Thursday, 2 April 2009

The Monty Hall problem

Another problem posed to Alan Davies (see below) is quite a bit trickier. It was seen on an American TV show and because of that it is called the Monty Hall problem. There are three paper cups covering two farmyard animals and one toy car. You want to choose the car. If you are old enough think of Tommy Cooper and the ‘bottle glass, glass bottle’ routine. You make a choice but then you are given more information and you are shown a farmyard animal from under one of the other cups. Do you stay with your original choice or do you change? The answer is that you change.

There's a model car under one of three cups. You pick a cup. There's a one in three chance that this is the cup with the car, If you decide in advance that you're going to stay with your first choice no matter what, then when the hidden object is revealed there's obviously still a one in three chance that it's the car.

Then the presenter of the trick lifts one of the other cups, showing you what's under it. Since he doesn't want you to find the car, he's going to lift a cup which has a model animal under it. One third of the time he will have a choice - he can lift either of the cups you didn't choose, because there's an animal under both. In those instances, switching results in you losing. But two thirds of the time he has no choice. He has to choose a cup which is different from the one you chose, and which has an animal under it - and there's only one cup fitting that description. In those instances, switching results in winning. So switching results in winning 2/3 of the time, and losing only 1/3 of the time.

That sums it up

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