Probability Expert Explains That Brooklyn Nine-Nine Brain Teaser

Probability Expert Explains That Brooklyn Nine-Nine Brain Teaser

During a recent re-watch of Brooklyn Nine-Nine I stumbled across the episode where Captain Holt and his husband were fighting over a brain teaser.

Known as the Monty Hall Problem, it’s quite a well known probability puzzle that involves cars and goats. It’s also quite difficult to wrap your head around, so we asked a probability expert to step in.

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We called upon the expertise of Dr Stephen Woodcock, Senior Lecturer in Mathematical Sciences at University of Technology Sydney to help with this one.

Dr Woodsock is also an Associate Investigator of the ARC Centre of Excellence for Mathematical and Statistical Frontiers (ACEMS) and frequently writes for The Conversation. You can find one of his articles on probability right here.

What is the Monty Hall problem?

Named after the host of U.S. game show, Let’s Make A Deal, it was kicking around long before it appeared on Brooklyn Nine-Nine.

It goes a little something like this:

You’re on a game show where you’re presented with a choice of 3 doors.

Behind 1 of the doors there’s a car, behind the other 2 doors are goats.

Let’s say you pick door number 1 but don’t open it yet. The host, who knows what is behind each door, opens door number 3, which reveals a goat.

The host then asks if you want to switch your choice to door number 2. Is it to your advantage to change your choice?

What does the show say?

In Brooklyn Nine-Nine everyone but Holt says that you have a better chance of winning the car if you switch your choice.

But some of us less mathematically minded people (including myself) had a little trouble wrapping their heads around why.

So why is it better to switch your choice?

This is what Dr Woodcock had to say about it:

“The critical fact at the heart of many people’s confusion is that the host knows where the car is. As such, the host’s response to the initial choice is giving the contestant additional information.

If one door randomly popped open instead there would be no advantage to switching choice. In this alternate case, we would be correct to stick with our choice 1/3 of the time, we would be correct to switch 1/3 of the time and the game would have been ruined (by the car being revealed) 1/3 of the time.

But this is not the case in the Monty Hall problem. The host’s decision of which door to open mean that the game is never ruined by revealing the location of the car.

The shortest explanation is the simplest. The only way that sticking with your original choice is the right option is if you chose correctly initially. This happens 1/3 of the time. If you chose a losing option initially, the right option is to switch. This happens 2/3 of the time.”

I want a more in-depth explainer

If you’re looking for a longer explainer, Dr Woodcock is more than happy to oblige.

He was kind enough to provide a number of scenarios to further demonstrate why switching offers you a 2/3 chance of winning, while sticking with your original choice only gives you a 1/3 chance.

“An issue people sometimes overlook is that the host of the show has two options if the contestant chooses the car initially (either of the two goats could be revealed) but only one option if the contestant chooses a goat initially.

The host needs some way of deciding between these two options. One way to picture this is for the host to label the goats as Goat A and Goat B and always open the door to reveal Goat B when both options are available.

With three options (Car, Goat A, Goat B) behind the three doors, there are six equally likely permutations for the layout – three choices for behind the first door, multiplied by two remaining choices for the second door, multiplied by one remaining choice for the third door.”

When the contestant chooses one door and the host opens one losing door, we see that three of the six options are eliminated. This also reveals that there is only one scenario where sticking with the original choice is correct.

Say, for example, that the contestant initially chooses Door One and the host then opens Door Two. This eliminates both Scenario Three and Scenario Four (as we have seen a goat behind Door Two and hence it does not contain the car) and also eliminates Scenario Six – as the host would not have revealed Goat B when the option to show Goat A still existed.

Of the three equally likely scenarios left, two of them have switching as the better option.”

So there you have it – while you still could win the car by sticking to your original choice, sheer probability dictates that you’re more likely to win if you switch.

A big thanks to Dr Stephen Woodcock and UTS for the explainer and ensuring the my eyes no longer glaze over when I try to work this out in my head.

This post was originally published on November 6, 2019.

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