Question 3.13: A miner is trapped in a mine containing three doors. The fir......
A miner is trapped in a mine containing three doors. The first door leads to a tunnel that takes him to safety after two hours of travel. The second door leads to a tunnel that returns him to the mine after three hours of travel. The third door leads to a tunnel that returns him to his mine after five hours. Assuming that the miner is at all times equally likely to choose any one of the doors, what is the expected length of time until the miner reaches safety?
Let X denote the time until the miner reaches safety, and let Y denote the door he initially chooses. Now,
E[X] = E[X|Y = 1]P{Y = 1} + E[X|Y = 2]P{Y = 2}
+ E[X|Y = 3]P{Y = 3}
=\frac{1}{3}(E[X|Y = 1] + E[X|Y = 2] + E[X|Y = 3])
However,
E[X|Y = 1] = 2,
E[X|Y = 2] = 3 + E[X],
E[X|Y = 3] = 5 + E[X], (3.7)
To understand why this is correct consider, for instance, E[X|Y = 2], and reason as follows. If the miner chooses the second door, then he spends three hours in the tunnel and then returns to the mine. But once he returns to the mine the problem is as before, and hence his expected additional time until safety is just E[X]. Hence E[X|Y = 2] = 3 + E[X]. The argument behind the other equalities in Equation (3.7) is similar. Hence,
E[X]=\frac{1}{3}(2 + 3 + E[X] + 5 + E[X]) or E[X] = 10