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Math Help - Dice probability statistics pondering

  1. #1
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    Dice probability statistics pondering

    Okey, I'm a little stubborn with these kinds of questions in that i HAVE to understand them from every way...Bit silly really since i have the answer buts its reeeeally bugging me i can't get it another way.

    4 Dice, what's the probability that at least two or more are the same number?

    Well i've done it by considering the fact its 1 - P(they are all different numbers) leaving the probability that at least one must be the same. I started working it out a different way, which didn't work, so settled for the above.

    Anywho, on my google travels, i came across this little table:

    ------------------------------------------Instances -- Probability
    All four match ---------------------------- 6 ----------0.46%
    two pairs of different numbers ---------90 --------- 6.94%
    three dice match-------------------------120 -------- 9.26%
    no matches------------------------------360 ---------27.78%
    one pair matches ----------------------720 ------55.56%
    TOTAL ---------------------------------1,296--------100.00%

    Now, this 'one pair patches' is bugging me. How the hell do you get 720?

    Because i was thinking:
    4 dice.
    Now if we choose any 2 of these dice (4 choose 2 = 6)
    As there are 6 different ways to get pairs in two dice {(1,1), (2, 2)...etc}
    There must be 6(number of ways to get 2 dice) x 6 instances.

    What is it in my thinking that is incorrect?

    Thankyou
    Last edited by AshleyT; January 17th 2011 at 10:53 PM. Reason: better description of thread
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  2. #2
    MHF Contributor matheagle's Avatar
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    I assume you want EXACTLY one pair.
    I would look at it as 4 slots, each a toss.
    Putting the match first there are 6 ways of having a pair in the first two positions.
    (1,1).... (6,6) then for the last two tosses, there are 5 choose 2 numbers to select which is 10.
    Now switching the positions, so that the pair need not land in the first two spots
    there are 4!/(1!1!2!) ways of rewriting this sequence

    6{5\choose 2} 4!/2!=(6)(10)(12)=720
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  3. #3
    MHF Contributor Unknown008's Avatar
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    Order is important here.

    The ways to get exactly one pair is for example:
    1 1 2 3
    1 1 3 2
    1 2 1 3
    1 3 1 2
    1 2 3 1
    1 3 2 1
    2 1 3 1
    3 1 2 1
    2 1 1 3
    3 1 1 2
    2 3 1 1
    3 2 1 1

    This is only to show the pairs you get with 1 and two other numbers 2 and 3

    Now, there is for every single number 1 through 6!
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