Let XY denote the event that a message passes from node x to node y. Event AD can be written in disjunctive normal form in terms of events for adjacent nodes

AD = (AB ∩ BD) ∪ (AB ∩ BC ∩ CD) ∪ (AC ∩ CD)

(In contrast to the formula by Sjoerd, there are no complements but the disjunctions are not disjoint)

Using the inclusion-exclusion principle, a generalized version of the formula P(N ∪ M) = P(M) + P(N) – P(M ∩ N), the probability for AD is

P(AD) = P(AB ∩ BD) + P(AB ∩ BC ∩ CD) + P(AC ∩ CD) – P((AB ∩ BD) ∩ (AB ∩ BC ∩ CD)) – P((AB ∩ BD) ∩ (AC ∩ CD)) – P((AB ∩ BC ∩ CD) ∩ (AC ∩ CD)) + P((AB ∩ BD) ∩ (AB ∩ BC ∩ CD) ∩ (AC ∩ CD)) =

P(AB ∩ BD) + P(AB ∩ BC ∩ CD) + P(AC ∩ CD) – P(AB ∩ BC ∩ BD ∩ CD) – P(AB ∩ AC ∩ BD ∩ CD) – P(AB ∩ AC ∩ BC ∩ CD) + P(AB ∩ AC ∩ BC ∩ BD ∩ CD)

and since the events are independent (adjacent nodes!), the result is

P(AD) = 2(1/3)^2 + (1/3)^3 – 3(1/3)^4 + (1/3)^5 = 55/243 ]]>

Pr(a->d)= Pr(a->d|

There are four parts to be computed:

1. Pr(

2. Pr(

3. Pr(a->d|

4. Pr(a->d|

Thus, Pr(a->d)=(1/3)*(25/81) + (2/3)*(17/81)= 59/243.

My extra 4/729 is from the possibility that packets are forwarded a->b->c->d and a->c, but the a->c packet is dropped. Your term (bd|cd) conflates this case with a->b->c->d and a->c->d.

]]>`(ab&ac) & (bd|cd) + (ab&!ac) & (bd | bc&cd)) + (!ab&ac) & cd`

which should be 55/243? ]]>