Paul M - ProSouth wrote:
This is not correct:
If you dont switch, you have a 1/3 chance of car and a 2/3 chance of goat If you switch, you have a 2/3 (!) chance of car and 1/3 chance of goat.
Why do you have a 2/3 chance of the car, and a 1/3 chance of the goat, when you only have two doors to choose from?
All that you know is that if you had chosen a different door (the one that the host revealed to have a concealed goat), you would definately have been incorrect. Now you have two doors, and hence a 50/50 chance.
This one always causes lots of discussion :)
Work it out as a permutation problem. You have three doors, labelled 1, 2 and 3. You have two goats (G) and one car (C). So the possible unique permutations (since we don't care /which/ goat) are:
123 CGG GCG GGC
So that tells us that for any one door, we have a 1/3 chance of there being a car behind it and a 2/3 chance of getting one of the goats. Let's choose a door and work some numbers:
If we chose car (at 1/3 probability):
Host opens door 2 or 3 (50% each)
Change - lose 1/3
Stay - win 1/3
If we chose either goat (at 2/3 probability):
Host opens door for other goat (always)
Change - win 2/3
Stay - lose 2/3So, for each combination of 'change/stay' and 'win/lose' we have:
Change:
win = 2/3
lose = 1/3
Stay:
win = 1/3
lose = 2/3So, by ALWAYS changing we win 2/3. If we NEVER change, we win 1/3.
Does that clear it up for everyone? :>
-- Corey Murtagh The Electric Monk "Quidquid latine dictum sit, altum viditur!" _______________________________________________ Delphi mailing list [EMAIL PROTECTED] http://ns3.123.co.nz/mailman/listinfo/delphi
