Extra exercise 6
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A brief solution is given only; the Wikipedia page gives extensive
details.

The sample space of the experiment consists of the possible
configurations of car and goats. The car can be behind any of the three
dors, so we could let S=(1,2,3) give the door number for the car, or we
could let S=(CGG,GCG,GGC) denote the content of all three doors. In any
case, S has three elements, and the natural probability model is equal
probabilities for the three situations.

If the contestant chooses door 1, the probability of winning the car is
therefore 1/3. The actions of the host do not change the basic
probability distribution.

If the contestant chooses to switch doors, we will have to examine the
implications of the actions of the host. 
* In the scenario where the car is behind the initially chosen door 
(say door no. 1), the contestant will switch to a door with a goat 
and will not win. 
* If the car is behind door no. 2, the host will display the
goat behind door no. 3, and therefore the unopened door to which the
contestant switches will contain the car. So in this case it's a win.
* If the car is behind no. 3, the host will display the goat behind 
door no. 2, and therefore the unopened (and hence chosen) door will be no. 3 
and have the car. So this is also a win.

That is, for both of the configurations where the car is not behind the
initially chosen door, the contestant will win the car by the switch
door strategy. Hence the probability of winning is 2/3.

It is therefore favourable to switch doors: it doubles the chance of
winning the car. This may at first seem counterintuitive because at the
point the contestant is given the choice of staying with or switching
doors, there are only two possible locations of the car so one might
intuitively think that the two strategies have equal probability (0.5)
of winning the car. As we have demonstrated, this is not the case.