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Tossing a fair coin

  1. Dec 18, 2011 #1
    1. The problem statement, all variables and given/known data

    A fair coin is tossed repeatedly and X is the number of tosses before the first head appears. You independently repeat the experiment, and Y is the number of tosses before the first head appears in the second sequence of tosses.
    a. Give the probability mass function of X
    b. Find P(X>n), for n≥1
    c. Find P(X=Y)
    d. Find P(X>Y)

    2. Relevant equations

    Geometric pmf: p(x) = p(1-p)x-1

    3. The attempt at a solution

    a. I believe this should be P(X=x) = (1/2)x+1. Is this correct so far?

    b. P(X>n) = 1 - P(X≤n) = 1 - [itex]\sum \limits_{k=1}^n[/itex](1/2)x+1

    I feel like a closed form for this is expected. I know there's some kind of formula for sums of powers, but we never looked at it in class. So, I'm wondering if there's another way to do this question.

    c. P(X=Y) = [itex]\sum \limits_{k=1}^∞[/itex](1/2)2x+2

    Is this correct? If so, same issue with explicitly evaluating the sum as with b.

    d. Since the problem is symmetric with respect to X and Y, presumably the answer is:
    P(X>Y) = 0.5*[1-P(X=Y)]
    Thus, I need the closed form solution from c. This solution relies on the fact that X and Y have the same distribution. Is there a more general method for this?
     
  2. jcsd
  3. Dec 18, 2011 #2

    Dick

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  4. Dec 18, 2011 #3
    Baha oh my head is stupid tonight, it seems. Aside from that, are my answers correct?
     
  5. Dec 18, 2011 #4

    Ray Vickson

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    Yes, except for your not defining what, exactly, you mean by x and what is its range. You see, x is different in the geometric distribution p(x) = p(1-p)x-1 and in your P(X=x) = (1/2)x+1. Defining what your symbols mean is an important part of the solution process.

    RGV
     
  6. Dec 18, 2011 #5
    Good point, I've let myself get a little sloppy. Thanks Dick and Ray!
     
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