Is it possible to calculate variance on stochastic variable using multiple formulas?

Question

Trying to calculate what the variance is when throwing two dice simultaneously. To calculate the variance i use the formula

$$\frac 1 n \sum_{i = 1}^n(x_i - \overline{x})^2 = \frac 1 n \sum_{i = 1}^n (x_i - \mu)^2 $$

Yet, I'm also aware of the other way to do the calculation: $$\operatorname{Var}(X) = \sigma^2 = \sum_{i = 1}^n P(X=x_i)\cdot (x_i - \mu)^2$$

But when i plot the numbers in to the formula i dont get the same answer.

Here i have included my worksheet made in excel to show how i calculated it all. I tried to make everything as organised as possible.

Now here my variance following the first formula will total $110$, and i should therefore divide i by $36$ to get my final answer. This yields $3,05$ which is not the correct answer. In my workbook from school they have included the original formula with the answers, and for the total variance they get $210$ instead of $110$. What am i doing wrong?

Here are pictures of my workbook. With the formula that yields the right answer. With the formula that yields the right answer. And the one i used.

Answer 1

Your total of $110$ assumes each of the numbers $2$ through $12$ occurs only once in your list. But this list is this: $$ 2,3,3,4,4,4,5,5,5,5,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,8,9,9,9,9,10,10,10,11,11,12 $$ I.e. $x_1=2,$ $x_2=3,$ $x_3=3,$ and so on.

So the sum is $210$, and $\dfrac{210}{36} = 5.833333\ldots\,.$