7.
Define L1 as being a hop away from the origin, L2 as being 2 hops away
from origin, L3 as being 3 hops away.
Thus we are looking for E[X|L1] since originally the ant is only a hop
away.
So, E[X|L1]=1/3(1)+2/3(E[X|L2]+1) since it may return to the origin
with chance 1/3 or go away another level with chance 2/3
Then, E[X|L2]=2/3(E[X|L1]+1)+1/3(E[X|L3]+1) since it may get closer
with chance 2/3, or further.
Finally, E[X|L3]=E[X|L2]+1 since it has no choice but to get closer.
Hence, E[X|L2]=2/3(E[X|L1]+1)+1/3(E[X|L2]+2) = 2/3(E[X|L2])=2/3(E[X|
L1]+1)+2/3=E[X|L2]=E[X|L1]+2
Thus, E[X|L1]=1/3(1)+2/3(E[X|L1]+3)= E[X|L1]=7
On Nov 2, 11:14 pm, ankur aggarwal ankur.mast@gmail.com wrote:
An ant stays at one corner of a cube. It can go only along the side with
equal probability, and taking one minute to get to another corner. What is
the expect minutes for the ant coming back to the original position?
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algoge...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
