好好聽的英式英文哦! 同學應該把它耐心地聽完, 練練聽力, 你會發現作者 James Newcombe 與 Ron Clarke 對 Monty Hall problem 的直覺解釋方式, 與我在課堂上說的一模一樣!

下面是英文的劇本全文, 同學對照著看, 練練聽力吧!

The Monty Hall Problem  

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 prize, a car. Behind the other two doors are booby prizes, two goats. You have no way of knowing which door conceals which item and whichever door you pick you’ll receive the prize 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’d 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’ll give you ten 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 it makes no difference whether you swap or not. Now this sounds perfectly sensible, however it is not correct.  

The Monty Hall Problem is a puzzle about probability. The problem is simple to understand, but the answer is counter-intuitive. 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 easiest 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 chose not to swap. At the start of the game you are asked to pick a door. Since there are three doors and only one hides the car the probability of you picking the car is one-in-three, or about 33%. And since there are 2 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 door the host opens, because you are sticking with your first choice and the chance that you’ve 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.  

So what about if you pick a goat first time? Well here’s is the crux of the problem. This time there is only one goat the host can reveal. The host opens the only other goat 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’ll 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?

ps. 這段影片由作者 James Newcombe 先生 (from England)  親自提供, 非常感謝!