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Math Help - dice

  1. #1
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    dice

    hello everyone!

    If i roll two dice, how many times(on average) will I have to roll it in order to get at least one 6?

    here's what i did, the probability is 1/36 for getting at least one 6 from two dice. So one divided by 1/36 gives 36, so I will have to roll them 36 times on average.
    Yes?
    thanks!
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  2. #2
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    Is the probability of getting AT LEAST one 6 1/36?

    There are three disjoint possibilities:
    EVENT A: Dice-1 gets a 6 and Dice-2 doesn't,
    EVENT B: Dice-2 gets a 6 and dice 1 doesn't,
    EVENT C: Both dice get a 6.

    So the probability of getting AT LEAST one 6 is the sum of the probabilities of these three events. P(A) + P(B) + P(C) = 1/6*5/6 + 5/6*1/6 + 1/6*1/6 = 11/36.

    You can get this same number by considering the only other possibility which is:
    EVENT D: Neither dice rolls a 6.

    This probability is P(D) = 5/6*5/6 = 25/36, and of course we want 1-P(D) = 11/36.
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  3. #3
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    Quote Originally Posted by iknowone View Post
    Is the probability of getting AT LEAST one 6 1/36?

    There are three disjoint possibilities:
    EVENT A: Dice-1 gets a 6 and Dice-2 doesn't,
    EVENT B: Dice-2 gets a 6 and dice 1 doesn't,
    EVENT C: Both dice get a 6.

    So the probability of getting AT LEAST one 6 is the sum of the probabilities of these three events. P(A) + P(B) + P(C) = 1/6*5/6 + 5/6*1/6 + 1/6*1/6 = 11/36.

    You can get this same number by considering the only other possibility which is:
    EVENT D: Neither dice rolls a 6.

    This probability is P(D) = 5/6*5/6 = 25/36, and of course we want 1-P(D) = 11/36.
    Thank you so much, I must say that was really very well explained. I understood it straightaway!!

    So, 1/ (11/36) will give me the average number of times I need to throw the dice, which is approx 3.3
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  4. #4
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    I like that average better. Whenever you arrive at a result I think it is useful to ask yourself if your asnwer is reasonable. 36 throws of two dice before you see ANY 6's seems hard to believe.
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