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Probability of getting a streak

  1. Dec 17, 2012 #1
    1. The problem statement, all variables and given/known data

    If a coin is tossed 200 times, what is the probability of getting 5 heads in a row?

    2. Relevant equations



    3. The attempt at a solution


    I know that the probability of getting 5 heads in a row for the first 4 flips is 0. I don't know what to do with that information.

    1 - [(0.5)^5 * (0.5)^195] ?

    I used this streak simulator:
    http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html#ev-runs

    And the probability is 0.965903381259682.
     
  2. jcsd
  3. Dec 17, 2012 #2

    haruspex

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    Are you familiar with Markov chains?
     
  4. Dec 17, 2012 #3
    No, I am not. :frown:

    We are starting binomial distributions in our stats class this week.
     
  5. Dec 18, 2012 #4

    haruspex

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    Then this rather an advanced problem. See e.g. http://math.stackexchange.com/quest...ngth-of-the-longest-run-in-n-bernoulli-trials, http://mathworld.wolfram.com/Run.html.
    You can get a crude lower bound by chopping the 200 into 22 blocks of 9 (and 2 left over). (9, because 9 = 5*2-1, so there cannot be two separate runs of 5 in the same block.) This allows us to break it into 5 disjoint cases:
    - HHHHHxxxx
    - THHHHHxxx
    - xTHHHHHxx
    - xxTHHHHHx
    - xxxTHHHHH
    for a tot prob of 3/32.
    The prob that none of the 22 blocks have such a run is therefore (1-3/32)22 ≈ 0.11. But 0.89 is some way short of the 0.96 you found.
     
  6. Dec 18, 2012 #5
    Oh you posted this before I was going to edit my first post.
    I asked this question earlier today and someone answered it. Would you mind explaining his solution to me?
    http://answers.yahoo.com/question/i...BRxA.2Lty6IX;_ylv=3?qid=20121217171124AAPB2E5


    What is (1/32)(1/2).
    Is this the probability of getting 5 heads for the first 5 flips multiplied by the probability of getting a head the next flip?
     
    Last edited by a moderator: May 6, 2017
  7. Dec 18, 2012 #6

    haruspex

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    It considers the last 5 toss outcomes at any point. If you've not had a run of 5Hs yet, the mutually exclusive possibilities can be categorised as:
    xxxxT
    xxxTH
    xxTHH
    xTHHH
    THHHH
    the probs being 1/2, 1/4, 1/8..., 1/32. It treats this as a 'steady state', i.e., so long as you've still not had a run of 5Hs this always represents the current state. (It's clearly not quite right because it does not add up to 1.)
    The prob of completing a run of 5Hs at the next toss is therefore 1/64. If we do not complete such a run, either because we toss a T or were not yet in the '4H' state, the assumption is that we remain in the 'steady state' distribution. So the prob of getting no run is (1-1/64)196 ≈ 4.6%
     
  8. Dec 18, 2012 #7

    Ah I see. Thank you.

    [STRIKE]
    So let me summarize what I get out of this. And then would you correct me?

    If we have not completed a run of 5H, then it can be assumed that we remain in 'steady state' distribution meaning it can be either of those mutually exclusive categories (P(0)-P(4)).

    And 1/64 is the probability of completing a run. So to find out the probability of not completing a run, we subtract 1/64 from 1. And take it to the power of 196 because we subtract the first 4 flips which have a probability of 0 of flipping 5 heads consecutively.
    [/STRIKE]


    I'm confused here:

    xxxxT_ The probability of getting a head next: 1/2
    xxxTH_ The probability of getting a head next: 1/4
    xxTHH_ The probability of getting a head next: 1/8
    xTHHH_ The probabilit of getting head next: 1/16
    THHHH_ The probability of getting a head next: 1/32

    And the probability of getting a run of 5H is 1/64.

    But isn't the probability of getting 5 heads .5^5? Or is it because you flipped a Tails first.
     
    Last edited: Dec 18, 2012
  9. Dec 18, 2012 #8

    haruspex

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    No, those are the probabilities of being in those states. In each case, the prob of a head next is 1/2.
     
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