Hm. With 52 choices for the first character and 62 thereafter, a 4 character is capable of producing 52 * 62**3 different strings:12,393,056

And yet:

C:\test>junk 4
57147Dup after 57283 reps at C:\test\junk.pl line 18.
57283
C:\test>junk 4
51003Dup after 51478 reps at C:\test\junk.pl line 18.
51478
C:\test>junk 4
2Dup after 23961 reps at C:\test\junk.pl line 18.
23961
C:\test>junk 4
4Dup after 44858 reps at C:\test\junk.pl line 18.
44858
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A tiny part of that is explainable by your using rand( $#array ) instead of rand( @array ) and therefore never picking the final character in those array, giving 51 * 61**3 := 11576031. But the earlyness of the repeat--as low a 24,000--is way too early to be explained even by the Birthday paradox.

Minorly tweaked version of your code used above:

#! perl -slw
use strict;

sub random_string {
    my $n    = shift;
    my $result = '';

    my @letters   = ('A'..'Z', 'a'..'z');
    my @letternum = ('A'..'Z', 'a'..'z', '0'..'9');

    $result .= $letters[ rand $#letters ];
    $result .= join "", map { $letternum[ rand $#letternum ] } 1..$n;

    return $result;
}
my %h;

printf("\r$_"), $h{ random_string( $ARGV[0] ) }++ and die "Dup after $
+_ reps" for 1 .. 1e6;
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