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The probability of rolling a 1 after one roll is $1\over6$.

To figure out the probability of rolling a 1 with a 6-sided die after 2 throws, I've listed all possible ordered pairs of the elements of $\{1, 2, 3, 4, 5, 6\}$, such as $(1,1), (1,2), ..., (6,6)$. I've counted 36 of them. There are eleven ordered pairs that contain a 1. This seems to suggest that the probability of rolling a 1 with a 6-sided die after 2 throws is $11\over36$. Is this correct?

Is there a way to figure out the probability of rolling a 1 after $n$ rolls without listing the sample space and counting the events?

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There is, using the law of complementary probabilities: Take 1 minus the probability that you don't get a specific number. In this case, there were $5/6$ numbers that would give you that result for the first row, and then another $5/6$ for the second row, so you would have $1-\left(\frac{5}{6}\right)^2=\frac{11}{36}$.

In general, for $n$, throws you would have $1-\left(\frac{5}{6}\right)^n$.

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