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Dice-How many throws?

  1. Nov 15, 2004 #1
    Dice---How many throws?

    How many throws of a pair of 6-sided dice are required (on average) for all points (2 through 12) to appear?

    My simulation program says between 55 and 60, but I am more interested in knowing how to derive the answer by calculating probabilities.
  2. jcsd
  3. Nov 15, 2004 #2


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    This is a variation of the so-called Coupon Collectors Problem (see, e.g. http://rec-puzzles.org/sol.pl/probability/coupon [Broken]). Your problem is harder because the probabilities are not all the same for the different "coupons."
    Last edited by a moderator: May 1, 2017
  4. Nov 17, 2004 #3
    I do not think that the average is so high.If X is the event representing the number of points which appear when a pair of dice is thrown then we have the distribution (it is the sum of 2 distributions for single die throws) :


    2 ---> P[2]=1/36

    3 ---> P[3]=2/36

    4 ---> P[4]=3/36

    5 ---> P[5]=4/36

    6 ---> P[6]=5/36

    7 ---> P[7]=6/36

    8 ---> P[8]=5/36

    9 ---> P[9]=4/36

    10 ---> P[10]=3/36

    11 ---> P[11]=2/36

    12 ---> P[12]=1/36

    If N is the number of throws the mean value for the apparition of different outcomes 'i';i=from 2 to 12 is N*P.Thus the average number of throws required for the appearance,at least once,of all outcomes is given by the outcomes having the least chance of appearance (sum 2 and 12) from the inequation: N/36 ≥ 1,that is the required N is 36.
    Last edited: Nov 17, 2004
  5. Nov 17, 2004 #4


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    There's a lot of handwaving in that argument, and it's wrong. Try applying it to a simlpe case: on average, how many flips of a coin before you get both a heads and a tails?
  6. Nov 17, 2004 #5
    My time,not yours,I do not appreciate the side comments.And yes you are right,I'm wrong overall (but not the distribution I have written).
    Last edited: Nov 17, 2004
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