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Math Help - dice throwing

  1. #1
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    dice throwing

    Hi,

    I would like to calculate the probability of getting the same outcome twice when throwing two dice in the following way. First die i throw 4 times and second die 3 times. So the total sample space is

    S = 6^4 \times 6^3

    the question is how many casaes there are that have exactly 2 matching pairs of outcomes in S.


    know this i probably a simple problem but I just cannot wrap my head arround it.

    thank you
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  2. #2
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    Re: dice throwing

    Quote Originally Posted by baxy77bax View Post
    I would like to calculate the probability of getting the same outcome twice when throwing two dice in the following way. First die i throw 4 times and second die 3 times. So the total sample space is
    S = 6^4 \times 6^3
    the question is how many casaes there are that have exactly 2 matching pairs of outcomes in S.
    I for one do not understand the setup.
    Tossing a die three times generates a triple. Tossing a die four times generates a 4-tuple.
    Now, triples cannot match 4-tuples. So what does "exactly 2 matching pairs" mean"?

    Does it mean for example (2,3,4,3)~\&~(2,3,3). There are exactly 2 matching pairs of threes??

    Please try to explain this more clearly.
    Thanks from baxy77bax
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  3. #3
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    Re: dice throwing

    Please try to explain this more clearly.
    sorry and thnx for replying.

    Does it mean for example  (2,3,4,3)~\&~(2,3,3). There are exactly 2 matching pairs of threes??
    no. in this particular example i have 3 pairs (2,2),(3,3) and (3,3)

    also in the case [TEX] (3,3,4,3) \& (2,3,3) [\TEX] i have only two matches of 3's since the smallest set dictates the max number of pairs because there will always be possible to maximally create the total of two pairs between set 1 and 2, given [TEX]S_{1} = {3,3,4,3} \mbox{ and } S_{2}={2,3,3} [\TEX] [TEX] (S_{1}[0]=S_{2}[1],S_{1}[1]=S_{2}[2]) [\TEX] is equal to [TEX] (S_{1}[1]=S_{2}[1],S_{1}[3]=S_{2}[2]) [\TEX]

    on the other hand given two sets (2,3,4,3) and (1,3,5), here i only have 1 pair (3,3) and this is something i do not care about.

    is it more clear now
    Last edited by baxy77bax; January 2nd 2014 at 04:46 AM.
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  4. #4
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    Re: dice throwing

    Quote Originally Posted by baxy77bax View Post
    I would like to calculate the probability of getting the same outcome twice when throwing two dice in the following way. First die i throw 4 times and second die 3 times. So the total sample space is
    S = 6^4 \times 6^3
    the question is how many cases there are that have exactly 2 matching pairs of outcomes in S.
    Thank you for the clarification. It is a counting nightmare.
    Start with the triple. Say (1,1,1), now to have exactly 2 matching pairs of outcomes the 4-tuple must contain exactly two 1's and two non-1's.
    There are \binom{4}{2} places for the two 1's and 5^2 ways to get two non-1's.

    But there are six ways to have a triple with three of the same entry.
    Thus in that case there are 6^2\cdot 5^2 ways to have exactly two matching pairs.

    I did say it is a counting nightmare. Say the triple is like (1,3,3).
    Now what can the 4-tuples look like?
    Thanks from baxy77bax
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  5. #5
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    Re: dice throwing

    well yes, in that case 4-tuple needs to have two threes and two non 3's including non 1, or 1 and a 3 plus two outcomes that do not contain 3 nor 1. yes i know it is a counting nightmare that is why i decided to ask for help since there is noone in my proximity that can help me with this. Therefore , Thank you !!
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