An even stronger result takes into account the birthday problem (<a href="http://en.wikipedia.org/wiki/Birthday_problem" rel="nofollow">http://en.wikipedia.org/wiki/Birthday_problem</a>).<p>Claim: <i>No two properly shuffled decks have ever been the same.</i><p>Proof: For convenience, set <i>upper bound on # of shuffles</i> = <i>n</i> = 10^23 and <i>number of possible shuffles</i> = <i>N</i> = 10^68. Then the probability that all shuffles are <i>not</i> unique is<p><i>q</i> = 1 X (1 - 1/<i>N</i>) X (1 - 2/<i>N</i>) X ... X (1 - (<i>n</i>-1)/<i>N</i>).<p>Since <i>n</i> << <i>N</i>, even (<i>n</i>-1)/<i>N</i> is small, so we can approximate <i>q</i> as<p><i>q</i> = 1 X <i>e</i>^(-1/<i>N</i>) X <i>e</i>^(-2/<i>N</i>) X ... X <i>e</i>^(-(<i>n</i>-1)/<i>N</i>)
= <i>e</i>^(-<i>n</i>^2/(2<i>N</i>))<p>using the series expansion <i>e</i>^<i>x</i> = 1 + <i>x</i> for <i>x</i> << 1.<p>Then the probability that any two decks have ever matched is<p><i>p</i> = 1 - <i>q</i> = 1 - e^(-n^2/(2N)).<p>Now,<p><i>q</i> = <i>e</i>^(-5 * 10^(-22)) = 1 - 5 * 10^(-22),<p>to good approximation, so<p><i>p</i> = 1 - <i>q</i> = 5 * 10^(-22)<p>which is zero for all practical purposes. Thus, no two properly shuffled decks have ever been the same. QED