>1. Probabilities. The probability of something happening is
>dependent on constraints. Given a ‘free' - unconstrained -
>choice between two actions, there is a probability of one
>half of either one being chosen. These probabilities may
>change if some constraint is put on the choice.
>

Yup, if Monty places the car AFTER the door is opened
but not before.

>2. In the Monty Hall problem, the contestant chooses on two
>occasions. In the first, unconstrained, he stands a one
>third probability of choosing the door hiding the car, two
>thirds of not doing so. In the second, unconstrained, he has
>a probability of one half of picking the car, one of the two
>other options having been eliminated.

No, only one half if the car is placed AFTER the door is
opened. The car is distriuted BEFORE the door
is opened. You are opening then placing. You cannot do
that! Place the car THEN open the doors in ALL combinations.
Place then open. Do not open then place! Do you understand?,
you are driving me mad.
>

>3. If the contestant constrains his second choice, this last
>probability may change. If he decides always to stick to his
>first choice, the second ‘choice' is now forced.

No sir, the probabilty on the second door is 2/3 even if the
choice is made after one door is eleminated. Run a simulation
and see, but make sure to label the same doors with the same number
each round.

He has
>essentially eliminated the probability of one half of
>winning on the second choice; he wins only if his first
>choice was correct Thus the probability of winning overall
>remains one third.
> If he decides always to switch, then he has anulled the
>one third probability of being right first time. He wins if
>his first choice was wrong with a probability of two thirds,
>since Monty has eliminated the other possible wrong first
>choice. If his first choice was right - with a probability
>of one third - he has given it up, and loses. (In this case,
>it might appear that he can lose in two ways, depending on
>Monty's choice of door, but neither of these possible events
>alter the contestant's choices, and hence lie within the one
>in three chance of the his first choice being wrong.)
>
>4. If the contestant decides before his first choice to
>switch at his second chance, he converts his two chances of
>choosing wrongly to winning choices, since Monty has to
>eliminate the other wrong choice. He does this at the
>expense of losing his one in three chances of having been
>right first time.
>
>5. So decide to switch every time, and never change your
>mind at the second choice. If you do, half the time you'll
>be wrong.

NO, That is wrong, wrong, wrong. If the door would have
been 2/3 by "keeping" your switch choice after the door was open,
then not switching is 1/2? 1/2+2/3=1? You are crazy