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Math Help - a probability question

  1. #1
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    a probability question

    THANK YOU IN ADVANCE
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  2. #2
    ddt
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    Quote Originally Posted by Narek View Post
    THANK YOU IN ADVANCE
    You just have to calculate the probabilities of both cases: rolling 8 with 2 dice, and rolling 8 with 3 dice.

    Starting with the 2 dice: which combinations of dice rolls yield 8?
    - 2 with dice1, and 6 with dice2
    - 6 with dice1, and 2 with dice1
    (these are two different events!)
    ...
    Count them, and count how many different combinations of dice rolls (with two dice) are possible overall.

    Before you start with the 3 dice, it helps to look at the probability distribution of 2-dice-rolls, as there's a simple pattern to it.
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  3. #3
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    here is my answer

    we consider P1 for rolling 2 dices:
    ( we roll dices together )

    so we have 5 pairs of numbers which if summed up, makes 8
    so the probability is:

    P1 = 5 / 6^2 = 5/36

    for 3 dices we have 18 pairs of numbers which if summed up, makes 8
    so the probability is:

    P2 = 18 / 6^3 = 18/216

    from here it is obvious that P1 > P2
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    Am I true? any mistake?
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  4. #4
    ddt
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    Quote Originally Posted by Narek View Post
    we consider P1 for rolling 2 dices:
    ( we roll dices together )

    so we have 5 pairs of numbers which if summed up, makes 8
    so the probability is:

    P1 = 5 / 6^2 = 5/36
    Correct.

    Quote Originally Posted by Narek View Post
    for 3 dices we have 18 pairs of numbers which if summed up, makes 8
    so the probability is:

    P2 = 18 / 6^3 = 18/216
    I think you made a counting mistake.
    D1 = 1: then D2 + D3 = 7, therefore: 6 triples
    D1 = 2: then D2 + D3 = 6, therefore: 5 triples
    ..
    D1 = 6: then D2 + D3 = 2, therefore: 1 triple
    Total: 21 triples

    Quote Originally Posted by Narek View Post
    from here it is obvious that P1 > P2
    __________________________________________________ ___________

    Am I true? any mistake?
    That conclusion still stands
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