Re: Advanced Monty Hall Problem with N door and M cars
From: Reef Fish (Large_Nassau_Grouper_at_Yahoo.com)
Date: 02/23/05
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Date: 22 Feb 2005 17:19:11 -0800
Richard Ulrich wrote:
> I'll try again.
> Even though Bruce seems to say it fine,
> I thought my first attempt was fine, too.
Actually what Bruce said is not exactly right either. No simulation
is required to understand that the conditional probability of a
change or no-change to win the car will BOTH be 1/2, on the given
trial.
>
> On 22 Feb 2005 08:23:33 -0800, "Reef Fish"
> <Large_Nassau_Grouper@Yahoo.com> wrote:
>
> [ ... ]
> >
> > No, it doesn't, GIVEN that the opened door has no car.
> >
> > The contestent has chosen a door which MAY or MAY NOT be a winner.
> > If Monty ALWAYS shows a losing door, before offering a switch,
> > then whether the contestent switch or not, the probability of
winning
> > a car is 1/2, because the only relevant information is that one of
> > the three doors has been ruled out to be the winner, and there are
> > two doors left: one has the car, and the other doesn't.
>
> Three doors.
Three doors, but one that Monty opens is ALWAYS a losing door, no
matter
whether the Guesser has chosen the right or the wrong one. Therefore,
if the Guesser randomly switches or stays, his probability of winning
the car is 1/2. But the random switching is an unnecessary red
herring.
>
> How is any information added to
> that original "2/3"? "Should you change?"
The original chance of winning was "1/3". But rare events DO happen.
So, the Guesser could have chosen the door with the car, even though
the chance was only 1/3. Monty Hall merely changed the CONDITIONAL
probability of winning to 1/2. On a single trial basis, it matters
not whether he changes or not. The conditional probability of either
door being the winner will be 1/2.
-- Bob.
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