replace nothing with goat for this example


/e and it's called the 'Monty hall problem"

This post was edited by majorblood on Jun 15 2010 09:33am

Removal of one of the wrong choices increases the chance of winning, with no other new information changing which door was chosen does not actually increase your chance. He already has 1/2 of the doors chosen, hence choosing the other of the 1/2 doors does not actually increase the chance, the removal of one of the wrong choices did.

The whole problem lies with the incorrect wording.

If simply one of the choices was removed that is fine, if it is specifically a wrong choice being removed and it isn't one of the options first chosen then the above code does not work.

This post was edited by raw blood on Jun 24 2010 11:35am

This is how the doors relate to blackjack:

3 doors. Each with one prize randomly, a car, and 2 lemons. The car is the desired win, while the lemons are considered losses.
You choose door A.
The host has to open a door, but he has to open a "wrong" door (ie. the door w/o the car). So he opens say, door C and then asks you if you would like to switch your selection to door B.
You originally made your decision out of 3 choices, so you had a 1/3 or 33% chance of getting it right.
Now, using the knowledge that he opened the door that wasn't the car, and the door you chose had 33% on it.
But also now, the number of doors is reduced, and so, adds a % chance to the door left alone, door B. So the host just gave you an extra 33%, which is worth taking.

The one thing i learned in blackjack is its always worth it to take the switch. It's better to try and improve your odds then play it safe. The problem you stated though is called Monty something, i forgot though, long time ago but i recently read a math proof that used matrices to prove it was worth switching doors.

Woo monty hall problem.

LOL

It's really not a hard concept to grasp.