The problem is as stated: "How many times do we need to roll four fair dice to get a better than even chance of at least one pair of sixes?"
Using this formula for getting a specific value in exactly $m$ out of $n$ rolls:
$$\binom{n}{m}\cdot\left(\frac16\right)^{m}\cdot\left(\frac56\right)^{n-m}$$
I calculated the probability of favorable outcomes - 1 pair of 6 or 2 pair of 6 as being 11.57% and 0.077%, respectively. Together, that gives us a 11.647% chance. Now, following this logic, how do I get the number of throws of 4 dice that would give us more than a 50% chance of a favorable outcome? I can only think of doing: $$x*11.647>50$$ $$x>4.2$$ So then $x$ would be equal to 5. I don't think this is the right answer tho I have no proof. Can anyone help me with this regard?
If $X$ is the number of sixes in a throw of four dice,
$P(X = 0) = \left(\dfrac{5}{6}\right)^4$
$ \displaystyle P(X = 1) = {4 \choose 1} \cdot \frac{1}{6} \cdot \left(\dfrac{5}{6}\right)^3$
$P(X \leq 1) = P(X=0) + P(X=1) = \dfrac{125}{144}$
You want to find number of throws $n$ such that,
$ \displaystyle [P(X \leq 1)]^n \leq \dfrac{1}{2}$
i.e, $\displaystyle \ n \ln \left(\dfrac{125}{144}\right) \leq \ln (0.5)$
Can you take it from here?