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Probability of a run

  1. Dec 3, 2009 #1
    What is the probability that a run of exactly L consecutive heads (or tails) appears in N independent tosses of a coin?

    Please help me with this one... I 've searched everywhere but I can't find a general answer, for example P(L,N) = ....
     
  2. jcsd
  3. Dec 3, 2009 #2
    [tex]2^{-n}[/tex]
     
  4. Dec 3, 2009 #3
    That's not correct.

    Let's suppose we toss a coin 3 times (N = 3) and we want a run of exactly 2 heads (L = 2). Then the combinations that include runs of HH are only two: THH and HHT
    The total combinations are 2N=3=8

    So, P(2,3) = 2/8 = 1/4

    Your answer gives 2-N= 1/8
     
  5. Dec 3, 2009 #4
    Hi there,

    I am sorry. I misunderstood your question. I thought you asked what is the probably of having N consecutive heads, on N toss.

    I'll look into it a bit deeper, and give you a more precise answer.

    Cheers
     
  6. Dec 3, 2009 #5
    Thank you
     
  7. Dec 3, 2009 #6

    statdad

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    Homework Helper

    You can look in probability texts for discussions of "runs". There is other information here

    http://mathworld.wolfram.com/Run.html

    - with a "formula" that gives probabilities as coefficients from a particular generating function.
     
  8. Dec 3, 2009 #7
    Well, I 've already seen that but I don't understand how these formulas work. Could you explain a little bit more if you can understand them?
     
  9. Dec 3, 2009 #8
    Hi , i think this is the answer P(n,l) = C[tex]^{l}_{n}[/tex] / 2[tex]^{n}[/tex]
     
  10. Dec 3, 2009 #9
    Thank you for your answer but what exactly is C?
     
  11. Dec 3, 2009 #10
    Oh , i made a mistake , sorry. Let me fix it . C[tex]^{l}_{n}[/tex] = [tex]\left(^{l}_{n}\right)[/tex]
     
  12. Dec 3, 2009 #11
    HOpe i'm not wrong this time .

    P ( N , L ) =
    [tex]\left\{2/2^{N} , if N = L + 1 [/tex]

    [tex]\left\{( 2^{ N-L+1} + 2^{N-L-2} + 2 ) / 2^{N} , if N = L + 2 [/tex]

    [tex]\left\{( 2^{N-L-1} + 2^{N-L-2} + \Sigma^{N-L-2}_{K=1}( 2^{N-L-2-K} * 2^{K} ) + 2^{N-L-1} ) / 2^{N}[/tex]
     
  13. Dec 4, 2009 #12
    Sorry mate but both your solutions are wrong. You can easily prove this if you try to find P(2,5) or P(1,5) or whatever you want...

    Does anybody know if such a formula even exists?
     
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