On Fri, 30 Jan 2009 04:22:27 -0500, "Eisboch"
wrote:
Tom was correct, although I am not sure why.
Only if the problem is expressed with the understanding that "Monty
knows."
IOW, if Monty knows, and *always* picks a door with goats behind it,
it changes the odds
Her probability of winning increases from 33.3% to 66.6% by changing her
initial pick. The majority of people do not get this right and don't
understand why.
This is a classic mathematical probability exercise that has been bashed
about for years by everyone from math PhD's to grammar school whiz kids.
When she first picked a curtain, she had a 2 in 3 chance of being wrong.
That didn't change after the first curtain with goats was opened. But it
does change if she changes her choice, and it can only increase her chances
of being correct.
That makes sense - but only when Monty always changes the randomness
with his knowledge. If Monty didn't know, 1/3 of the time he would
pick the car, making the contestant a loser
OTOH, if he didn't know, and picked the goats, it wouldn't do any good
for the contestant to change doors.
That's how I took the problem.
Good explanation by the guy below.
And it's well explained here
http://en.wikipedia.org/wiki/Monty_Hall_problem
Now, having proved I was right given how the problem was expressed and
my assumption of randomness, I'll admit I *might* have been fooled if
was expressed as "Monty knows." I'll never know now. But since it
wasn't, I'll just say I was right and leave it at that (-:
--Vic
http://www.tugbbs.com/forums/showthread.php?t=68845
(post #8)
"I'll try to explain it in words, without going through the math.
When you first select a door, there is a one-third chance that you
picked the correct door, and a two-thirds chance the correct door is
one of the two you did not pick.
Now, let's imagine a slight variation of the game. The rules are the
same, except that when Monty shows you a door he also has no idea what
is behind the door he picks. If Monty's door has the car, you lose and
the game is over. If Monty's door has a goat, you have the choice of
keeping your door or selecting another. Note that the only difference
between this and the "real" game is that Monty does not know what is
behind his door. Since Monty doesn't know where the car is, one-third
of the time he picks the car and you lose.
The other two-thirds of the time - when Monty's door has a goat - the
car is equally likely to be behind either door.. In that case it makes
no difference if you switch. Note that in this variation of the game
your odds of winning are one-third no matter what strategy you employ.
One third of the time you lose when Monty opens his door and you
aren't even given a chance to switch. Of the remaining two-thirds, you
win half of the time.
***
But Monty doesn't play the game that way. Monty knows what door has
the car, and when he picks a door he never picks a door that has a
car. IOW - Monty eliminates the one-third of outcomes where you lose
without having a chance to pick a door. (If you're a craps player,
it's like playing a craps game in which 2, 3 or 12 aren't craps - you
can never lose on the come out roll.)
Think about how that changes the game. Go back to when you first
picked a door. At that point there was a two-thirds chance that you
picked the wrong door. That situation remains. When Monty opens his
door, you now know that:
There is a two-thirds chance that the car is behind one of the doors
you didn't pick.
The car is not behind the door that Monty picked - which means that
there is a two-thirds chance the car is behind the door Monty did not
pick.
****
I think that what hangs up many people on the Monty Hall situation is
that they don't appreciate the significance of the fact that Monty
does not open a door randomly. In the first variant I laid out, Monty
does open a door randomly; in that case door switching doesn't make
any difference. But when Monty doesn't select a door at random, the
odds change.
__________________
Steve Nelson"