I recently came across Marilyn Vos Savant's Monty Hall problem, which states: You are at a game show, and you are presented three doors, behind one of the doors there is a car and the other two there are goats. You pick a door, and the gameshow host opens a door that you did NOT pick which contains a goat, you have a better chance of winning the car if you switch your door to the other remaining door. I did not believe this statement when I first heard it, and I actually made a bet that it wasn't true. I then spent some time thinking about it and realized it in fact was true and wrote a little perl program to prove it, you actually have a 66.666...% chance of picking the right door if you switch your answer.. anyways here's the code..
#!/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";
UPDATE: Following Joost's recommendation I have removed my own random wrapper and used int rand instead


daN.

In reply to Marilyn Vos Savant's Monty Hall problem by mutated

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