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4 flips of a fair coin

  1. Feb 6, 2008 #1
    Given 4 flips of a fair coin, what is the probability of {H,H,H,H}?

    I thought since flips of a fair coin are independent, then P(H&H&H&H) = P(H)P(H)P(H)P(H) = 1/(2^4) = 1/16. Am I close?
  2. jcsd
  3. Feb 6, 2008 #2


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    Since these are 4 independent trials, yes.

    You can verify this & similar problems by figuring out how many ways 4 coins can "land."
  4. Feb 6, 2008 #3
    thanks; you're right. I used the tree-notation and got N(ways HHHH)/N(all ways) = 1/16. Thanks again. I'm not entirely sure why that works.

    Is it possible to compute the probability that one pattern comes before the other one? Like THHH will happen before HHHH?
  5. Feb 6, 2008 #4


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    Each of the 16 possible total outcomes (HHHH through TTTT) is equally possible, so there is no way to predict which will occur before another.

    - Warren
  6. Feb 6, 2008 #5
    Thanks. I should have known that.
  7. Feb 7, 2008 #6


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    Remember to distinguish the probability that "THHH" comes before "HHHH" in a series of discrete tuples of four flips from the probability that the subsequence "THHH" comes before "HHHH" in a continuous series of individual flips.
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