A fair die is rolled many times. Find the expected number of rolls made until the sum of all the points rolled reaches 2010 (i.e., becomes no less than 2010).
Let $X_{n}$ be the number of rolls needed to reach a sum of $n$. Introduce the indicator $I_{k}: I_{k}$ is 1 if the first roll gives $k$ points, and 0 otherwise. Clearly, $E I_{k}=1 / 6$.

It is clear that $X_{n}=X_{n-1} I_{1}+X_{n-2} I_{2}+\ldots+X_{n-6} I_{6}+1$. The quantity $X_{n-k}$ does not depend on the result of the first roll. Therefore, the quantities $X_{n-k}$ and $I_{k}$ are independent. Transitioning to expectations, we get: $\mathrm{E} X_{n}=\mathrm{E} X_{n-1} \mathrm{E} I_{1}+\mathrm{E} X_{n-2} \mathrm{E} I_{2}+\ldots$ $+\mathrm{E} X_{n-6} \mathrm{E} I_{6}+1=1 / 6\left(E X_{n-1}+E X_{n-2}+\ldots+E X_{n-6}\right)+1$.

Denoting $\mathrm{E} X_{k}=e_{k}$ for brevity, we write the recurrence formula: $e_{n}=1 / 6\left(e_{n-1}+e_{n-2}+\ldots+e_{n-6}\right)+1$.

This formula is valid for all $n \geq 1$, if we set $E X_{k}=0$ for all $k \leq 0$.

From (1), we can derive another formula. Write the equality (1) for $e_{n-1:} e_{n-1}=1 / 6\left(e_{n-2}+e_{n-3}+\ldots+e_{n-7}\right)+1$.

Now, subtracting the second equality from the first term by term, we get: $e_{n}-e_{n-1}=1 / 6\left(e_{n-1}-e_{n-7}\right)$, from which $e_{n}=7 / 6 e_{n-1}$ $-1 / 6 e^{n-7}$.

In deriving this formula, it was already assumed that $e_{n-1}>0$, so we can calculate $e_{n}$ only starting from $n=2$.

Since $e_{1}=1$, using formula (2) we sequentially compute: $e_{2}={ }^{7} / 6 \cdot 1-1 / 6 \cdot 0=7 / 6 ; e_{3}=7 / 6 \cdot{ }^{7} / 6-1 / 6 \cdot 0=49 / 36$; $e_{4}=7 / 6 \cdot 49 / 36-1 / 6 \cdot 0=343 / 216, \ldots$,

$e_{2010} \approx 574,761904$

## Answer

$\approx 574,761904$.

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