probability

A miner is trapped in a mine containing 3 doors.The first door leads to a tunnel that will take him to safety after 3 hours of travel,The second door leads to a tunnel that will return him to the mine after 5 hours of travel. The third door leads to a tunnel that will return him to the mine after7 hours of travel. If we assume that the miner is at all times equally likely to choose anyone of the doors, find the expected time for him to get to safety.

 

it is approximately between 12.75 and 13 hours.!
is it ??

 

Reminds of me a old problem when I did when I was preparing for CT4 a couple of years back - so concepts maybe a bit edgy

Let X = expected time to go out of mine

X = (1/3)*(expected time to go out of mine from door 1 = 3 hours) + (1/3)*(X + 5) + (1/3)*(X + 7)

X + 5 stems from the fact that if he chooses door 2, he will have to get back to the mine but will have an additional 5 hours wasted (same with door 3)

=> X = 15 hours.

 

15 is correct.

MATLAB Code to solve the same using Monte Carlo:

Code:

n = 100000;
total_hours = zeros(n,1);

for i = 1:n
    exit = 0;
    hours = 0;
    while exit == 0;
        Door = ceil(rand*3);
        switch Door
            case 1
                hours = hours + 3;
                exit = 1;
            case 2
                hours = hours + 5;
                exit = 0;
            otherwise
                hours = hours + 7;
                exit = 0;
        end
    end
    total_hours(i,1) = hours;
end

answer = mean(total_hours)

 

Is the answer "9" ??