Question

There are three way mazes.

1) Take an hour to escape the maze

2) 2 hours back

3) 3 hours back

What is the expected time to get out of the maze?

Answer 1

Let X denote the time to get out of maze, and let Y denote the way initially chosen. Since, nothing is known initially, we assume that P(Y=y) =1/3 for y=1,2,3.

Then E[X] = E[X|Y = 1]P{Y = 1} + E[X|Y = 2]P{Y = 2}+ E[X|Y = 3]P{Y = 3}………..(*)

Note that, E[X|Y = 1] = 1 as the first way gives an escape in 1 hour.

However, E[X|Y = 2] = 2 + E[X], because if the second way is chosen, then 2 hours is lost in the maze

and ends in returning to the original position. But once the return to the original position is made, the procedure starts afresh and hence the expected time until go out of the maze becomes   E[X].

Hence E[X|Y = 2] = 2 + E[X] as 2 hours are already lost in return to the original position.

In a similar way, E[X|Y = 3] = 3 + E[X]

Using all these in (*), we get

E(X)=(1/3)*{1+E(X)+2+E(X)+3}

and solving we get E(X)=6 hours.

for query in above, comment.