TL;DR

Some reflections about the Monty Hall problem.

The Monty Hall problem is a nice “paradox” that isn’t. It’s easy to fall and fail, and it’s easy to also get it wrong in explaining how it works.

This is one of the most common formulations (from Wikipedia, which comes from Game Show Problem by Marilyn vos Savant):

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

The host knows what’s behind the doors and always opens a door with a goat. It’s always possible, because there are two doors left by the player and there’s only one car, so one of the two doors MUST have a goat behind.

There is a definite advantage in switching the choice, which leads to a win in 2 out of 3 times (odds 2:1), so the answer to the question is a yes. Many people think that, after the revelation, it’s a 1:1 odds situation though.

My no-brainer go-to solution to this apparent paradox is this alternative game: there are two players, the first can pick whatever door they want, the other player takes all of the remaining. There is no switching. Which player would you like to be?

Your answer, I think, will be the second player. They get two doors instead of one, so double the chances to a win, right? OK, unless you’re a magician and know what’s behind the doors, of course.

In picking the two-doors alternative, anyway, you already know fore sure that at least one of them is a goat. Again, you have two doors and there’s only one car, so one of the doors MUST have a goat behind.

At this point, the host of the show starts opening your doors, revealing the goat that you know is somewhere (and the host knows where it is). Is your situation any worse than before? No! Because you already knew that there was (at least) one goat on your side, the host only told you which door it was out of the two.

If you’re more Perl inclined, Let’s Make a Deal:

#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
no warnings 'experimental::signatures';
use List::Util 'shuffle';

our $WIN = 'car'; our$LOSE = 'goat';

my ($player_class,$monty_class) = @ARGV;
my $player =$player_class->new;
my $monty =$monty_class->new;
my $total = 1_000; my$wins = 0;
for (1 .. $total) {$wins++ if monty_hall_round($player,$monty);
}
my $percentage = sprintf '%.1f%%', 100 *$wins / $total; say "What a season! Players won$percentage of times!";

sub monty_hall_round ($player,$monty) {
my @door_names = ('Door A', 'Door B', 'Door C');

# build a scenario
my %prize_behind;
@prize_behind{@door_names} = shuffle($WIN,$LOSE, $LOSE); # let the player choose my$player_choice = $player->initial_choice(\@door_names); say "Well! Player chose$player_choice...";

my ($revealed,$unrevealed) = $monty->reveal( \%prize_behind,$player_choice, \@door_names);
say "Look at this! A $LOSE behind$_!" for $revealed->@*; if ($player->swaps_with($unrevealed)) { say "Player swaps$player_choice with $unrevealed!";$player_choice = $unrevealed; } else { say "Player keeps$player_choice!";
}
say '';

return $prize_behind{$player_choice} eq $WIN; } ## end sub monty_hall_round package Player; sub new { bless {}, shift } sub initial_choice ($self, $alternatives) { my @alts =$alternatives->@*;
$self->{alternatives} = \@alts;$self->{initial} = @alts[rand @alts];
}
sub swaps_with ($self,$unrevealed) { ... }

package StubbornPlayer;
use parent -norequire => 'Player';
sub swaps_with ($self,$unrevealed) { return 0 } # never swaps

package MathsPlayer;
use parent -norequire => 'Player';
sub swaps_with ($self,$unrevealed) { return 1 } # always swaps

package RandomPlayer;
use parent -norequire => 'Player';
sub swaps_with ($self,$unrevealed) { return int rand 2 }

package ABCPlayer;
use parent -norequire => 'Player';
sub swaps_with ($self,$unrevealed) {
for my $alternative ($self->{alternatives}->@*) {
next if $alternative eq$self->{initial};
return $alternative eq$unrevealed;
}
}

