You've estimated a proportion of $\hat{p}=0.833564$ from $n=1000000$ trials. You can for example construct a $95\%$ confidence interval: $$[\hat{p}-1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\hat{p}+1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}]=[0.8329,0.8343]$$ So with $95\%$ confidence you can say that you are off by less than $0.00073$ i.e. inside the above interval.
BTW the true answer is $\frac{5}{6}=0.8333...$ as you can see by computing a simple integral and therefore in reality you are off by just $0.00023$.