An interesting dice question on counting probability 
How many rolls of two dice are needed so that the probability of
double six occurring at least once is at least $50 \%$?

Attempt.
Notice that probability of double six $\frac{ 1+1}{6^2} = \frac{1}{18}$.
Let $n$ be the number of rolls of the dice. So,
$$ P(at \; least \; one \; double \; six \; in \; n \; throws) = 1 - P(no \; double \; 6) = 1 - (17/18)\cdot(17/18)\cdot...\cdot (17/18) = 1- (17/18)^n$$
Therefore,
$$ 1 - (17/18)^n \geq 0.5 \iff 0.5 \geq (17/18)^n \iff \frac{\ln .5 }{ ln (17/18) } \leq n \iff n \geq 12 $$
So, atleast 12 are needed.
Is this a correct approach??

Your approach is fine, but the probability of a double $6$ is $\frac 1{36},$ not $\frac 1{18}$.  When you roll the first die, the probability of a $6$ is $\frac 16$.  Assuming you get that the chance of another is also $\frac 16$, so by the multiplication principle the chance of a double $6$ is $\frac 1{6^2}$
