One Last Try on Monty Hall

My assertion that it doesn't matter if Monty Hall knows has garnered little acceptance even among the highest IQ people.  I woke this morning with my best proof yet.  Here it is.

There is a set that contains three types of members
A= games where I selected the car door and it is always 1/3 of the members.  Monty must open a goat door
B= games where Monty opens a goat door and I did not select the car door
C= games where Monty opens a car door

B= 2/3-C

If Monty knows, the assumption is that C=0 and B, therefore, equals 2/3.  The advantage in switching is B/A or 2.

If Monty doesn't know, it is assumed that C=1/3 and B/A=1.  There is no advantage or disadvantage in switching.

We see that it doesn't matter if Monty knows or doesn't know.  All that matters is the value of C.  If C=1/2, then B=2/3-1/2=1/6 and B/A=1/2 and there is an advantage in not switching.

This is all Bayesian friendly.

In the case as stated, the set has one member.  In other words there is no reason to assume that this game belongs to a set of games that has more than one member.  In this set C=0, B, therefore equals 2/3 and B/A=2.  One should switch.

Now, if the problem stated that several games were to be  played, THEN you would need to know the likely value of C.  But that is not the problem as stated.

I think I am done with the Monty Hall Problem.