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I know how to calculate the probability of throwing at least one die of a given face with a set of dice, but can someone tell me how to calculate more than one (e.g., at least two)?

For example, I know that the probability of throwing at least one 4 with two 6-sided dice is 27/216, or 1 - (3/6 x 3/6 x 3/6). How do I calculate throwing at least two 4s with four 6-sided dice?

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    The probability of at least one four with two dice is $1-(\frac 56)^2=\frac {11}{36}$2012-09-09
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    @RossMillikan Why is that? Could you please explain.2013-04-03
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    @ihatetoregister: The chance you don't get a $4$ with the first die is $\frac 56$. Of that, the chance that you don't get a $4$ the second time is again $\frac 56$. So the chance you get no $4$ is $(\frac 56)^2$. The chance you get at least $1 \ \ 4$ is the rest of the time. Many of these probability questions are easier to answer if you count the other probability and subtract from $1$.2013-04-03
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    @RossMillikan Thank you, now I understand. But I still find it weird that the "same" logic does not get me the same answer if you count the chances to _get_ four. Eg $ \frac 1 6 \frac 1 6 \ne (11/36) $. I'm obviously missing something but I'm not sure of what.2013-04-03
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    Think I got it: The change to get four on first throw + not getting four on first throw but on the second, right? $ \frac 16 + \frac 56 \frac 16 = \frac{11}{36} $2013-04-03
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    @ihatetoregister: that is correct. Now if we want the chance of at least one $4$ in four throws, I can do $1-(\frac 56)^4$. Your approach would be $\frac 16 +\frac 56 \frac 16 +\frac 56 \frac 56 \frac 16 +\frac 56 \frac 56 \frac 56 \frac 16$. They get the same place. Sometimes one is easier, sometimes the other. Sometimes it is a matter of taste.2013-04-03

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