#!/usr/bin/perl
my $wins = 0;
my $lose = 0;
my $goat;
for (my $i=0;$i<1000000;$i++) {
  my $car = int rand(3);
  my $choice = int rand(3);
  do {
    $goat = int rand(3);
  } while (($goat != $car) && ($goat != $choice));
  if ($choice != $car) {$wins++} else {$lose++};
}
print "WINS = $wins!\nLOSE = $lose!\n";
[download]

takes about 10 wallclock seconds at my machine, while the original takes more than 70 seconds.

Hey cool thanks!


daN.

Monty Hall declared, “Her answer’s right. You should switch.” But, after thinking about it a bit, Monty qualified his response: “If the host is required to open a door all the time and offer you a switch, then you should take the switch. But if he has the choice whether to allow a switch or not, beware. Caveat emptor. It all depends on his mood.”

This is a falacy. It ignores the possibility that the host will pick the door with the car and therebye deprive you of the option of switching your original choice.

Try this simulation and see if you still agree?

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

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

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

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
    my $opened = ( grep{ $_ != $choice } 0 .. 2 ) [ rand 2 ];

      ## If the host picks the car, I lose anyway.
    $loseEitherWay++, next if $doors[ $opened ] eq 'car';

      ## Count my  wins if I stick or switch
    $doors[ $choice ] eq 'car' ? $stick++ : $switch++;
}

printf "Odds of
    Losing either way: %.3f
    Win if you don't switch: %.3f
    Win if you do switch: %.3f\n", 
    $loseEitherWay / $DIVISOR, $stick / $DIVISOR, $switch / $DIVISOR;

__END__
P:\test>384288
Odds of
        Losing either way: 33.370
        Win if you don't switch: 33.307
        Win if you do switch: 33.324

P:\test>384288 -TRIALS=5000000
Odds of
        Losing either way: 33.318
        Win if you don't switch: 33.336
        Win if you do switch: 33.346
[download]

Any bias between the outcomes is statistical variation only.

It runs very quickly because it doesn't loop trying to pick a number that hasn't already been picked. The grep removes the number I picked from the choices, and rand 2 picks between the 2 that are left.

That's nice, but the stipulation of the Monty Hall problem is that he ALWAYS reveals a goat, never the prize.
_____________________________________________________
Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart

I've never seen the Monty Hall problem presented with that stipulation explicit.

That extra information is an assumption that is supposed to be based on your knowledge of how game shows work. In fact based on Monty's knowledge and motivations it is possible that you want to switch (Monty knows where the car is and always shows a goat), it doesn't matter (Monty doesn't know where the car is), or you should stick with your choice (Monty knows where the car is and is trying to keep you from getting it).

That fallacy is the point of the exercise. If you assume that the host will never pick the door with the car (because the host shows you what's behind the door), your chance of picking the car is doubled if you change your mind after seeing the goat-door the host picks.

------
We are the carpenters and bricklayers of the Information Age.

Then there are Damian modules.... *sigh* ... that's not about being less-lazy -- that's about being on some really good drugs -- you know, there is no spoon. - flyingmoose

I shouldn't have to say this, but any code, unless otherwise stated, is untested

your chance of picking the car is doubled if you change your mind after seeing the goat-door the host picks.
Not exactly--the chance goes from 1 in 3 to 1 in 2.

It is stipulated that he will always open a door that you did not pick, reveal a goat, and allow you to try again. Of course if you vary from the proposed problem you will obtain a different solution. I still think it's interesting that changing your choice between to unknowns effects your odds. I understand why, it just seems odd...


daN.

$goat = rand(3);
[download]

$goat = int rand(3);
[download]

Yup I guess I really didn't pay much attention to that part of the code as it doesn't really do anything, it's fixed now regardless though, I could remove the entire do while loop and the code still works fine..


daN.

If you stay with your first choice, you have 1/3 chance of getting the car.

If you switch, there is 2/3 chance of getting the car:
2/3 chance you will pick a goat on the first pick, he reveals the other goat, and then you switch to the car.
1/3 chance you will pick a car, he reveals a goat, and you switch to the other goat.

Your right, Rand really isn't neccessary, and I can remove the do while loop as well, but I think that having them in there helps connect the word problem with the code..after all once someone has thought the problem through enough to realise that the do while is not neccesary they don't need a demonstration program any more to prove it ;).


daN.