It's funny that when people "tackle" the Monty Hall problem with a program, they tend to write one that uses a simulation, running thousands (if not more) of test runs. And that while it's easy to brute-force it, as there are at most 108 cases to examine (3 places for the main prices, 3 initial picks, 2 possible doors for Monty to pick, switch/not-switch).

use 5.010;

my @doors = (1 .. 3);   
my $switch_wins = 0;
my $switch_losses = 0;
my $noswitch_wins = 0;
my $noswitch_losses = 0;

#
# Place a price  
#
foreach my $price (@doors) {   
    #
    # Pick a door
    #
    foreach my $pick (@doors) {
        #
        # Monty's pick
        #
        my @monty_choices = grep {$_ != $price && $_ != $pick} @doors;
        foreach my $monty_pick (@monty_choices) {
            #
            # Switch or not?
            #
            foreach my $switch (0, 1) {
                #
                # If switching, find final door
                #
                my $final_pick;
                if ($switch) {
                   ($final_pick) =
                      grep {$_ != $pick && $_ != $monty_pick} @doors;
                }
                else {
                    $final_pick = $pick;
                }   
                my $prob = 1 / @monty_choices;
                if ($switch) {
                    $switch_wins += $prob if $final_pick == $price;
                    $switch_losses += $prob if $final_pick != $price;
                }
                else {
                    $noswitch_wins += $prob if $final_pick == $price;
                    $noswitch_losses += $prob if $final_pick != $price
+;
                }
            }
        }
    }
}
        
printf "Switching:    win percentage = %.2f%%\n",
           100 * $switch_wins / ($switch_wins + $switch_losses);
printf "No switching: win percentage = %.2f%%\n",
           100 * $noswitch_wins / ($noswitch_wins + $noswitch_losses);

__END__
Switching:    win percentage = 66.67%
No switching: win percentage = 33.33%
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