Imagine the scenario; you find yourself on a garish and brightly-lit game show set. Before you stand three identical doors. A beaming game-show host informs you that behind one of the doors is a brand new sports car. Behind the other two, some rather obnoxious farm-yard livestock. The game-show host asks you to select a door…

The situation sounds fairly straightforward until the game-show host opens one of the doors to reveal a goat. He then asks you would you like to switch doors…

Whilst the choice may seem like it could be described by a simple 50/50 selection, the truth is actually far more non-intuitive and forms the backbone of one of mathematics’ most devilish and divisive puzzles.

The garish set from television’s ‘Let’s make a Deal’ hard to believe that one of mathematics’ most devilish problems arose from this studio

On 30th September 2017, legendary US television host Monty Hall passed away in his Beverly Hills home aged 96. Although he was the host of several popular game-shows including “Let’s make a deal”, “Split Second” and “Chain Letter” and producer of a multitude more, Hall’s legacy ultimately may lie in the field of mathematics rather than that of entertainment. This is due to his association with “The Monty Hall Problem” or the “stay or switch dilemma” a probability puzzle based on his show “Let’s make a deal”, which caused much controversy amongst mathematicians and statisticians throughout the 1990’s as a result of its deeply counter-intuitive nature.

First introduced in 1975 in a statistical study, the Monty Hall problem was popularised by Marilyn Vos Savant in her “Ask Marilyn” newspaper column. The problem places you in the shoes of a hypothetical contestant on “Let’s make a deal” and challenges you to use mathematics to ensure you walk away with a star prize, whilst avoiding two booby prizes.

Let’s make a deal: The Monty Hall problem

You are faced with three doors, let’s label them A, B and C. Behind two of the doors is a goat, behind the other a top of the line sports car. Obviously, unless you’re quite odd, you want to win the car. Let’s assign probabilities to each door and your chances of a successful result if you were to select that door. It’s easy to see that the chance of success is 1/3 for each door. You are given the choice of doors and you select door A. It doesn’t matter which door you choose, at this stage each is as likely of hiding the car.

So far so simple. Here’s the catch.

At this point, our host Monty, opens one of the other doors, let’s say C, to reveal a goat. The thing to take from this is that at this stage the host always opens an unselected door and this door always reveals a goat.

Now, it gets fun; Monty offers you a choice. You can stick with your initial choice (A) or you can switch to the remaining unopened door (B). So, what do you do?

Stick or switch?

Most people conclude at this stage that they are faced with a similar choice to that they were offered initially, except this time the odds of correctly selecting the automobile is equally distributed across the two remaining doors. You have got a fifty-fifty chance of success if you switch, the same if you stick. There is a ½ chance of the car being behind A and the same for B. It’s just common sense, right?

Marilyn Vos Savant disagreed with this seemingly common sense answer, pointing out that a contestant was more likely to win the car if they switched their choice to the other door (B in our scenario). In fact, they were twice as likely to be driving away in a new car if they changed their choice of door. This created outrage in her readership. Thousands of readers wrote into the paper to criticise or correct Vos Savant, many of these responses were from PhD holders in mathematics and science. In a latter column, Vos Sant stated that 92% of the correspondence was against her in this matter.

Some of the choice responses are below:

“You’re in error, but Albert Einstein earned a dearer place in the hearts of people after he admitted his errors.”-Frank Rose, PhD. University of Michigan

“I have been a faithful reader of your column, and I have not, until now, had any reason to doubt you. However, in this matter (for which I do have expertise), your answer is clearly at odds with the truth.”-James Rauff, PhD. Millikin University

“I am sure you will receive many letters on this topic from high school and college students. Perhaps you should keep a few addresses for help with future columns.”-W. Robert Smith, PhD. Georgia State University

“You are utterly incorrect about the game show question, and I hope this controversy will call some public attention to the serious national crisis in mathematical education. If you can admit your error, you will have contributed constructively towards the solution of a deplorable situation. How many irate mathematicians are needed to get you to change your mind?”-E. Ray Bobo, PhD. Georgetown University

“You made a mistake, but look at the positive side. If all those Ph.D.’s were wrong, the country would be in some very serious trouble.”-Everett Harman, PhD. U.S. Army Research Institute

But they were wrong and Marilyn was right. Here’s how.

