Question

Suppose there are two cells, A prisoner is trapped in a cell containing 3 doors. The first door leads to a tunnel that returns him to other cell after 2 days’ travel. For the other cell, there are two doors, one door of which leads to freedom after 3 days of travel, and the other door leads back to the prisoner’s original cell after 3 days of travel. Each door is equally likely to be chosen. What is expected number of days until the prisoner reaches freedom?

Answer 1

The first door leads to a tunnel that returns him to other cell after 2 day's of travel.

Now at this cell there are two doors each with (1/2) probability of getting select.

Case 1: He took right gate and got freedom. So, days taken = 3

Total days = 2+3 = 5

Case 2: He took wrong gate and return to original cell after 3 days. Then again tunnel then he will chose the right door because he knows the one he selected before is wrong.

So, total days = 2 + 3 + 2 + 3 = 10

Each case has (1/2) probability.

So, E(x) = (1/2)*5 + (1/2)*10 = 15/2 = 7.5 days.

Hence expected number of days until the prisoner reaches freedom is 7.5 days.

Please comment if any doubt. Thank you.