Calculating the probability of a message distributed in a graph I want to calculate the probability of a message being distributed in a weighted directed graph. Consider the following example:

$G=(V,E)$
$V=\{1,2,3,4,5,6\}$
$E={(1;2;0.8),(1,3;0.6),(2;4;0.9),(3;5;0.5),(4;5;0.8),(4;6;0.3),(5;6;0.95)}$
The edge weight represents the probability of a message being transferred over this edge. In the example a message shall be delivered from node $1$ to node $6$. There are 3 possible paths:


*

*$Path 1: 1\to2\to4\to6$

*$Path 2: 1\to3\to5\to6$

*$Path 3: 1\to2\to4\to5\to6$


I have no problem calculating the probabilities for each path independently or for the whole graph when there are no edges used multiple times. However I can't determine how to do it if there are multiple paths using the same edges. In the example these are the edges $1\to2, 2\to4$ and $5\to6$.
To make it easier, it is assumed, that the graph contains no cycles and that each node sends a copy of the message over each outgoing edge with the given edge probability. Also it is assumed, that a node does only send a message once, should it get the same message as input (in the example that could be node $5$). It is unimportant, how many messages arrive at node $6$ if at least one gets delivered.
For testing I wrote a simulation of the described process and ran it a few thousand times. As it turns out, the probability of a message delivered from $1$ to $6$ should be around 0.71. How can I calculate this probability without the usage of simulation? Is there a general solution or are there flow algorithms that solve the problem?
Thx for your answers in advance. 
 A: Not an answer: but a bit too big for a comment.
There are some things to iron out here e.g. does the graph have cycles? But I think to get around your problem of repeated edges you might be successful considering all the "broadcast trees" that don't reach the desired node, summing up their probabilities, and subtracting from one. They should be finite, assuming no cycles in the graph. By "broadcast trees" I mean a tree whose leaves are "terminal" nodes that don't broadcast any more and whose root is the initial node. Edges in the tree correspond to those in the graph. The probability of a given tree is the product of the probabilities of all its edges- (assuming they're independent), multiplied by the product of termination at each leaf. Termination at a given leaf is in turn given by the product of $(1- p_e)$ for all of its edges $e$.

Update: Due to a lot of shared structure of the trees, you can probably optimise the formula rather than calculate each probability separately. E.g. I'd imagine some sort of formula using nesting would do the trick.
A: (Edited. The first solution had erroneously assumed certain independencies.)
Denote by $p_k$ the probability that the message reaches vertex $6$, given that it has reached vertex $k$. Then
$$\eqalign{p_6&=1\cr p_5&=0.95 p_6\cr p_4&=0.3 p_6+0.8 p_5-(0.3 p_6\cdot 0.8 p_5)\cr
p_3&=0.5 p_5\cr p_2&=0.9p_4\cr p_1&=0.8 p_2+0.6 p_3-(0.8 p_2\cdot 0.6 p_3)\cr}$$
The solutions are
$$p_6=1,\quad p_5=0.95,\quad p_4=0.832,\quad p_3=0.475,\quad p_2=0.7488,\quad p_1=0.713314\ .$$
The probability we are looking for is $p_1=0.713314$.
