Question

A prisoner is trapped in a cell containing 3 doors. The first door leads to a tunnel that returns him to his cell after 2 days’ travel. The second leads to a tunnel that returns him to his cell after 4 days’ travel. The third door leads to freedom after 1 day of travel. If it is assumed that the prisoner will always select doors 1, 2, and 3 with respective probabilities .5, .3, and .2, what is the expected number of days until the prisoner reaches freedom?

Answer 1

12 days

Explanation:

`E[X]=E[E[X|Y]]`

`E[E|Y=1]P{Y=1}+E[E|Y=2]P{Y=2}+E[E|Y=3]P{Y=3}`

`(2+E[X])(1)/(2) +(4+E[X])(3)/(10)+1((2)/(10) )`

Solving for E(x) we get 12