You're correct. I didn't test the code well, and seem to have gotten several things backward or out of whack. Here's properly tested code (I hope):

#!/your/perl/here
use strict;
use warnings;

my $str="17:43:33:21:23:19:27:6";      
my %hash=split/:/,$str;                
my $count;
my %ad_lookup;
foreach my $k(keys %hash)              
{              
    $count += $k;                        
    $ad_lookup{$count} = $hash{$k};
}                        

my @ad_lookup_sorted_keys = sort {$b <=> $a} keys %ad_lookup;

my %seen;
for my $rand ( 1..$count )
{
#    my $rand = rand($count);    
    my $adid;
    foreach my $key ( @ad_lookup_sorted_keys )
    {
        if ( $rand <= $key )
        {
            $adid = $ad_lookup{$key};
        }
        else
        {
            last;
        }
    }
#    print "$adid\n";
    $seen{$adid}++;
}
foreach my $k ( sort {$a <=> $b} keys %seen )
{
    print "\$seen{$k}: $seen{$k}\n";
}
[download]

Which outputs:

$seen{6}: 27
$seen{19}: 23
$seen{21}: 33
$seen{43}: 17
[download]

[There would be a slight shift going back to rand, because of the 0..99 and 1..100 difference. I believe you just need to change:

    $ad_lookup{$count} = $hash{$k};
[download]

to:

    $ad_lookup{$count-1} = $hash{$k};
[download]

]

I should note there are several ways to do this and get it right, and this is but one. Also, the key sort is only done once. A binary search on the keyspace would be faster, if there are many keys, but I'd estimate you would need a hundred keys or so to make it worth while. [Feel free to test that if you like]