Quote:
Originally posted by Seraph:
There are only three possibilities, and they are all equally as likely.
The player picks empty locket one. The game host picks empty locket number two. Switching will win the car.
The player picks empty locket number two. The game host picks empty locket number one. Switching will win the car.
The player picks the prize. The game host picks either of the empty lockets. Switching will lose.
You win 2/3 of the time by switching, it is pretty straight forward.
Draw a probability tree, use the Bayes' Theorm, or (or better yet 'and') search "Monty Hall Problem" on google and Wikipedia if you don't belive me.
|
Solution #1
There is a 1/3 chance that you'll hit the prize door, and a 2/3 chance that you'll miss the prize. If you do not switch, 1/3 is your probability to get the prize. However, if you missed (and this with the probability of 2/3) then the prize is behind one of the remaining two doors. Furthermore, of these two, the host will open the empty one, leaving the prize door closed. Therefore, if you miss and then switch, you are certain to get the prize. Summing up, if you do not switch your chance of winning is 1/3 whereas if you do switch your chance of winning is 2/3.
Solution #2
After the host opened one door, two remained closed with equal probabilities of having the prize behind them. Therefore, regardless of whether you switch or not you have a 50-50 chance(i.e, with probabilities 1/2) to hit or miss the prize door.
from your search (the #2 answer on google) Aragorn and Myself are arguing #2, you're arguing #1, both are valid.
[ 06-22-2005, 04:34 PM: Message edited by: Morgeruat ]