Monty Hall Problem

This problem is actually a little tricky and counterintuitive, but really fun to think about (as a brain teaser).

Here is the problem (I’ll just copy it from Wikipedia, lazy evil being :)

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?

At first sight, it appears to be a fifty-fifty chance to win without switching, as there’re only two doors left, and behind one there’s the car. Most people do think so (me too). But there is a piece of important information that people may not notice, the host knows what’s behind each door. On your choosing (a 1/3 probability of the car and a 2/3 probability of a goat), he purposely opens a door that has a goat. Actually, when you choose a door, you only have a 1/3 chance to win the car. The car behind one of the other two doors has a 2/3 chance. This won’t change with or without the host telling ya one of the other two has a goat. If the host opens a door with a goat, that means the other door left has a 2/3 chance to have the car. Thank you for that extra 33.3%!!

Mathematically speaking, we denote Ga, Gb the two goats. When choosing one door, say door 1, it has a 1/3 chance that the car is right behind it. But more likely (a 2/3 probability) it’s a goat behind it. It can be Ga or Gb. When it’s Ga, the host opens one of the other two doors behind which Gb is. When it’s Gb, the host opens the door where the one remaining goat Ga is. If now you decide to switch, then you get the car. There’s a 2-in-3 chance of happening (on choosing Ga or Gb, the car may be behind one of the other 2 doors, thus we have 2 x 2 = 4 cases, divided by the total number of permutations which is P(3, 3) = 6 cases). But if you were lucky and picked the door with the car (with a 1/3 probability) at the very beginning, and then decide to switch to the other remaining door, you lose. But there’s just a 1-in-3 chance (1 x 2 = 2 divided by P(3, 3) = 6). So, statistically, if you continue to play this game for, like 300 times, you’ll probably win about 200 times and lose only about 100 times. Overall, the odds are in your favor if you do the switch. (I also found a nice picture on Wikipedia, so again, I’ll shamelessly steal it)

But if there’re, say 100 doors, and only behind one of them has a car. And you’ve just picked your lucky number, door No. 13, but still, you are pretty much damn sure about it that the chance you win will be very very slim (1/100). Now if the host opens the other 98 doors, all with goats (why the hell would he do so??), and let you decide if you will switch your choice. Ha, this is a no-brainer, of course, I’ll f*$#ing do it, just switch!

Anyway, the most important thing is the host has the information (the power you want :) If he doesn’t know where the car is and randomly opens a door you didn’t pick, then you’ll get a 50% chance of winning. But that’s not how the game plays, he may open a door with the car (Ooops)! Also, it’s different from that first let the host opens a door with a goat and then let you choose one (you just have two choices then), it’s really a 50% win.

Interestingly and historically, this problem was made popular in a letter from reader Craig F. Whitaker, quoted in Marilyn vos Savant‘s “Ask Marilyn” column. There’s a great youtube video about it, see

Recommended watching & reading:

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