The Monty Hall Problem
I first saw this problem when I was watching the movie ‘21’. There is a short classroom scene where Professor Micky Rosa tests Ben Campbell with what sounds like a simple probability puzzle.
This is their conversation.
Professor: Ben, suppose you're on a game show and you are given a chance to choose from three different doors. Now behind one of the doors is a new car and behind the other two – goats. Which door would you choose?
Ben: Door number 1.
Professor: Ben chooses door number 1! The game show host, who by the way knows what's behind all the other doors, decides to open another door. Let's say he chooses door number 3… behind which sits a goat. Now, Ben – the game show host comes up to you and says, “Ben, do you want to stay with door number 1 or go with door number 2”? Now is it in your interest to switch your choice?
Ben: Yeah.
Professor: Well, wait – remember the host knows where the car is so how do you know he's not playing a trick on you – trying to use reverse psychology to get you to pick a goat?
Ben: I wouldn't really care. I mean my answer is based on statistics; based on variable change.
Professor: Variable change? But he just asked you a simple question!
Ben: Yeah, it just changed everything.
Professor: Enlighten us…
Ben: Well, when I was originally asked to choose a door I had a 33.3% chance of choosing right but after he opens one of the doors and then re-offers me the choice it's now 66.7% if I choose to switch. So yeah, I’ll take door number 2 and thank you for the extra 33.3%.
Professor: Exactly! People, remember if you don't know which door to open always account for variable change. Now see most people wouldn't take the switch out of paranoia, fear, emotion, but Mr. Campbell kept emotions aside and let simple math get his ass into a brand new car!
Here is the analysis:
Initially, each of the doors has a 1/3rd chance – 33.3% probability – of a car being behind it. Then, assume that you have chosen door number 1 like Ben and that the car is behind door number 2. When the host opens door number 3 behind which there is a goat and asks you if you want to switch, one might think that as door number 3 is out of the picture, there is 50-50% chance of the car being behind door number 1 or door number 2.
But this misses the point that the host always knows what is behind each door, and in opening door number 3, the host will always eliminate one “goat door”. Therefore, if the car is actually behind door number 2, and you picked door 1, the host will open door 3, and you would be better off switching to door 2 to get the car. If instead, you had picked door number 3, the host would open door number 1, and again you would be better of switching to door number 2 to get the car. Only if you had picked door 2, and the host would then pick either door number 1 or door number 3, would you would be better off sticking to door number 2.
So sticking would be a better choice only 1/3rd or 33.3% of the time. But 66.7% of the time, you would be better off switching.
The mathematical way of looking at this is using Bayes’ Theorem:
P(A|B) = P(B|A) x P(A)
P(B)
P(A|B): Probability of A given B
P(B|A): Probability of B given A
P(A): Probability of A
P(B): Probability of B
Let’s name the probabilities and events:
The probability of the car being behind door 1 = C1
The probability of the car being behind door 2 = C2
The event HD3 is the Host opening door 3
Then, P(C1|HD3) = P(HD3|C1). P(C1)
P(HD3|C1).P(C1) + P(HD3|C2).P(C2)
P(HD3|C1) = ½ because the host knows what’s behind the doors and would therefore only open door 2 or door 3 if the car was behind door 1
P(C1) = 1/3 as there are 3 doors and the car could be behind any of them
P(C2) = 1/3 as there are 3 doors and the car could be behind any of them
P(HD3|C2) = 1. Because you have opened door 1 which has a goat and he will only open the other door that has a goat, which is door 3.
P(C1|HD3) = 1/2 x 1/3
1/2 x 1/3 + 1x 1/3
= 1/3
The Probability of a car being behind door 2 is therefore 1- probability of the car being behind door 1 given that door 3 has a goat, which is equal to 2/3 or 66.7%
The intuitive 50:50% chance of car behind door number 1 and door number 2 does not apply because that fact that the host will only open a ‘goat door’ turns a case of simple probability to one that requires one to account for variable change.
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