# Skyfire Cycle: Brooklyn Nine-Nine gives me a reason to explain the Monty Hall Problem

I’ve been waiting years for this. Full disclosure, this episode(Season 4, Episode 8) aired on 11/29/2016, but I only saw it this last week.

First, for those who might not be super familiar with the episode:

Captain Holt and his husband Kevin ask Amy to settle an argument centered on “an inane math problem” that Kevin brought up at dinner. The Monty Hall Problem gets its name from the TV game show, ** Let’s Make A Deal**, hosted by Monty Hall 1. The scenario is such: you are given the opportunity to select one closed door of three, behind one of which there is a prize. The other two doors hide “goats” (or some other such “non-prize”), or nothing at all. Once you have made your selection, Monty Hall will open one of the remaining doors, revealing that it does not contain the prize. 2. He then asks you if you would like to switch your selection to the other unopened door, or stay with your original choice.

Here is the problem: **Should you switch?**

Short Answer: **Yes, Yes you should!**

Long Answer as follows:

Captain Holt thinks that once the Host opens one of the doors your odds are even since there are only two options left. Kevin believes that it is 1/3-2/3 and that the odds are “locked in” once you make your original choice. Kevin is right. Although, it is purposefully difficult to understand this, due to the way that the game is set up. That is to say, only 3 total options(doors). The best way to think through this problem is to visualize 100 doors!

Now we can walk through the problem again, with Kevin’s (correct) thought process. There are 100 hundred doors, and you get to choose 1. From the Law of Total Probability (the foundation of Bayesian Statistics), we know that the total probability of any event must be equal to 1(100%). You have picked 1 out of 100 doors. Since the act of choosing is the event, your choice sets (or ‘locks in’ as Kevin would say)the probability of you winning. That probability is 1/100 or 1%. And, since there are no ties, the probability of you losing is 99/100 or 99%. So now you have a 1% chance of winning, slim odds certainly. What if you were offered the chance to switch sides? What if you could exchange your door for all of the others? What if you could trade your 1% for the 99%? Well, that’s the Monty Hall problem.

The Monty Hall problem breaks down into a choice of one door/option vs **all of the other doors/options**.

The genius of the game is that, by opening all the doors but one, it convinces you that your odds are even at the end. And it’s a masterpiece of psychological manipulation. Even in our exaggerated example, if, after the choice was made, we removed 98 out of the remaining 99 doors the 50–50 illusion would still persist at the end. And therein lies the solution. If you want to really understand why Kevin is right, imagine having the chance to switch with **all the doors still in play**. Would you want one chance at winning, or two?

# Closing thoughts

1)Brooklyn Nine-Nine is a hilarious show that everyone should watch.

2)There should be more Math puzzles in primetime sitcoms.

3)It was extremely difficult writing about this episode without giving any spoilers.

4)Always switch when Monty Hall gives you the choice!