Video: The Monty Hall Problem

In this video we use probabilities to help us devise an optimal strategy for winning a car in The Monty Hall Problem.

06:39

Video Transcript

In this video, we’ll be looking at the famous Monty Hall problem, which was based on a TV game show called Let’s Make a Deal. The show started in the United States in 1963. But it’s been running in various formats around the world ever since. It was originally hosted by a chap called Monty Hall. And that’s how this problem got its name.

The format is that during the show, people are picked to make trades to win a prize. And they usually end up with a chance to trade a small prize for a much bigger more valuable one, hidden behind a curtain or a door or in a box.

The thing is they may end up getting a fabulous prize, like a new car, or they may end up with a dud prize, like a pile of fake money or in item of dodgy clothing or even a live animal which would be very tricky to take home.

These dud prizes are called zonks. Now the Monty Hall problem takes one of these trade scenarios and asks if you can come up with a strategy to help you to decide how to play. It goes like this.

You’re invited to play the big deal of the day. The host stands you in front of three closed doors and tells you that behind one of them is the brand new car of your dreams and behind the other two are zonks, in this case goats.

The doors are numbered one, two, and three. And all you have to do is choose one of the doors and you’ll get the prize or zonk that lives behind that door. Easy! So you make your choice. For example, you say, “Door Two!”

Now the host knows which door the car is behind. And instead of letting you open door two, he dramatically opens one of the other doors to reveal a goat calmly chewing on some grass [goat bleating]. Now here’s the twist, he says, “Okay, would you like to stick with door two or would you like to switch to the other closed door?”

Well can you change your odds of winning the car by sticking or switching? Does it make a difference? There are two doors left to choose from. And one has a goat, and the other has a car. It’s 50/50, right? Well pause the video now and think about what you’d do before I explain the maths.

Okay, this problem caused a great deal of controversy when it was first published. And lots of people disagreed about the best strategy. So let’s go through it step by step. It seems that you have a choice of two doors. And one’s got a goat. And one’s got a car behind it.

Surely it’s obvious that it makes no difference which one you pick. They’re both equally likely to have the car or the goat behind them. But this isn’t taking the whole story into account. When you announced your choice of door, the host opened a different door that had a goat behind it.

Now the host knows where the car is. So if you chose the door with the car, then they could open either of the other doors to reveal a goat. However, if you chose the door with a goat behind it, then the host had to be very careful to pick the other door with a goat behind it to open.

And this sequence of events has left us in the situation where if you were right about which door had the car behind it to start with, then if you switch doors you will now be wrong. And if you were wrong about which door had a car behind it to start with, then if you switch doors you’ll now be right.

At the beginning, when you are faced with three doors and you chose one at random, there was a 33 and a third percent chance you’d get the car and a 66 and two-thirds percent chance that you get a goat.

One of the three doors has a car behind it. And the other two out of the three have goats. That means if you were to star in the show every week and stick with your original choice every time, then you’d get a car one-third of the time.

However, if you were to star in the show every week and switch your choice, then the times that you are originally wrong about which door hid the car become times you get the car. That’s two-thirds. Now the times that you were originally right about which door hid the car become times you get the goat. That’s one-third.

So the best strategy for getting the car statistically is to switch. You’re increasing your chances of winning a car from 33 and a third percent to 66 and two-thirds percent. That’s a 100-percent increase in your chances of winning.

But we can talk about percentages another time. Now let’s use a probability tree to write all of this down. First let’s consider the strategy where you stick with the door that you originally chose. First you got to choose the door. Now in a third of cases, that door will have a car behind it. And two-thirds of cases, it will have a goat behind it.

Now given that you pick the car first time, if you stick with that choice, there’s a probability of one you’ll keep the car and there’s a probability of zero that you’ll get a goat. But if the door you originally chose had a goat behind it and you stick with that choice, there’s a zero probability that you’ll end up with a car and absolute certainty that you’ll end up with a goat.

Now if we multiplied these conditional probabilities together along the branches, we find that the probability that we started off choosing a car and ended up with a car on this strategy of sticking is a third. The probability of starting off choosing a car and ending up with that goat is zero. The probability of starting up with a goat ending up with a car is zero. And the probability of starting off with a goat ending up with a goat is two-thirds.

This means the overall probability of ending up with a car is one-third plus zero, which is a third. And the overall probability of ending up with a goat is zero plus two-thirds, which is two-thirds. So let’s make a note of that over here.

If we stick, the probability of getting a car is a third and the probability of getting a goat is two-thirds. Now let’s consider the strategy of switching. When you choose a door at random, there’s still a probability of one-third of that being a car behind it and two-thirds of that being a goat behind it.

Now because the host revealed the other door, it means if the door that we chose had a car behind it and we switch, we’ll now definitely have a goat and we definitely won’t get the car. And if we chose the door with a goat behind it and we switch, we definitely will get the car and we definitely won’t end up with a goat.

So with the switching strategy, the probability of ending up with a car is zero plus two-thirds and the probability of ending up with a goat is one-third plus zero, which is one-third. So the strategy of switching leaves you with the best probability of ending up with a car.

Now does this strategy mean that you’ll always win the car? Well no it doesn’t. But it does mean that you win one two- thirds of the time rather than only one-third of the time. So that’s gotta be worth doing. To sum it all up then, a little bit of mathematical thinking can help you devise the best strategy to win a car [goat bleating].

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