The Monty Hall solution

There are multiple ways of addressing the Monty Hall problem, the easiest way being to first establish why the chances of finding the car behind the two doors isn’t a straight 50/50 split. To do this let’s go back to before door C was opened to reveal a rather jolly little goat. This time let’s not just give the probability that the car is behind the door (marked in black) but also the probability that the car is NOT behind the door in question (marked in red).

At this point the switch shouldn’t seem attractive, there’s no benefit to changing. That all changes when Monty opens door C. This action means that there is then still a 2/3 chance that the car isn’t behind door A. This means there is a 2/3 chance the car is behind B or C. The opening of door C hasn’t affected the probabilities associated with door A but it has affected those associated with door B. It’s almost as if door B must carry the full load of probabilities that was previously held by both B and C. This is because C isn’t chosen randomly. There is a 2/3 chance the car is behind the B&C pairing, that burden is left to B. Let’s look at that diagrammatically representing B & C as a pair with joint probabilities.

Now let’s have Monty open door C again and see how that affects the probabilities.

Leaving you with the choice as below. Door A has a 1/3 chance of success and a 2/3 chance of failure. Door B then has a 2/3 chance of success and a 1/3 chance of failure. So a contestant has twice the chance of selecting the car if he or she switches doors from A to B.

Still not convinced? Let’s consider a more extreme case, the game show “Let’s make a deal” from an alternative world where the fundamental difference is TV shows are much longer!

The Hundred Doors Solution to The Monty Hall Problem.

In this hypothetical version of the problem the contestant chooses not from 3 doors but from a hundred, for convenience we’ll label these doors 1–100 and say our contestant selects door 1. She has a 1/100 of selecting the door hiding the car and a 99/100 chance of not selecting a car. Let’s represent that with a stripped-down diagram. Again, black denotes the chance of selecting the car. Red the chance of not selecting the car.

Now alternative universe Monty works his way through doors 2–100 careful revealing bearded goats until the bored contestant is left with two doors. Door 1 which she initially chose and door 2 the last remaining unchosen door. Here’s how the odds stack up at this point.

As you can see it’s almost certain in this case that when Monty asks this contestant to “stick or switch” they would be foolish to stick. The odds of success with the original choice are 1/100 compared to 99/100 chance of success when switching.

But can we evidence this? Let’s leave this very smelly TV studio in our alternative universe and return to our original example to find out.

Proving the Monty Hall problem solution.

One of the easiest ways of verifying that a contestant who switches doors is twice as likely to achieve a successful outcome is to use the mathematical formula known as Bayes Theorem which is used to analyse conditional probabilities, the probability of an event happening given another event has already occurred and is carried out for us in the linked article.

Bayes theorem allows us to define that the probability event (A) will happen given that second event (B) has happened, is equal to the probability of event B will happen given that event A has happened. This is multiplied by the probability of event A occurring divided by the probability of B occurring.

Using this formula gives the success rate of 2/3 for door B as predicted above and by Vos Savant. Also, the results of the original run of the show demonstrated the pattern predicted by Vos Savant, which will please the experimentalists amongst you. If a more formal test is your cup of tea, MIT ran a study in 2005 which found that subjects that switched doors were successful 66.6% of the time, whilst subjects who stuck with their original choice succeeded only 33.3% of the time. The 2/3 versus the 1/3 distribution predicted by Vos Savant and by Bayes Theorem.

Try it yourself

Fortunately, it’s pretty easy to run a simulation of the Monty Hall problem yourself, although if you are performing it in the living room I recommend forgoing the goats. Run the experiment with three bowls and a toy car for example and a subject of course. Don’t let the subject know the nature of the test and document your results in the comments section.

I’m sure you’ll enjoy playing quizmaster almost as much as Monty did.