Ron Clarke: Hello and welcome to the Monty Hall Problem, with me, Ron Clarke. Imagine you are on a game show, the game show host shows you three doors, behind one of the doors is the star price, a car, behind the other two doors, are booby prices, two goats. You have no way of knowing which door conceals which item, and whichever door you pick, you will receive the price behind it. You are asked to pick a door, but before it is opened, the game show host opens one of the other two doors.
Now the host knows where the car is, and he always opens a door to reveal a goat, you are then asked whether you would like to swap your chosen door for the one remaining closed door? The question is, should you swap? Should you stick with your original choice, or does it make no difference, what you do? Which would give you the greatest chance of winning the car? I will give you 10 seconds to think about it.
So what do you think? Now most people will say, that it makes no difference whether you swap or not, behind one closed door is a goat, and behind the other closed door is the car. Therefore the chances of choosing the car are 50-50 so that makes difference whether you swap or not. Now this sounds perfectly sensible, however it's not correct. The Monty Hall Problem is a puzzle about probability, the problem is simple to understand, but the answer is counterintuitive.
So what should you do? The answer is, you should always swap as this gives twice the chance of winning the car, why? Well, there are many different ways to explain why, but perhaps the easies,t is to examine what your chances of winning the car are for the two strategies, swapping and not swapping. Let's start by looking at what happens if you choose not to swap. At the start of the game you are asked to pick a door.
Since there are three doors, and there is only one hide to the car, the probability of you picking the car is one in three, or about 33%, and since there are two goats, the probability of you picking a goat is two in three, or about 66%. Now if you don't swap your door, it doesn't matter which other goat's door the host opens, because you are sticking with your first choice, and the chance that you have already picked the car, is 33%, and the chance that you've already picked a goat is 66%.
So by not swapping, you have a 33% chance of winning the car, and a 66% chance of winning a goat. Now let's look at the consequences of swapping. Let's consider what happens if by luck, you pick the car first time, a 33% chance. It's obvious that if you pick the car on your first go, and then swap, you are going to end up with a goat. So if you swap, you are going to win a goat at least 33% of the time. What about if you pick a goat first time? Well, here is the crux of the problem, this time there is only one goat, the host can reveal.
The host opens the only other goat's door, and then you swap to the remaining closed door, the car. In fact, every time you pick a goat door first time and then swap, you will win the car, and the chances of you picking a goat first time, are 66%. So by swapping you have a 33% chance of winning a goat by picking the car first time and a 66% chance of winning the car by picking a goat first time. So you should always swap to the remaining door, why? Because if you do, you will have a 66% chance of winning the car, and only a 33% chance, if you don't. And that is double the chance.
I hope this explanation makes sense, and that you can see the truth behind the Monty Hall Problem. The only remaining question is, do you actually want to win a car, or would you rather win a goat?
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