Expected value of length of time trapped in mine

[stgermaine]

Homework Statement

A miner is trapped in a mine containing three doors. Door 1 leads to safety after 3 hours. Door 2 leads back to the mine in 5 hours. Door 3 leads back to the mine after 7 hours. What's the expected length of time until he reaches safety?

Homework Equations

X = amt of time until the miner reaches safety, Y = number of door that he initially chooses.

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

The Attempt at a Solution

The solution given says that
E[X|Y=1] = 3 (i)
E[X|Y=2] = 5 + E[X] (ii)
E[X|Y=3] = 7 + E[X] (iii)
as part of its solution.

Eqn (i) i can understand but not ii and iii. The textbook justifies by saying that if the miner chooses the second or third door, he spends five and seven hours, respectively, walking back to the mine, but once he reaches the mine, his problem is same as before.

However, if he is an intelligent miner, he should remember which door he walked into, so when he returns to the mine after choosing door 2 or 3, he has now a 1/2 chance of selecting door 1 rather than door 2 or 3, which does not lead to safety.

The possible values of X are 3, 8, 10, and 15, with respective probabilities 1/3, 1/6, 1/6, 1/3, if the miner doesn't choose the door he previously selected.

Should I have assumed that he would not remember the door he selected?

 

[Fredrik]

I agree with you. I would assume that he remembers what doors he has tried already.

 

[clamtrox]

I think the problem is much more interesting if you assume the miner is stupid.

 

[Ray Vickson]

Homework Statement

A miner is trapped in a mine containing three doors. Door 1 leads to safety after 3 hours. Door 2 leads back to the mine in 5 hours. Door 3 leads back to the mine after 7 hours. What's the expected length of time until he reaches safety?

Homework Equations

X = amt of time until the miner reaches safety, Y = number of door that he initially chooses.

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

The Attempt at a Solution

The solution given says that
E[X|Y=1] = 3 (i)
E[X|Y=2] = 5 + E[X] (ii)
E[X|Y=3] = 7 + E[X] (iii)
as part of its solution.

Eqn (i) i can understand but not ii and iii. The textbook justifies by saying that if the miner chooses the second or third door, he spends five and seven hours, respectively, walking back to the mine, but once he reaches the mine, his problem is same as before.

However, if he is an intelligent miner, he should remember which door he walked into, so when he returns to the mine after choosing door 2 or 3, he has now a 1/2 chance of selecting door 1 rather than door 2 or 3, which does not lead to safety.

The possible values of X are 3, 8, 10, and 15, with respective probabilities 1/3, 1/6, 1/6, 1/3, if the miner doesn't choose the door he previously selected.

Should I have assumed that he would not remember the door he selected?

The book's solution is for what might be called the "forgetful miner" problem: every time the miner returns to the mine he forgets what he did the last time, so starts again. Your solution is for the "non-forgetful" miner. I'm not sure it makes sense to speculate about the "correct" way for a mythical, non-existent miner to behave, so the book's version is perfectly OK, as is yours also.

RGV