What are the odds of rolling a 1 and a 6 in four dice throws?

Question

There are $\binom42 = 6$ different ways to arrange the two desired rolls in a sequence of $4$ dice, and $6^2$ possible results for the remaining two dice. There are $6^4$ possible total outcomes, so the odds of getting a 1 and a 6 should be $6^3/6^4 = 1/6$, but I get the feeling this is not correct. I think my logic is flawed because I am double-counting some configurations somewhere, but I don't know how to do the computation correctly.

Answer 1

There are $6^4$ rolls, of which $5^4$ have no $1$'s and $5^4$ have no $6$'s. However, we've counted the rolls with neither $1$'s nor $6$'s twice. The probability of both a $1$ and a $6$ is $$\frac{6^4-2\cdot5^4+4^4}{6^4}=\frac{151}{648}\approx.23302469$$

Answer 2

It should be permutation $P(4,2)=12$, not combination $C(4,2)={4\choose 2}=6$.

Once you fix the two desired dice ($1$ and $6$), there are $4^2$ arrangements for other two dice.

Hence, the probability of one $1$ and one $6$ in four rolls is $$\frac{12\cdot 4^2}{6^4}=\frac4{27}\approx 0.148.$$