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Math Help - Probabilities in games of chance - help me solve this mystery

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    Probabilities in games of chance - help me solve this mystery

    I am having difficulty solving this problem...

    What is the probability of getting more 6's than 5's on four
    tosses of a fair six-sided die?

    I'm not sure how to set up the different cases...
    P(6-6-6-5), P(6-6-5-x) and ???? How do I factor in other case such
    as rolling one 6 and no 5's? The answer is 0.325 but I don't know how
    to arrive at this answer. Thanks, Nancy
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    Quote Originally Posted by nancydrew View Post
    I am having difficulty solving this problem...

    What is the probability of getting more 6's than 5's on four
    tosses of a fair six-sided die?

    I'm not sure how to set up the different cases...
    P(6-6-6-5), P(6-6-5-x) and ???? How do I factor in other case such
    as rolling one 6 and no 5's? The answer is 0.325 but I don't know how
    to arrive at this answer. Thanks, Nancy
    You could have 1 six and no fives in 4(4*4*4) way: 4 places for the six 4 other numbers in three places.

    You could have 2 sixes and no fives or 1 five in 6[4*4+2*4]: 6 places to put 2 sixes etc.

    You could have 3 sixes and no fives or 1 five: 4[4+1]

    You could have 4 sixes: 1 way.

    Add those up and divide by 6^4 and see what you get.
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    Aha!

    Plato is worthy of his/her name! Merci!
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    Quote Originally Posted by Plato View Post
    You could have 1 six and no fives in 4(4*4*4) way: 4 places for the six 4 other numbers in three places.

    You could have 2 sixes and no fives or 1 five in 6[4*4+2*4]: 6 places to put 2 sixes etc.

    You could have 3 sixes and no fives or 1 five: 4[4+1]

    You could have 4 sixes: 1 way.

    Add those up and divide by 6^4 and see what you get.
    Here is another way to proceed. Let a be the probability that there are more 6s than 5s, b be the probability that there are exactly as many sixes as fives (2 each), and c be the probability that there are more 5s then 6s.


    Then b = \frac{4!}{2! 2!} (1/6)^4
    and by symmetry, a = c.

    Now use a + b + c = 1 to solve for a.
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