###
PUZZLE #2: THE MONTY HALL PROBLEM (SOLUTION)

(1) MH asks P to pick a door from a row of three that are on the stage.
Behind one of the doors is a prize; behind each of the other two, a brick.
P obligingly picks a door (say, door 1).
(2) MH now opens one of the doors not picked by P (say, door 3)
and purposely reveals a brick.

(3) Finally MH gives P the opportunity to switch the original pick to
the other door (which is now door 2). Before the right door is
opened, P must choose either to stick with the original pick or
to switch. (The crowd goes wild giving advice at this point.)

The problem is to decide which, if either, strategy is most
likely to get P the prize.

The following is a more "common sense" approach to convince the
skeptical that P is actually better advised to switch than to stick
with the door originally picked. A more formal solution, using
Bayes' theorem, may be found by looking
here.

*Let's suppose P decides to stick with the original pick. Then
the probability of success is 1/3; i.e., if the "stick" game were to be
repeated 300 times, P would make the right pick approximately 100
of those times. Now MH, knowing where the prize is, always reveals one
of the doors hiding a brick.
This means that the prize is behind the remaining door approximately
200 out of the 300 trials. Now suppose there is another player,
a kibbitzer Q playing over P's shoulder, who follows P's first pick
but who then plays the "switch" strategy. Then Q picks the prize
exactly when P doesn't; i.e., Q wins the prize approximately 200
out of the 300 trials. Thus Q's chance of success is a whopping 2/3.
*

*
The moral of the tale is that a person playing the "switch" strategy
has twice the chance of winning the prize as a person playing
the "stick" strategy. *