package MontyHall;
sub new { bless {}, shift }
sub unchosen ($self,$scenario, $player_choice,$alternatives) {
my (@wins, @loses);
for my $alternative ($alternatives->@*) {
next if $alternative eq$player_choice;
if ($scenario->{$alternative} eq $WIN) { push @wins,$alternative;
}
else {
push @loses, $alternative; } } return (\@loses, \@wins); } sub reveal ($self, $scenario,$player_choice, $alternatives) { ... } package RandomMontyHall; use List::Util 'shuffle'; use parent -norequire => 'MontyHall'; sub reveal ($self, $scenario,$player_choice, $alternatives) { my$n_unchosen = $alternatives->@* - 1; my ($unchosen_loses, $unchosen_wins) =$self->unchosen($scenario,$player_choice, $alternatives); my @loses = shuffle($unchosen_loses->@*);

# reveal exactly n-1 of the unchosen doors!
my @revealed = splice @loses, 0, $n_unchosen - 1; my ($unrevealed) = (@loses, $unchosen_wins->@*); return(\@revealed,$unrevealed);
}

package OrderedMontyHall;
use parent -norequire => 'MontyHall';
sub reveal ($self,$scenario, $player_choice,$alternatives) {
my $n_unchosen =$alternatives->@* - 1;
my ($unchosen_loses,$unchosen_wins) =
$self->unchosen($scenario, $player_choice,$alternatives);
my @loses = $unchosen_loses->@*; # NO SHUFFLING HERE!!! # reveal exactly n-1 of the unchosen doors! my @revealed = splice @loses, 0,$n_unchosen - 1;
my ($unrevealed) = (@loses,$unchosen_wins->@*);
return(\@revealed, $unrevealed); }  This program implements different strategies on the player and on the host side. There are four players: • StubbornPlayer: they always stick to their initial choice; • MathsPlayer: they always switch; • RandomPlayer: they switch half of the times, randomly; • ABCPlayer: they adopt a strategy to switching. There are two hosts: • RandomMontyHall: if one of the two unchosen door contain a car, they reveal the other one; if they both contain a goat, they reveal either one randomly; • OrderedMontyHall: if one of the two unchosen doors contain a car, they reveal the other one; if the both contain a goat, they reveal the “first” one in door order. It’s easy to see, in the code, that the initial choice of the player is always out of three doors, so their odds of taking the car is 1:2 (1 in 3 chances). This does not change whatever the host that is chosen, because the array of unchosen alternatives is scanned to make sure that only goats are revealed. Hence, what remains at the end is the distillation of unchosen possibilities. Running the simulation with the stubborn and the maths inclined players and a truly random Monty Hall confirms our initial guess: switching is beneficial: $ ./monty-hall.pl StubbornPlayer RandomMontyHall
...
What a season! Players won 35.8% of times!

$./monty-hall.pl MathsPlayer RandomMontyHall ... What a season! Players won 68.2% of times!  People that think that odds have gone down to 1:1 after the revelation are “rightish” in the sense that a random choice between the two doors yields a car half of the times: $ ./monty-hall.pl RandomPlayer RandomMontyHall
...
What a season! Players won 50.2% of times!


This happens because at that point there are only two doors, one with a car and one with a goat (by construction), so a random 1:1 choice between them gives a car 1:1. I’d argue that this is throwing away all the information that we have that lead to how these two doors were actually selected to be closed at the end, which is a shame because this information can definitely give an edge.

Last, it’s important that the choice of the door to open by the host when the player chose the car is random. If the host decides to always open the “first” one (whatever order can be given to the doors), then the player might use this information the their advantage and choose to switch whenever this reveals something:

\$ ./monty-hall.pl ABCPlayer OrderedMontyHall
...
What a season! Players won 67.4% of times!


Contrarily to the other case, in this case the player only switches in 1 out of 3 cases when they’re sure about the car position, and keeps in the other two so, in a sense, they’re more stubborn than inclined to change. BUT they know exactly when it’s best to change, shifting the odds in their favor by taking advantage of the deterministic way of choosing the door to open by the host.

Well… this is my take on the Monty Hall problem, and I’ve expressed in code so there is no ambiguity as to what I mean about the actions of the different actors!