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Probability of consecutive elements in set

  1. Sep 17, 2008 #1
    Lets say i have some number 'N' of numbers - in a particular order.
    I then remove some fraction 'F' of those numbers.

    I want to know the probability of there being (some number) 'Y' consecutive initial-numbers remaining.

    Any ideas?

    Would it just be F^Y ?
  2. jcsd
  3. Sep 17, 2008 #2


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    If the numbers are removed independently with probability p, then the chance that Y numbers follow some given sequence member (not too close to the end) is (1-p)^Y.

    If one element is removed uniformly at random from the sequence, then a second, a third, and so on until FN have been removed, then (for large N) this approximates the above with p = F.

    Thus for each remaining member, the chance that it is followed by Y-1 members is approximately (1-F)^(Y-1).

    Since we're assuming N is large, then very few members are close to the end, so we'll assume all N(1-F) remaining members are far from the end.

    Now the chance that a given member is not followed by Y-1 others is, of course, 1 - (1-F)^(Y-1) in our approximation. Thus the chance that all N(1-F) remaining members are not followed by Y-1 others is about
    (1 - (1-F)^(Y-1))^(N(1-F))
    and so the chance that at least one not-removed member is followed by Y-1 others
    1 - (1 - (1-F)^(Y-1))^(N(1-F))

    For example, removing half of a million members, the chance that 20 in a row remain is about 62%.

    If N is not large, the formula falls apart. I'm not sure how large you'd need, but maybe N > 50Y would be good enough.
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