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Single event probability equivalent to that of its permutations?

  1. Mar 1, 2007 #1
    Is the probability for a particular event, out of a set of events, equal to the total [normalized?] probability for all permutations of events from the set, including the particular event?

    Say you have a 2 x 2 square with cells numbered 1 to 4. I am asking if the probability for square 1, p(1) is equal to the total probability for its permutations, p(1,2)+p(1,3)+p(1,4)+p(1,2,3)+p(1,2,4)+p(1,3,4)+p(1,2,3,4). Or should I divide this by the number of permutations?
     
    Last edited: Mar 1, 2007
  2. jcsd
  3. Mar 1, 2007 #2

    matt grime

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    The answer is almost surely no, whatever question you meant to ask. Just suppose that something has zero probability.

    This doesn't make sense. I have a square with 4 numbers. This is nothing to do with probabilities.


    How is P(1,2) a permution of P(1)?
     
  4. Mar 3, 2007 #3
    Consider this, demonstrable no doubt by the binomial theorem:

    Given a set of events occurring with equal probability, the total number of permutations including a given event is one more than the total number of permutations without that event.
     
  5. Mar 4, 2007 #4

    matt grime

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    I'll say it again: the total number of permutations of what?

    There are 6 permutations of the numbers 1,2,3 and 4 permutations of 1,2. You know, n!, right? And again, probability has nothing to do with this fact.
     
  6. Mar 4, 2007 #5
    matt,

    It turns out, upon further inspection, that what I was trying to describe is a redundancy. Sorry for the wild goose chase.
     
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