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Homework Help: Counting Outcomes - Probability Question

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Z plays a game where independent flips of a coin are recorded until two heads in succession are encountered.

    Z wins if 2 heads in succession occurs.
    Z loses if after 5 flips, we have not encounter two heads in succession.

    1) What is the probability that Z wins the game?

    2) Suppose coin is fair. Z plays twice. What is the probability that both games have the same outcome?


    2. Relevant equations

    N.A.

    3. The attempt at a solution

    I get very confused by questions related to probability. :(

    There are 5 flips at most so the sample space has 2^5 = 32 possible outcomes. When seen as 5 "slots", as long as two adjacent slots are filled with H (for "heads") then Z wins.

    P(Z wins) = 1 - P(Z loses)

    Let X = number of heads. Then, Z loses when X = 1. Z loses in some cases when X = 2,3.

    X = 1 : There are 5 possible outcomes.

    X = 2 : The two heads are either 1 slot apart (3 choose 1 = 3 outcomes) or 2 slots apart (2 choose 1 = 2 outcomes) or 3 slots apart (1 outcome)

    X = 3 : The three heads need to be in slot 1,3,5 => 1 outcome.

    So total outcome that results in a loss = 1 + 5 + 3 + 2 + 1 = 12.

    Therefore, P(Z win) = 1 - 12/32 = 20/32 = 5/8.

    Question : Is this method correct? Are there "cleaner" methods? I have no idea how to start part 2.
     
  2. jcsd
  3. May 16, 2010 #2

    D H

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    Are you sure you have accounted for all 32 outcomes?
     
  4. May 16, 2010 #3
    Doh!

    X = 0.

    Thats one more outcome. :(

    But my approach is correct?
     
  5. May 16, 2010 #4

    D H

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    Staff Emeritus
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    Yes.
     
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