Probability – Monty Hall problem
February 27, 2011 Leave a comment
Our friend Monty Hall is a game show host. Every week he asks one person from his audience to come onto the stage. On the stage are 3 closed doors. Behind one is a brand new car, and behind the other 2 are goats. The car is equally likely to be behind any of the 3 doors. The contestant selected from the audience obviously wants to win the car.
Monty Hall explains the rules of the game to the contestant — In front of you are 3 closed doors. Behind one of them is a car, and there are goats behind the other 2. I know what is behind each door. Your first task is to select a door.
Our contestant selects door 1.
Monty now again starts expounding on probability — You have selected door 1. Now, I will select either door 2 or 3 to open. Note that I know what is behind each of the doors. I will decide which door to open based on the following rules:
– If door 2 has a car behind it, I will open door 3.
– If door 3 has a car behind it, I will open door 2.
– If neither of the above conditions hold true, i.e. both doors 2 and 3 have goats behind them, I will randomly select one among these 2 doors with probability 0.5 and open that door.
Now, having explained thus to our contestant, the omniscient Monty goes and opens door 3, and reveals to the world (including our contestant) the goat standing behind it.
Now, Monty turns to the contestant for the endgame and speaks thus — As you can can see, door 3 has a goat behind it. Doors 1 and 2 are still closed. Whichever door you select, you will get whatever is behind that door. You can either decide to stay with your original choice i.e. door 1, or you could decide to switch i.e. select door 2. What do you want to do — stay or switch?
As the music in the recording studio rises to a crescendo, our dear contestant is faced with this question on probability to answer. What should he do?
At first glance, it does seem that it is immaterial as to what the contestant chooses. One could argue that there are 2 doors (doors 1 and 2) one of them having a goat behind it, and the other having a car behind it. The car is equally likely to be behind either of the 2 doors. So, the contestant could as well flip a fair coin, and choose which door he wants to go with.
However there is a flaw in the above reasoning.
— The probability that there is a goat behind door 1 is 2/3, and the probability that there is a car behind door 1 is 1/3.
To see why this is so, consider the following argument — Initially there were 3 doors. Two of them had goats behind them and one of them had a car behind it. Door 1 was one among these 3 doors. Therefore, the probability that door 1 has a car behind it is 1/3 and the probability that it has a goat behind it is 2/3.
Now, you could argue against the above explanation this way : ” What you said about the probability of a car being behind door 1 being 1/3, and that probability of a goat being behind door 1 being 2/3 was true initially, and not after Monty went and opened door 3, and showed everyone the goat standing behind it. Once Monty did that, we were left with 2 doors, and 1 goat and 1 car, and hence the probability that there is a car behind door 1 is 1/2 and the probability that there is a goat behind door 1 is also 1/2.”
But again, the argument that you presented above is not wholly correct. We can see it this way — we agree that initially i.e. before Monty opened door 3 and showed a goat behind that, the probability of these being a car behind door 1 is 1/3. My claim is that this remains the same even after what Monty did.
To better understand that, consider the following situation —
Assume that there are 100,000 doors instead of just 3. Also, behind 99,999 of these doors are goats and behind a single one is a car. The car is equally likely to be behind any of the 100,000 doors. So, what we have here is a 100,000-door version of Monty’s 3-door problem.
Now, you select a door, say door 385.
Now, Monty has been asked to open 99,998 doors, i.e. he has been asked to keep 2 doors closed. It is also specified in the rules that one of these closed doors should be door 385 since that was the one you selected. Further, Monty knows what is behind each of the doors, and Monty has been specifically instructed that the 99,998 doors that he opens should all reveal goats behind them.
So Monty goes and opens 99,998 doors. He leaves door 385 and one other door, say door 6724 closed. Now, as we said earlier, behind each of the 99,998 doors that Monty opened, was a goat.
Now, what do you think is the probability that there is a goat behind your selection door 385?
It is 1/100,000 and not 1/2.
The reason as to why door 385 remained closed even when just 2 doors remained (i.e. 385 and 6724) was because Monty had been instructed not to open 385 as part of the rules.
There is 1/100,000 probability that door 385 contained a car behind it, and Monty selected door 6724 uniformly at random from the remaining 99,999 doors for keeping closed. In such a situation, staying with your original selection (i.e. door 385) would win you the car.
However, it is much much more likely (99,999/100,000) that your initial selection (385) had a goat behind it, and that Monty knowing that 6724 had a car behind it, opened the remaining 99,998 doors. In this situation, switching (i.e. selecting door 6724) would get you the car.
Coming back to our 3 door version, we can argue that switching would win you the car with probability 2/3, while staying with your original selection would get you the car with a probability of 1/3.
Hence, one who has a working knowledge of probability should advise the contestant to switch.