I was reading a probability problem that asked the following:
How many rolls $n$ do we need to make the triple $6$ event favorable on a three-dice roll?
I found out the answer to be $150$ by my own methods, but the answer mentions the answer being $216 \cdot \log(2)$ and I was wondering where does the logarithm formulation come from? Any help would be appreciated.
With $p=\frac{1}{216}$ your calculation got you the smallest integer $n$ with $(1-p)^n < 0.5$. Since the left hand side is decreasing in $n$ you can look for the solution to $(1-p)^r = 0.5$ in real numbers $r$ and then choose $n =\lceil r \rceil$. That equation has the solution $r=\frac{-\ln 2}{\ln(1-p)}$ (I'm using "$\ln$" here to donate the natural logarithm with base $e$). With the approximation $\ln(1+x) \approx x$ for $|x| \ll 1$ we get the formula you were given: $r \approx \frac{\ln 2}{p} = 216\ln2$.