I watched the movie "21" the other night and although I thought it was rubbish, I was slightly intrigued in the explanation to a question posed in the movie, which turns out to be the Monty Hall Problem.
Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?
It turns out that I've always hated this type of math despite generally enjoying using math to solve engineering problems, having a pretty good understanding of No-Limit Texas Hold'em and perpetually trying to crack all those math tricks that make you do a series of seeded numerical operations to magically reveal some significant number. For me the probability math in quantum physics was bewildering and in statistical process control it was incredibly boring as well.
Asked this question last week and I would probably have meekly replied, "Umm, maybe 50%?" and mumbled something about "stupid probability questions" then maybe attempted to draw out some permutations before giving up with absolutely no confidence in my answer.
Having no confidence in my answer may have been wise in that case, but in his following post about the answer to a similar problem, Jeff made a point that resonated with me:
You don't need to be a mathematician to prove this. I'm just a crappy programmer, and even my crappy code can brute force the answer by simulating results from thousands of games.
As soon as I read that I decided I was going to give the Monty Hall Problem the same treatment, at that stage I'd only heard the explanation in the movie which didn't make much sense to me.
import random; class GameShow: def __init__(self): self.prizeDoor = random.randint(1,3); print "\nCreating Game, Prize Door is ", self.prizeDoor; def SelectDoor(self, door): self.firstSelection = door; print "Contestant Selects Door", self.firstSelection; def HostOpenDoor(self): options = filter(lambda m: (m != self.prizeDoor) and (m !=self.firstSelection), range(1,4)) self.hostOffer = options[random.randint(0, options.__len__()-1)]; print "Host opens door", self.hostOffer; def ChangeDoor(self, swap): self.swap = swap; print "Contestant", self.swap and "Takes Swap" or "Keeps Original Choice"; def OpenDoor(self): self.alternativeDoor = filter(lambda m: (m != self.hostOffer) and (m!= self.firstSelection), range(1,4)); self.selectedDoor = (self.swap and self.alternativeDoor or self.firstSelection); self.wins = (self.selectedDoor == self.prizeDoor); print "Contestant Selects", self.selectedDoor, "and", self.wins and "Wins Prize" or "Looses"; def RunGame(swap): game = GameShow(); game.SelectDoor(random.randint(1,3)); game.HostOpenDoor(); game.ChangeDoor(swap); game.OpenDoor(); return game.wins; def RepeatRun(swap, repeats): wins = 0; for i in range(repeats): if (RunGame(swap)): wins += 1; return wins; if __name__ == "__main__": iterations = 1000; swapWins = RepeatRun(True, iterations); stayWins = RepeatRun(False, iterations); print "\nOverall Results\n"; print "Swap Wins",swapWins/float(iterations),"%"; print "Stay Wins",stayWins/float(iterations),"%";
After some fooling around with Python I got my brute force results that proved to me making the swap was the right choice. Actually I'd proved it to myself by the time I'd written the code that works out which doors the host can open. It turns out that if you select a door with a goat behind it first, the host can only open one door to reveal a goat, the other is the car. And that is the key; If your first selection is a goat which is a 2/3 chance, then the switch will always yield a car (this is better than the 1/3 chance first selection).
I've satisfied my interest in counter intuitive probability problems for now, which is good because my friends are sick of me talking about them.