# The Monty Hall Problem: You Really Need to Switch

**Story**

Feel free to skip next five paragraphs.

This last Thanksgiving at the turkey table, someone thought it would be fun to quiz people on their game shows knowledge. And it got engaging as soon as it started. As this trivia went on, I searched my mind for a game show I could ask about, which is funny because I don’t really watch game shows. I couldn’t think of many. In fact, I could only think of one. It’s called *Let’s Make a Deal*.

I knew *Let’s Make a Deal* because of the Monty Hall problem, a brain teaser loosely based on the game show*. *The problem* *was something I had been introduced to four years prior in a probability class. At the dinner table, I mentioned the show, described the Monty Hall problem, and asked what everyone would do. Switch doors or nah? Some said they wouldn’t. Some said they would, correctly anticipating it was a trick question.

As I opened my mouth to explain why switching is the way, I remembered something. Actually, two somethings: i) I was rusty on the logic and the math, and ii) not everyone had a mathy background. So when I finished, my explanation was fuzzy and people were even more confused. ‘Why isn’t it just a 50% chance?’, someone asked, a very common mental block for a lot of people that attempt to understand the problem.

On-and-off, I spent the next day (maybe two, I forget) reading on the problem and explaining it in all sorts of ways. I became unreasonably invested in it (I am blogging about it, ain’t I?). Finally, after sitting on my knees for too long, pointing and scribbling on a piece of paper, somebody’s face brightened. Someone finally got it! It’s not a 50–50 chance of winning.

My non-mathy explanation made sense, so I decided to share it here with everyone else. Hopefully it makes sense to you too :)

## Description of the problem (straight out of Wikipedia)

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say №1, and the host, who knows what’s behind the doors, opens another door, say №3, which [always] has a goat. He then says to you, “Do you want to pick door №2?” Is it to your advantage to switch your choice?

## Let’s reason with 100 marbles (no math!)

A lot of people that come across the Monty Hall problem have a hard time understanding why the chance of winning is not 50%. That’s the first thing they think of. I know I did. To see why it’s not, let’s not think about 3 doors. Let’s do 100 doors. Or actually, let’s think in terms of marbles.

Our 100 marbles are **identical in shape and size,** but one is red and the rest are green. We put all the marbles in an opaque large bag so we can’t see the colors. I then ask you to draw one marble and put it in another smaller opaque bag. You do this without looking, so you don’t know the color of the marble in the smaller bag.

Which of the two bags would you pick if you wanted the bag that was more likely to have the red marble? The large bag or the smaller bag? Hopefully you see why it’s ridiculous to pick the smaller bag (with just one marble!). The smaller bag has a smaller chance of having the red marble (not 50%).

If you see where I am going with this, you can try finish the reasoning on your own and convince yourself. Otherwise, read on :)

Let’s say I (pretend I am the host, ok?) have the larger bag and you’re keeping the smaller bag. I ask you to draw another marble from the bigger bag (now with 99 marbles) and swap it with your older marble. Would you do it? Let’s say you’re not sure. Or let’s pretend you decide not to. (FYI: Swap them.)

To make it more interesting, I try something else: I take the larger bag, turn around so you can’t see, remove and throw away **98** marbles **that are not red**. Each of the two bags has one marble now. Would you give up your older marble for the one marble in the larger bag now? Hint: I am trying to help you win! Reminder: you want the red marble.

We have established that the larger bag has a higher chance of containing the red marble. Me removing 98 marbles that are not red and letting you pick the last remaining marble is essentially the same thing as you picking the larger bag, opening it, and looking for the red marble (with your eyes open!). Again, I am only removing marbles that are not red, so you should be pretty confident that you’re more likely to end up with the red marble if you give up your older marble.

If we pretend for a sec that the red marble was indeed in the larger bag (again, very likely), me removing all non-red marbles leaves you with the red marble if you pick the larger bag.

You can apply the same logic to three doors in the original problem. You select a door. The host reveals the one that has a goat in it (always the goat one!). Should you switch to the third door? You can pretend that you and the host are on the same team. Your team picks two doors instead of one (larger bag). Your teammate, the host, tells you which door has the goat (removes the 98 marbles) and lets you pick the last unrevealed door (larger bag with one marble).

Hopefully, you can now see that the door you picked initially (or the smaller bag) have a lower chance of having the car (or the red marble). It’s not 50%. Ponder the explanation above a bit more if you need to. Or read on if you want an explanation with more numbers.

## Continuing with 100 marbles (with some math thrown in)

In the larger bag of marbles, any of the 99 remaining marbles could be the red marble. It’s a 99% chance that **any** of the marbles in the larger bag is red. That’s huuuge! On the other hand, there is only a 1% chance it’s in the smaller bag. If I, the host, take the larger bag and remove 98 marbles that are not red, I am making it easier for you to pick a red marble by removing the 98 that aren’t red.

The chance that the remaining marble in the larger bag is red is still 99%.

In terms of three doors, any of the remaining 2 doors could have your next car. Or it’s a 2/3 chance that you will win a car if you give up your first selection and take both of the two remaining doors. I am always opening the door with a goat, which is just like picking both doors and discarding the one without the car. On the other hand, you have a 1/3 chance of winning if you stay with your first door. It’s a 2/3 v. a 1/3 chance of winning. Not 1/2 v. 1/2.

## If this is not a 50–50 chance scenario, what is then?

I will leave this part to you, but I will ask a question to get you started. Consider a scenario where there is initially only two doors, one with a goat and one with a car. You pick your door. I then add a third door, open it, and reveal a goat. Should you switch to the second unopened door? Does it matter?

You can think of this in terms of initial probability distribution and how it affects results.

## If still not convinced:

- Simulate it yourself with code. Thinking through it with code might be the way you arrive at your own conclusion why the math makes sense.
- Play it with a friend! Take three index cards, each representing the three doors and what’s behind them (door number on one side and what is behind the door on the other side). Flip the cards to only show numbers and start playing. Play several times without switching and then several other times with switching. Compare probabilities of winning for the different cases. As you play, you will hopefully see why the math makes sense.