What are the chances... a famous math puzzle

Switch doors. There was (and still is) a 2/3 probability that you picked the wrong door. Therefore, you will win two out of three times by switching.

So the contestant, after seeing that the host has opened a door with a goat in it, has the choice to swap to one of the two doors left?

yup. this is sometimes called the Monty Hall problem, and Marilyn vos Savant of the high IQ made her rep giving the correct solution, switching. of course, if the contestant is psychic and knows what door the car is behind from the outset, switching would lose every time. :wink:

Holy crack! this is the first time I've seen this. I first thought that it should be 1/2 he gets the prize- it won't matter if he switches or not.

This site helped me get it though: http://www.grand-illusions.com/monty.htm

Here's how they explain it: Suppose that the contestant choosed door 1 and sticks with it (i.e. doesn't switch):

(the one in green is the door that monty opens)

Door 1 | Door 2 | Door 3
goat | goat | Car | LOSE
goat | Car | goat | LOSE
Car | goat | goat | WIN

So prob of winning = 1/3

But if the contestant switches every time:
Door 1 | Door 2 | Door 3
goat | goat | Car | WIN
goat | Car | goat | WIN
Car | goat | goat | LOSE

So prob of winning = 2/3.

WHOA!

When I first heard of it I thought you'd think it was 50 - 50, but seeing the question is asked lets think it thru carefully. I still was hedging on 50 - 50 when I glanced up outside - it was night and I saw the stars.

I had a eureka moment. I expanded the problem. I thought what if the host had the same number of door as there were stars, say for clarity a trillion and two (its actually more like 10^26 observable). Now imagine this host eliminated a trillion doors and left you with two, the one you initially picked and the one he leaves you, would you stay which your choice or his. I quickly realised his door has a one and one trillion chance in one trillion and two in being correct and your original choise has a one in one trillion and two choice - show you should stop. This approach can be worked backwards to a one eliminate in three doors.

The insight is the host has knoweldge so his selection is connected - you are still really choosing from three doors, not two, but if you invert your choice using his connectedness you double your chances to 2/3 of being right!