The Monty Hall Problem

<p>I'm sure some of you guys have seen this, but for the people that haven't, this should be interesting.</p>

<p>"This story is true, and comes from an American tv game show. Here is the situation. Finalists in a tv game show are invited up onto the stage, where there are three closed doors. The host explains that behind one of the doors is the star prize - a car. Behind each of the other two doors is just a goat. Obviously the contestant wants to win the car, but does not know which door conceals the car.</p>

<p>The host invites the contestant to choose one of the three doors. Let us suppose that our contestant chooses door number 3. Now, the host does not initially open the door chosen by the contestant. Instead he opens one of the other doors - let us say it is door number 1. The door that the host opens will always reveal a goat. Remember the host knows what is behind every door!</p>

<p>The contestant is now asked if they want to stick with their original choice, or if they want to change their mind, and choose the other remaining door that has not yet been opened. In this case number 2. The studio audience shout suggestions. What is the best strategy for the contestant? Does it make any difference whether they change their mind or stick with the original choice?</p>

<p>The answer to this question is not intuitive. Basically, the theory says that if the contestant changes their mind, the odds of them winning the car double. And over many episodes of the tv show, the facts supported the theory - those people that changed their mind had double the chance of winning the car.</p>

<p>Why should this be so? After all, the contestant doesn't know which door the car is behind, and so the chance of the car being behind any one particular door is one third, isn't it? So surely the chance of winning the car if they stick with their original choice is one third, and the chance of winning the car if they change their mind is also one third? How can the odds double?</p>

<p>The answer goes like this. When the contestant makes their first choice, the chance of them being correct is indeed one in three, or one third. And if, after the host has done his patter and opened another door, they stick with their original choice, then their chance of being correct is unaltered, it is one third. Indeed, how could this possibly change?</p>

<p>We get some pretty abusive e-mails here at Grand Illusions telling us we are talking rubbish. But before sounding off at us, take note - this is an old mathematical puzzle, and has been published in probability textbooks for over 100 years. However, as we said, it is counter-intuitive, and the maths that most people do in school does not cover this.</p>

<p>This is not an example of simple probability (suppose there are two doors, therefore there is a 1 in 2 chance of the car being behind either of the doors). This is an example of conditional probability: what is the chance of something happening, given that something else already has."</p>

<p><a href="http://www.grand-illusions.com/monty.htm%5B/url%5D"&gt;http://www.grand-illusions.com/monty.htm&lt;/a&gt;&lt;/p>

<p>...i don't see where you want this to go...</p>

<p>This is indeed an old problem and I am sure many people have heard of it...the probability of winning changes from 1/3 to 1/2 after the choice is switched. There are a myriad of ways to prove this. Is that what this post is about? Or is there something else that you wanted to say?</p>

<p>Actually it changes from 1/3 to 2/3. </p>

<p>I thought this may be interesting to people that have never seen this before. If you have seen it, no need to post.</p>

<p>Oops...lol...</p>

<p>it is easy... oh we're NOT supposed to prove it... haha</p>