A person is trapped in a room with 4 doors. What is the expected number of days

Question

A person is trapped in a room with 4 doors. The first door leads to a passage that returns him to the room after 2 days; the second also returns him to the room, after 4 days. The third door takes him to freedom after 1 day. Finally, the fourth door takes him back to the room after 5 days. Assuming he chooses between unexplored doors at random, what is the expected number of days to freedom? (The prisoner can mark or remember which doors he's already chosen, so he never chooses the same door twice.)

Answer 1

Hint: There is a $50\%$ chance that any given returning door will be chosen before the escaping door.

Answer 2

"The prisoner can mark or remember which doors he's already chosen, so he never chooses the same door twice."

This is crucial. There are only 24 ways of choosing 4 doors in some order and here there are even less scenarios since as soon as the prisoner picks door 3 the choosing stops. So you can make a list of all scenarios (given that there are less than 24 this is not so much work).

Next for each scenario compute the probability (see example below) and the number of days it costs. Then add 'probability times time' over all scenario's as in the definition of expectation.

Example: scenario Door 1 first, then Door 2 second then door 3 third. This has probability 1/4 times 1/3 times 1/2 = 1/24 because at first Door 1 has 1/4 probability of being chosen from four options, then door 2 has 1/3 probability of being chosen from 3 options and finally door 3 has 1/3 probability of being chosen from the two remaining options. Also this scenario takes 2 + 4 + 1 = 7 days.

Now repeat this type of reasoning for all possible scenarios