Question

Problem 5.A miner is trapped inside a mine and has access to 3 doors. The first door leads to a tunneland can take him to safety in 3 hours. The second door is a trap and brings him back to the mine againafter 5 hours through a tunnel. The third door also is a trap that brings him back to the mine after 7 hoursthrough a tunnel. The miner at all times is equally like to choose any of the doors. (In the darkness he canhardly distinguish the three). What is the expected length of time until he reaches safety

Answer 1

Expected time is 15 hours for him to get to safety.

Explanation:

We define X as the time that this miner would get to safety.

We define Y as the door he chooses initially.

P(Y= 1) = P(Y=2)=P(Y=3) = 1/3

We have E[X|Y=1] = 3

E[X|Y] = 5 hours + E[X}

E[X|Y] = 7 hours + E[X]

Then we have

E[X] = 1/3(3 + 5 + E[X] + 7 + E[X])

We cross multiply

3*E[X] = (15 + 2E[x])

3E[X] - 2E[X] = 15

E[X] = 15

So the time it would take to get him to safety is 15 hours