Sun, 2008-12-14 17:18
You are on a game show with three doors, behind one of the doors is a car and behind each of the other two is a goat (the host knows which door the car is behind). You initially pick door #1, then the game show host shows you a goat behind door #3, but before opening door number #1 he gives you a chance to change your answer. Is it advantageous for you to switch your answer?
Sun, 2008-12-14 17:55
#1
No. You're chances have just
No. You're chances have just improved to 1 in 2 - why change? Besides, this may be in Afghanistan where the goat would be much more valued than the car 
Sun, 2008-12-14 20:07
#2
If you stay with your first
If you stay with your first door, you still have only a one-in-three chance of winning. Seeing the goat after having made the choice does not change the original proposition. Only if you change doors do you create a new proposition at one-in-two.
The door you first chose loses two out of three times. After the reveal, the other door loses one out of two times.
Net effect is switching wins two out of three times, and loses once.
cheers,
gary
Sun, 2008-12-14 20:20
#3
gary.turner wrote:Net effect
Net effect is switching wins two out of three times, and loses once.
...and gary for the win!
http://en.wikipedia.org/wiki/Monty_Hall_problem#Why_the_probability_is_not_1.2F2
Sun, 2008-12-14 23:08
#4
Switching wins two times out
Switching wins two times out of three. Staying wins one time out of three. To maximize your winning chances, switch doors.
This old problem has even fooled statistics professors. The thing to remember is that the host always opens a door he knows is not the one with the prize. Probability problems, even very easy ones like this one, fool most of us most of the time. Apparently when we were in the jungle the ability to solve probability quizes did not improve our survival chances. Either that or, when God made us She decided not to give us a good sense of probability.