n a TV show a prize is hidden behind one of three doors (the first, the
second and the third). A player wants to guess that door. She randomly points
at one of the doors (suppose this happens to be the third door). Then the
host, who knows where the prize is, opens one of the other doors that does not
have a prize behind it (let it be the second door). Then the player may stay
with her choice (the third door) or switch to the other door (the first door)
that remains closed. Should she switch?
The intuitive answer would be as follows. Suppose the player uses the
switching strategy. She originally points to the right door with the
probability
and this is the only case when she loses. If the player uses the staying
strategy then she loses with the probability
if she originally points to the wrong door. Therefore, the switching strategy
has twice better chances of success.
Let us consider the problem with more rigor. Suppose the third door is
originally picked and the second door is opened. What is the conditional
probability that the first door has the prize behind it? We introduce the
notation
We would like to calculate the probability
Prob
.
Note that all probabilities of the form
Prob
are easily computable. Hence, we use the idea of the formula
(
Inversion_remark
). According to the
(
Bayes
formula
)
Using the (
Bayes formula
) again we calculate the
numerator
Using the independence of the prize location and the door picking we
continue
We calculate denominator using the (
Total
probability rule
) with
Prize:
Again, by the independence of the prize location and the door picking we
write
Note that we substituted
Prob
This means that the host opens a door at random when he has a choice.
Therefore,
Hence,
because the last two probabilities are complimentary. This brings us to the
same answer as in the former non rigorous argument provided that the host acts
randomly.
