Ones against fives

Idea. Count with indicators. Different rolls are independent, so only the pairs of indicators on the same roll contribute to the covariance, and on one roll a one and a five exclude each other.

Step 1: variances. $X$ is binomial with $n=60$ and $p=\frac{1}{6}$, so $\mathrm{Var}(X)=60\cdot \frac{1}{6}\cdot \frac{5}{6}=\frac{25}{3}$, and the same for $Y$.

Step 2: covariance. $\mathrm{Cov}(X,Y)={\sum }_{i,j}\mathrm{Cov}(A_i,B_j)$. Terms with $i\ne j$ vanish. For $i=j$, $A_i{B}_i=0$ (a roll cannot be both), so $$\mathrm{Cov}(A_i,B_i)=0-\frac{1}{6}\cdot \frac{1}{6}=-\frac{1}{36},$$ and $\mathrm{Cov}(X,Y)=60\cdot \left(-\frac{1}{36}\right)=-\frac{5}{3}$.

Step 3. $$\mathrm{corr}(X,Y)=\frac{-5/3}{25/3}=-\frac{1}{5}.$$

Remarks. The number of rolls cancels: the answer is $-\frac{1}{5}$ for any number of rolls. In general, for two categories with probabilities $p$ and $q$, the correlation of their counts is $-\sqrt{\frac{pq}{(1-p)(1-q)}}$. The correlation is negative because every roll that shows a one is a roll that does not show a five.