You have answers that do what you set out to do. However, I wonder if you realise that what you are doing probably isn't giving you the results that you might be after?

Reading between the lines of you question and the variable names in your code, you appear to be wanting to pick values 'randomly', but according to some predetermined distribution. The trouble is, that your distribution will go out of the window because you are requiring unique picks.

To demostrate what I mean, imagine that you want to pick from 'A' through 'F', with 'B' having double the chance of being picked than 'A'; and 'C' double the chance of being picked than 'B'; and so on up the list. The following code, which is just a variation on the other solutions posted, will attempt to do that.

#! perl -slw
use strict;

our $ITERS ||= 1000;
our $PICKS ||= 4;

sub genPicker {
    my $n = shift;
    my @indices = 0 .. $n - 1;
    return sub {
        return unless @indices;
        my $index = int rand @indices;
        my $choice = $indices[ $index ];
        $indices[ $index ] = $indices[ $#indices ];
        --$#indices;
        return $choice;
    };
}

my $n = 1;
my @data = map{ ( $_ ) x ( $n *=2 ) } 'A' .. 'F';

print "Required frequencies";
for my $value ( 'A' .. 'F' ) {
    printf "$value : %.1f\n", grep( { $_ eq $value } @data ) / @data *
+ 100;
}


my %chosen;
for ( 1 .. $ITERS ) {
    my $picker = genPicker ( scalar @data );
    my %picks;

    while( keys %picks < $PICKS ) {
        my $pick = $picker->();
        my $value = $data[ $pick ];
        $picks{ $value }++ ;
    }
    $chosen{ $_ }++ for keys %picks;
}

print "\nActual frequencies:";
printf "$_ : %.1f\n", $chosen{ $_ } / ($ITERS * $PICKS) * 100
    for sort{ $chosen{ $a } <=> $chosen{ $b } } keys %chosen;
[download]

And if you only pick one unique value, the frequencies of choices match those required within the bound of random variation:

C:\test>618798 -ITERS=1e4 -PICKS=1
Required frequencies
A : 1.6
B : 3.2
C : 6.3
D : 12.7
E : 25.4
F : 50.8

Actual frequencies:
A : 1.7
B : 3.1
C : 6.4
D : 12.4
E : 25.3
F : 51.1
[download]

However, when we pick 2 values repeatedly, the uniqueness criteria starts to distort the actual frequencies:

C:\test>618798 -ITERS=1e4 -PICKS=2
Required frequencies
A : 1.6
B : 3.2
C : 6.3
D : 12.7
E : 25.4
F : 50.8

Actual frequencies:
A : 2.1
B : 4.2
C : 8.1
D : 15.4
E : 29.5
F : 40.8
[download]

And as we increase the number picked, so the distortion grows rapidly:

C:\test>618798 -ITERS=1e4 -PICKS=3
Required frequencies
A : 1.6
B : 3.2
C : 6.3
D : 12.7
E : 25.4
F : 50.8

Actual frequencies:
A : 3.0
B : 5.7
C : 11.3
D : 20.2
E : 28.0
F : 31.7

C:\test>618798 -ITERS=1e4 -PICKS=4
Required frequencies
A : 1.6
B : 3.2
C : 6.3
D : 12.7
E : 25.4
F : 50.8

Actual frequencies:
A : 4.5
B : 8.9
C : 16.1
D : 21.6
E : 24.0
F : 24.9
[download]

Until, by the time you are picking one of each value, the distribution is flat. (as it has to be!):

C:\test>618798 -ITERS=1e4 -PICKS=6
Required frequencies
A : 1.6
B : 3.2
C : 6.3
D : 12.7
E : 25.4
F : 50.8

Actual frequencies:
F : 16.7
A : 16.7
D : 16.7
C : 16.7
E : 16.7
B : 16.7
[download]

In order for the distribution to remain as you require it, it would be necessary to use a different set of data and distribution values at each level of pick.

Maybe you know this already, or maybe it doesn't matter for your application, but as I noticed the problem when attempting to verify my iterative picker, I thought it worth pointing out.