Find the variance of the digits sum

Question

We have picked (uniformly) a random number $x$ from the set $[0,\dots 6^{10}-1]$.
Random variable $\xi$ denotes sum of digits of $x$ when $x$ converted to base 6.
Task is to find $Var(\xi)$.

Solution draft
$\xi = \sum_{i=0}^9 c_i$; $c_i = \{0,1,\dots 5\}$
$E\xi = \sum_{i=0}^9Ec_i$ Since $Ec_i= \frac {1 + 2 + 3 + 4 + 5} 6$ we may write $E\xi = 10 \cdot 2.5 = 25$

$Var(\xi) = E\xi^2 - (E\xi)^2$
$E\xi^2 = \sum_{i=0}^9Ec_i^2$ where $Ec_i= \frac {1 + 4 + 9 + 16 + 25} 6 = \frac {55} 6$ then $E\xi^2 = \frac {275} 3$

Finally, $Var(\xi) = E\xi^2 - (E\xi)^2 = \frac {275} 3 - 25^2$ and that gives a nonsensical result.
I've made simulation with python script, which proved that $E\xi$ is indeed $25$, and $Var(\xi)$ must be $~25$ as well.
(But my analytical result is somewhat squared of what I need to get)
Probably, I am misunderstanding something here.
Thank you for your suggestions!

Answer 1

I will use $X_i$ for the $i$-th digit, and $Y$ for the the digit sum $\sum_0^9 X_i$.

The $X_i$ are independent, so $\text{Var}(Y)=\sum_0^9 \text{Var}(X_i)$.

For the variance of $X_i$, we use your calculation. Note that $E(X_i)=\frac{5}{2}$ and $E(X_i)^2=\frac{55}{6}$. Thus $$\text{Var}(X_i)=\frac{55}{6}-\frac{25}{4}=\frac{35}{12}.$$ For the variance of $Y$, multiply by $10$.

Remark: The error in the OP is in the calculation of $E(\xi^2)$, or in my notation $E(Y^2)$. We have $Y=X_0+X_1+\cdots +X_9$, so one would need to expand $(X_0+X_1+\cdots+X_9)^2$ and take the expectations. We avoided that by using the fact that the variance of a sum of independent random variables is the sum of the variances. But expanding will also work, since $E(X_iX_j)=E(X_i)E(X_j)$.