Suppose you have a computer network with 5 code as following.
Packet can arrive at any node and any other node can be its destination equal uniform probability. Determine the average number of steps between the source node (begining node), and the destination node. If one of the link (one side of the pentagon) is broken, repeat again the average number of steps above.
Assume the source node is not equivalent to the destination node.
Attempt:
Say a packet arrive the top edge of the pentagon, then to get to a destination (say bottom left node), it could either travel from the left-direction (2 steps), or from the right-direction (3 steps). If we could only use 1 direction, then we could treat it as a uniform r.v, and the expectated number of steps just the expectation of a uniform r.v.
However, since we could get to the destination from 2 directions, can I say that we use conditional probability? My result shows that part 1 of the question is the same as part 2.
Part 1: Let $X=# of steps take to get to the destination$ Let $M=direction$
$E[x]=E[X|m=1]*P(m=1)+E[X|m=0]*P(m=0)$