| This three door problem supposedly came from an American game show and I first encounted it during a fun puzzles lesson in maths when I got the answer wrong. Having not recieved a suitable explaination for the correct answer I forgot about it until I reread the problem in 'How Long Is A Piece Of String' a brilliant popular maths book by Rob Eastaway and Jeremy Wyndham. I highly recommend you read this book as it explores many areas of life that are governed by mathematics and it isn't a very challenging read. The explaination I am using for the problem is not the one covered in this book but rather one I came up with when explaining the solution to someone else. However this is more than enough blab on with the problem. |
| You have just won a gameshow and are now on the last round to collect the prize. There are three doors, behind one is the prize, which for convenience I will say is a car and behind the other two is a piece of junk, say a dustbin. The doors are labelled 1,2 and 3. The host asks you to select a door and you choose door 3. To increase the tension he opens one on the doors you didn't choose revealing a dustbin. He then asks you if you want to switch doors. Should you? |
| At first glance it doesn't seem to matter if you switch doors. Since there are two doors and one prize the chance that it is behind the door you have chosen is 1/2 and the chance that it is behind the other door is 1/2. Don't worry if you can't see this. This is how I first oversimplified the problem and it is completely wrong. |
| Let us look at the problem from the beginning |
| There are three doors 1,2 and 3 and the car is behind one of them. If we list the possible arangments for the position of the car behind the doors we get: |
| Door number: 1 2 3 Arrangment 1 C B B Arrangment 2 B C B Arrangment 3 B B C |
| Where B represents bin and C represents car. |
| Now since we know nothing at the beginning is is fair to conclude that the probability that it is arrangment 1 is equal to the probability that it is arrangment 2 and that is equal to the probability that it is arrangment 3. Since there are 3 equally likely outcomes and it is certain that is one of them the probability of each one is 1/3. |
| Now we get into the game show You have chosen door 3. The probability that the car is behind the door you have chosen is 1/3. The probability that it is behind one of the other two doors is 2/3. The probability it is each arrangment above is 1/3. Now the host comes to open the doors. The host however is not an innocent bystander. He knows what door the car is behind. If he didn't the chance that the car is revelaed when he opens the door to increase the tension is 1/3. That would ruin the show. The car can't be revealed until the end. This is the important point that make the 1/2 chance theory incorrect. Let's return to the arrangments. If the car is in arrangment 1 the host will open door 2. No car will be revealed. If the car is in arrangment 2 the host will open door 1. No car will be revealed. If the car is in arrangment 3 the host can open either door 1 or 2. Still no car revelaed. Rembering you choose door 3 originally. If you decide to switch doors then you would win in arrangment 1 or 2, since the car is behind the other closed door. Both arrangments occur 1/3 of the time so the total proability of one of those two events occuring (and you winning if you switch doors) is 2/3 If you decide to stick to your door then you would win in arrangment 3. The probability of this occuring is 1/3 The proabilities are different. You are more likely to win the game if you switch doors. The answer of course is yes. |
| The Door Problem |