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Binomial Distribution.

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data

    A coin can be flipped a maximum of four times

    The following conditions exist:

    H(first) = $1
    H(second) = $2
    H(third) = $3
    H(fourth) = $4

    Where H = Heads
    And first, second, third and fourth, refer to what order one head is obtained.

    What is the expected gain
    2. Relevant equations

    Binomial Distribution.

    E(x) = Sum of ((x) . P(x))
    3. The attempt at a solution

    Drew up a binomial distribution:

    Combination are as follows:

    HHHH
    HHHT
    HHTH
    HHTT
    HTHH
    HTHT
    HTTH
    HTTT

    THHH
    THHT
    THTH
    THTT
    TTHH
    TTHT
    TTTH
    TTTT

    It ends up being (8/16) for each one which is equal to half which is the same for each one. Where have I gone wrong? Please help.
     
  2. jcsd
  3. Jan 28, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Nowhere. You said it yourself, it's a binomial distribution. That means - by definition - that each successive throw is independent of all the others, so each time you flip the coin you have 1/2 probability for heads (as you expect from unbiased flipping). Writing out all the [itex]2^4 = 16[/itex] possibilities just showed that explicitly.
     
  4. Jan 28, 2009 #3
    Compuchip:

    First, thanks a bunch :). But say this would have been a game of dice with four throws:

    e.g.


    3(first) = $1
    3(second) = $1
    3(third) = $3
    3(fourth) = $4

    The probability of success for getting a 3 is (1/6) and that of a fail is (5/6)

    It still ends up the same because it is just going to be: [(5/6)(5/6)(5/6)] . (1/6) each time just in a different order?
     
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