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Say I throw a fair k-sided dice N times. What is the probability that the specific value 1 will never appear, while the value 2 will appear at least once?

The probability for 1 never to appear is easy: $P = \left(1-\frac{1} {k}\right)^N$

For 2 to appear at least once: $P = 1- \left(1-\frac{1} {k}\right)^N$

But since they are dependent (if 1 appears in a roll, 2 does not), I cannot simply multiply them. Is there any easy way to find the combined probability?

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Count the number of sequences directly. The number of sequences possible is $k^N$. The number of sequences with no $1$'s and no $2$'s is $(k-2)^N$. The number of sequences with no $1$'s and no requirement on $2$'s is $(k-1)^N$. So... the number of sequences with no $1$'s and at least one $2$ is...

$(k-1)^N-(k-2)^N$

And so the probability is...

$$\dfrac{(k-1)^N-(k-2)^N}{k^N}$$

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  • $\begingroup$ Simple and elegant, thank you! $\endgroup$ – jarondl Aug 17 '17 at 12:51

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