I'm not sure of the relevance of this. Stats was never my strong suit but...

If the following simulation bears any resemblance to reality (of the re-stated version of the problem), then you only get a greater chance of winning if you always switch your choice after the host opens the first door on the goat.

If at that time you make another choice--either, whether to switch or not, or a random choice between the two remaining doors (which amounts to the same thing?)--then your odds of success remain at 33.33%.

#! perl -slw
use strict;
use List::Util qw[ shuffle ];

our $TRIALS ||= 1_000_000;
my $DIVISOR = $TRIALS / 100;

my( $stick, $switch, $newChoice );

for my $try ( 1 .. $TRIALS ) {
      ## Randomly hide the prizes behind the doors.
    my @doors = shuffle 'car', 'goat', 'goat';

      ## Pick a door
    my $choice = int rand 3;

      ## The host opens a door that I didn't pick
      ## and that isn't the car
    my $opened = grep{ $_ != $choice and $doors[ $_ ] ne 'car' } 0 .. 
+2;
    
      ## Count my wins if I stick or switch
    $doors[ $choice ] eq 'car' ? $stick++ : $switch++;
    
        ## Make a new choice from the remaining two
    my $new = ( grep{ $_ != $opened } 0 .. 2 ) [ rand 2 ];

      ## And if I make a completely new random choice.
    $doors[ $new ] eq 'car' and $newChoice++;
}

printf "Odds of
    Choose again: %.3f
    Win if you don't switch: %.3f
    Win if you do switch: %.3f\n",
    $newChoice / $DIVISOR, $stick / $DIVISOR, $switch / $DIVISOR;

__END__
P:\test>test
Odds of
    Choose again: 33.331
    Win if you don't switch: 33.286
    Win if you do switch: 66.714

P:\test>test
Odds of
    Choose again: 33.370
    Win if you don't switch: 33.353
    Win if you do switch: 66.647
[download]

Assuming that this simulation isn't at fault, I'd love to see a (preferably layman's terms) explaination for why a predetermined strategy (always switch) would have such an effect on the odds of success?