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Expected number of runs

  1. 1. The problem statement, all variables and given/known data

    Hi,

    Suppose we have a die with 3 colors on it.

    4 sides are blue => P(Z=Blue) = 2/3
    1 side is green => P(Z=Green) = 1/6
    1 side is red => P(Z=Red) = 1/6

    I throw it 20 times and have Z=(Z1,..., Z20). Now what is the expected number of "runs"?

    Run is defined as the number of times the color changes, or equivalently, as the number of consistent blocks of a color.

    For example: string "bbgrg" has 4 runs ( |bb|, |g|, |r|, |g| )

    2. Relevant equations



    3. The attempt at a solution

    Attempt #1:
    Change the representation of the sequence from "bbgrg" into a sequence of 1 and 0. One being a new color block (a success), 0 being just another ball of the previous color.
    "bbgrg" becomes 10111.

    In other words, P(Xi=1), if {Zi != Zi+1}.

    This is, however, only a restatement of the problem and doesn't solve the initial problem: how many "1" do I have in 20 throws?

    Attempt #2:

    The number of throws before a given color occurs is geometrically distributed (Geo(p)). Thus:

    E(number of throws until blue occurs) = 1/P(Z=Blue) = 3/2
    E(number of throws until green occurs) = 1/P(Z=Green) = 6
    E(number of throws until red occurs) = 1/P(Z=Red) = 6

    I also know E(# Blue) = n * P(Z=Blue) = 20*2/3 = 40/3
    E(# Green) = E(# Red) = 20/6

    I could maybe use those 2 pieces of information but I can't see how. Any comments are welcomed.



    Thank you for help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Ray Vickson

    Ray Vickson 6,101
    Science Advisor
    Homework Helper

    I would do it by an iterative method. If B(n) = expected number of runs in n tosses, given the first toss is Blue, and G(n), R(n) are defined similarly, I would get the answer in terms of B(20), G(20) and R(20). Then I would get recursions for B(n), G(n) and R(n) by noting how B(n) is related to B(n-1), G(n-1) and R(n-1) by looking at the next colour, etc.

    RGV
     
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