Tuesday, 03 August, 2004

Here's a little maths exercise called the Monty Hall Problem. Here's the set-up. There's three doors. Behind one of those doors is a great prize, behind the other two doors there's nothing. Monty asks you to select a door you think the prize is behind. Before telling you if you're right or wrong Monty opens a door which he knows the prize isn't behind. With two doors remaining he asks you if you want to stay with the door you selected initially or gamble and switch to the other remaining door.

The Question: Are you more, less or equally likely to win if you switch?

Initially, my instinct told me you were equally likely to win regardless of whether you switched or not. However, The result is that switching doors doubles your chance of winning. The key to understanding why this occurs is to realise that Monty always opens a door he knows is empty and by doing this he's communicating something about where the prize is.

Imagine a ridiculous version of the game where there's a million doors. Monty asks you to select your first door and then he opens every other door except for just one unopened door. Would you switch? Of course you would. Your chances of winning the prize on your first guess are one in a million (literally Emoticon: Wink) but with the second door things are different. You should ask yourself the question why would monty choose to leave that one door unopened out of the million? Well, first he can't open a door which has the prize behind it. Second he has to open 999,998 doors. Third he can't actually open the door that contains the prize. What does that tell you about the probability of winning if you switch?

Taking this insight back to the original problem consider your choices:

  1. Stick with your initial door.
  2. Switch doors after an empty door has been opened

The first option has a success rate of one third because you have a one in three chance of guessing correctly. The second option is a little more difficult to analyse. The best way to do this is to walk through the possibilities.

Imagine we have our three doors: 1,2,3 and I choose door 2 initially when the prize is behind door 3. After telling Monty that my selection is door 2 Monty only has the option of opening door 1. This is because he must show an empty door so he can't open door 2 because it's the door I just picked and he can't open door 3 because that is the door which the prize is behind. If I switched in this case, I win. If I don't I lose.

Next, imagine that I'm still picking door 2 initially but this time the prize is behind door 1. Monty only has the choice of opening door 3 this time since I picked door 2 initially and door 1 is the door the prize is behind. If I switch I win, If I don't I lose.

Finally, imagine that I'm still picking door 2 initially but this time the prize is actually behind door 2. This time Monty can choose either door 1 or 3 to open. If I switch in this case I lose and if I stick I win.

In two thirds of our three cases switching won the prize where-as sticking didn't. Granted, we only worked through the possibilities where I selected door 2 as my initial guess but there isn't anything supernatural about door 2. If you work through the problem with doors 1 and 3 as your initial guess you will find the probability of winning if you switch is still two thirds.

I can understand why people strugle with that because people generally aren't used to something that is counter-intuitive and reject it out of blind faith chiefly it doesn't feel right. It's just that when you do a lot of science you can of get in to bed and make love with that feeling Emoticon: LOL


18:38:56 GMT | #Maths | Permalink
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