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Percolation Problem

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  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data
    We have a 1-D lattice [a line] of ##L## sites. Sites are occupied with probability ##p##. Find the probability that a given site is a member of a cluster of size ##s##. (A cluster is a set of adjacent occupied sites. The cluster size is the number of occupied sites in the cluster)

    2. Relevant equations


    3. The attempt at a solution
    For some site ##x##, I'd say:
    $$ Pr( x \in s) = \begin{pmatrix} L \\ 1 \end{pmatrix} \ \frac{ s \times Pr( \text{Make cluster s}) }{L} = sPr(\text{Make Cluster s})$$

    ##Pr(\text{Make Cluster s})## is the probability that a cluster of size s exists. This is given by (supposing we are not near the ends of the lattice) :
    $$Pr(\text{Make Cluster s}) = (1-p)^{2}p^{s}$$

    Is this reasoning correct? My textbook also gets to the same answer, but simply states the result, so I am curious [being very rusty on anything to do with statistics] if I have actually done this correctly. (Apologies if this is the wrong forum - I'm aware it's elementary probability theory, but the rest of the text isn't)
     
    Last edited by a moderator: Oct 19, 2016
  2. jcsd
  3. Oct 19, 2016 #2

    mfb

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    The probability that a cluster of size s exists in the range depends on L. I guess your Pr(Make Cluster s) is the probability to have a cluster at a given place. In that case the derivation works, as the events are all mutual exclusive you can add the probabilities. This assumes that you are at least s steps away from the border.
     
  4. Oct 19, 2016 #3

    Ray Vickson

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    The probability that site ##x## is the first member of an ##s##-cluster is ##(1-p)^2 p^s##, because the site immediately before ##x## must be unoccupied, then sites ##x, x+1, \ldots, x+s## must be occupied and finally site ##x+s+1## must be unoccupied. You get the same probability if site ##x## is the second or third or fourth or ... or last site in the ##s##-cluster. So, indeed, the probability you want is ##s (1-p)^2 p^s##.
     
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