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Tossing a coin n times

  1. Nov 22, 2015 #1
    Suppose I toss a fair coin 100 times. If I consider the order of apparition, then obtaining all head, or fifty times head has the same probability, namely (1/2)^100 ?
     
  2. jcsd
  3. Nov 22, 2015 #2

    blue_leaf77

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    How did you get that answer for the second case, getting 50 heads?
     
  4. Nov 22, 2015 #3

    Krylov

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    No, because there is only one way in which you can toss all heads, but there are many more ways in which you can obtain 50 heads in 100 tosses.
     
  5. Nov 22, 2015 #4
    I'm considering one particular order hhhh...ttttt... fifty time each.
     
  6. Nov 22, 2015 #5
    Yep, in that case the probability is the same.
     
  7. Nov 23, 2015 #6
    But, intuitively, it seems to me that the serie htht... were more plausible than hhhh...ttt.... ?
     
  8. Nov 23, 2015 #7

    blue_leaf77

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    I'm a bit confused with your wording there, did you mean getting 50 consecutive heads followed by 50 consecutive tails and you want to know the chance of getting that particular order? If that's the case, it should be indeed (1/2)^100.
    Then you may want to refine your intuition, getting one particular order is as hard as getting another different particular order.
     
  9. Nov 23, 2015 #8
    Yes. I think I'm mixing with the statistics of the length of runs. The probability to obtain k times the same is 1/2^(k+1) hence if i write a program computing that statistics I should obtain an exponential decaying curve.

    But that's where my problem arises above, since run length that are longer have less chance to appear, I would expect the case hhhh....tttt.... were less likely to appear.

    I did a computer simulation of throwing 100 times a coin, and repeating 100'000 times the experiment, storing run length values :
    https://drive.google.com/file/d/0B9pGxyM9yy2fYlpIV08wMF9jVk0/view?usp=sharing
     
    Last edited: Nov 23, 2015
  10. Nov 26, 2015 #9

    DrClaude

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    I think you may be confusing the statistics of run length and the maximum run length of a given outcome. On average, you will get many runs of length 1, but you have only one possibility (if you start by h) that the longest run is of length 1.
     
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