probability of getting a double six ($2$ dice) rolling them $24$ times This is what I got. $\dfrac{1}{6} \cdot \dfrac{1}{6} = 2.78\% \cdot 24 = 66.72\%$
I believe that since it is a six sided dice, since you roll both of them simultaneously it would be $\dfrac{1}{6} \cdot \dfrac{1}{6}$. 
So since they are rolling them $24$ times, I would just multiply it by $24$, so $2.78\% * 24$ would be $66.72\%$, which would mean I have a  $67.7\%$ chance of rolling a double six.

Do you think this is correct? Am i doing this correctly? 

 A: We calculate the probability of rolling at least one double-six in $24$ rolls of two dice. The probability we roll a double-six is, as you point out, $\frac{1}{36}$. 
So, on any roll, the probability of not getting a double six is $\frac{35}{36}$.
The probability of "failure" $24$ times in a row is therefore $\left(\frac{35}{36}\right)^{24}$.  
So the probability of at least one "success" in $24$ rolls  is $1-\left(\frac{35}{36}\right)^{24}$.  
Remark: If we interpret the question as asking for the probability of exactly one double-six, the answer is different, and more complicated. In that case, we want the probability of $1$ success and $23$ failures. This probability is $\binom{24}{1}\left(\frac{1}{36}\right)\left(\frac{35}{36}\right)^{23}$.
However, the usual interpretation is "at least one six."
This problem goes back to the the seventeenth century beginnings of probability theory. The Chevalier de Montmort was curious about what was more likely, at least one six in $4$ rolls of a die, or at least one double-six in $24$ rolls  of two dice. Each has probability about $50\%$ and the expected number of successes is the same. At the time, expectation was the primary notion.
