Monty Hall Paradox explained

This is a well known problem based on a 1960s US television game show called Let’s Make a Deal. How closely it resembles that particular show, I don’t know, but it’s not relevant, because it’s easy to imagine. The show’s host’s stage-name was Monty Hall, hence the name of the puzzle.

 

In 1975, an American statistician and professor at the University of California, Berkeley, Steve Selvin, published a short article on the Monty Hall Paradox in The American Statistician, which he saw as a curiosity for a very select group who would appreciate its quirkiness and counter-intuitive answer. He received some criticism, which he easily countered.

 

Another totally unrelated (weekly) periodical, Parade magazine, with a circulation in the tens of millions, had a column called Ask Marilyn, who specialised in solving mathematical puzzles, brain teasers and logical conundrums sent to her by readers. She was Marilyn vos Savant, and entered the Guinness Book of Records in the 1980s as the woman with the highest recorded IQ (185). I obtained all this information from Jim Al-Khalili’s book, Paradox; The Nine Greatest Enigmas in Physics.

 

Someone sent Marilyn the Monty Hall puzzle and she came up with the same counter-intuitive answer as Selvin, but she created an uproar and was ridiculed by mathematicians and academics across the country. Al-Khalili publishes a sample of the responses, at least one of which borders on misogynistic. Notwithstanding, she gave a more comprehensive exposition in a later issue of Parade, emphasising a couple of points I’ll come to later.

 

Now, when I first came across this puzzle, I, like many others, couldn’t understand how she could possibly be right. Let me explain.

 

Imagine a game show where there is just a contestant and the host, and there are 3 doors. Behind one of the doors is the key to a brand-new car, and behind the other 2 doors are goats (pictures of goats). The host asks the contestant to select a door. After they’ve made their selection, the host opens one of the other 2 doors revealing a goat. Then he makes an offer to the contestant, saying they can change their mind and choose the other door if they wish. In the original scenario, the host offers the contestant money to change their mind, upping the stakes.

 

Now, if you were a contestant, you might think the host is trying to trick you out of winning the car (assuming the host knows where the car is). But, since you don’t know where the car is, you now have a 1 in 2 chance of winning the car, whereas before you had a 1 in 3 chance. So changing doors won’t make any difference to your odds.

 

But both Selvin and vos Savant argued that if you change doors you double your chances. How can that be?

 

I found a solution on the internet by the Institute of Mathematics, giving a detail history and a solution using Bayes’ Theorem, which is difficult to follow if you’re not familiar with it. The post also provides an exposition listing 5 assumptions. In common with Al-Khalili, the author (Clive Rix from the University of Leicester), shows how the problem is similar to one posed by Martin Gardner, who had a regular column in Scientific American, involving 3 prisoners, one of whom would be pardoned. I won’t go into it, but you can look it up, if you’re interested, by following the link I provided.

 

What’s important is that there are 2 assumptions that change everything. And I didn’t appreciate this until I read Al-Khalili’s account. Nevertheless, I found it necessary to come up with my own solution.

 

The 2 key assumptions are that the host knows which door hides the car, and the host never picks the car.

 

So I will describe 3 scenarios:

 

1)    The assumptions don’t apply.

2)    We apply assumption No1.

3)    We apply assumptions 1 & 2.

 

In all 3 scenarios, the contestant chooses first.

 

In scenario 1: the contestant has a 1 in 3 chance of selecting the car. If the contest is run a number of times (say, 100 or so), the contestant will choose the car 1/3 of the times, and the host will choose the car 1/3 of the time, and 1/3 of the time it’s not chosen by either of them.

 

Scenario 2: the host knows where the car is, but he lets the contestant choose first. In 1/3 of cases the contestant chooses the car, but now in 2/3 of cases, the host can choose the car.

 

Scenario 3: the host knows where the car is and never chooses the car. Again, the contestant chooses first and has a 1 in 3 chance of winning. But the host knows where the car is, and in 1/3 of cases it's like scenario 1. However, in 2/3 of cases he chooses the door which doesn’t have the car, so the car must be behind the other door. Therefore, if the contestant changes doors they double their chances from 1 in 3 to 2 in 3.