Probability Question

Discussion in 'Probability' started by mukesh tiwari, Aug 11, 2010.

  1. Hello all ,
    I am just trying to solve a problem and stuck on To solve
    this problem, i am generating the whole tree but it grows quite fast
    [factorial ] with level and till level 15 its almost impossible
    enumerate. Here is sketch of solution which i figured out. First turn
    there are two balls [B,R] , second level [B,R,R] and number of
    branches are 6, for third level [B,R,R,R] and branches are 24 and for
    fourth level [B,R,R,R,R] and branch are 120. Now to win in four turn ,
    the number of blue ball chose must be more than red so winning
    condition is in all four turn i chose all the 4 blue or 3 blue and
    1 red. I counted and got there are 11 turn in which number of blue
    balls are more than red but this method is not applicable to 15 level
    since 15! is too large to bruteforce. Kindly give me some hint in
    right direction as i am trying to learn probability.
    Thank you
    mukesh tiwari, Aug 11, 2010
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  2. mukesh tiwari

    Pavel314 Guest

    Use the binomial distribution.
    Pavel314, Aug 17, 2010
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  3. mukesh tiwari

    Henry Guest

    "A bag contains one red disc and one blue disc. In a game of chance a
    player takes a disc at random and its colour is noted. After each turn
    the disc is returned to the bag, an extra red disc is added, and
    another disc is taken at random. The player ... wins if they have
    taken more blue discs than red discs at the end of the game. If the
    game is played for four turns, the probability of a player winning is
    exactly 11/120, ... a single game in which fifteen turns are played."

    The total number of ordered cases of 15 red or blue discs is not 15! =
    1,307,674,368,000 but 2^15 = 32,768 which is certainly possible, for
    example on a spreadsheet. But even that is more than you need and not
    the best approach, and you should be able to get the number down to
    120 or less.

    For example, the probability of reaching 3 blue and 1 red in any order
    is 4/5 of the probability of reaching 3 blue and 0 red plus 1/5 of the
    probability of reaching 2 blue and 1 red, i.e.
    (4/5)*(1/24)+(1/5)*(1/4) = 1/12.
    Henry, Sep 13, 2010
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