Probability: Prison Escape

Question

A person is kept in a dark, underground prison that has three tunnels out of it. If he picks the first tunnel, he will escape to freedom after 30 minutes of walk. Taking the second tunnel would bring him back to where he is now after 1 hour of walk. He would escape to freedom after 1 and a half hours of walk by taking the third tunnel. He is equally likely to take any of the tunnels in an attempt, except that he will not take a tunnel in subsequent attempts if he was unsuccessful with it prior If the person is tireless in his attempt to escape, what is the expected length of the time until he escapes the prison?

Answer 1

The above answer certainly is correct, but here is a more thorough explanation using concepts and terminology from basic Probability.

The problem is about the Expected Value of a Discrete Random Variable: the time it takes to escape. There are four possible outcomes here (each outcome being a sequence of choices of Tunnels leading to escape):

1st Tunnel -> 30 minutes

2nd Tunnel, then 1st Tunnel -> 90 minutes total

2nd Tunnel, then 3rd Tunnel -> 150 minutes total

3rd Tunnel -> 90 minutes

Since each individual choice is equally likely, we immediately have P(1st) = P(3rd) = 1/3

The other two probabilities are computing using the Multiplication Rule. For instance:

P(2nd, then 1st) = P(2nd) * P(1st | 2nd) = (1/3) * (1/2) = 1/6

and similarly, P(2nd, then 3rd) = 1/6.

Multiply the probability of each outcome by the value of that outcome, and add these to get the Expected Value:

(1/3)*30 + (1/6)*90 + (1/6)*150 +(1/3)*90 = 80 minutes.

Answer 2

expected time is a weighted average, with each of the 3 alternatives weighted by 1/3

= (1/3)(half hour) + (1/3)(1.5 hours) + (1/3)[(1 hour + 1/2(half hour) + 1/2(1.5 hours)]

= (1/3)(1/2) +(1/3)(3/2) + (1/3)(1+ (1/2)(1/2) +(1/2)(3/2))

= 1/6+ 1/2 + 1/3 + 1/12+ 1/4

= 4/3 hour

= 1 hour and 20 minutes

= expected time to escape