An email is sent on the network in which the recipients (0,1,2,3,4,5} are in communication.
1 can send to 4 and 2
2 to 1,3,5
3 to 0,2,5
4 to 1, 5
5 to 0,2,4
0 to 3 and 5
If a message is sent to 2,3,4,5 it is forwarded randomly to a neighbour (even if this means a repeat). 0 and 1 never forward messages.
Let ek be the expected number of time that a message starting at k is passed on.
E(X) = [tex]\sum[/tex] E(X|A)P(A)
The Attempt at a Solution
I think I need to partition this but I'm unsure on the partition.
Let X be the number of times a message is sent on
E(X) = E(X|1st move is to 0)P(1st move is to ) + E(X|1st move is to 1)P(1st move is to 1) + E(X|1st move is to 2)P(1st move is to 2) + E(X|1st move is to 3)P(1st move is to 3) + E(X|1st move is to 4)P(1st move is to 4) + E(X|1st move is to 5)P(1st move is to 5)
I'm not sure how I'd work these out using a general k to start at.
If I assume I start at 4, as I'm trying to find e4 then I get
e4 = 0 + 1x(1/2) + 0 + 0 + e5 (1/2)
2e4 = 1 + e5
e5 = 1x(1/2) + 0 + e2 (1/3) + 0 + e4 (1/3)
I don't really think I'm going about this the right way, I would have thought I need to find a formula for starting at a general k but I don't know how.