You're trapped in a maze. There is a way out. Path junctions are all 3-way.
If you use the strategy of always taking the path going right, what will happen?
(Note: This problem is deliberarely vague.)
(In reply to
re(11): About the by friedlinguini)
ok well we're never going to agree, the 1/n characteristic is axiomatic
and for me is clear for a random topology. i really dont know how
to further support this.
nonetheless and even if for you it wont prove anything, the 1/n
characteristic applied to your abstraction yields:
expectation value for the number of paths in the starting loop is:
x such that the probability that an x path route comes back
to the initial path = 1/2
E(x) = x such that P(x) = 1/2
ie such that 1 - P(x) = 1/2
ie such that the probability of not coming back after x paths is 1/2
P(not coming back after x) =
N(x) = n-1/n * n-2/n-1 * n-3/n-2 * .... n-x/n-x-1.
cancelling each top term with the next denominator yields
N(x) = n-x/n
for E(x) we need N(x) = 1/2
1-x/n = 1/2
x = n/2
the average length of the loop youre placed is n/2 paths.
since the exit is one of those paths, there is 1/2 chance the exit is
in one of the loops you mention.
the above is by no means rigorous but im trying to answer
"Is there any reason to assume that picking the "exit" loop is more probable than picking any other loop? Is there any reason to assume that a large number of nodes will not create a large number of loops?"
by stating that the loops created in a random topology will be of
such a size that the probability of exit is 1/2. in other words, under
both pictures the reasoning is coherent.
re(12): About the
