16. Non-strict Backgammon (6 p. 9-11). A fair die is rolled many times. Find the expected number of rolls made until the sum of all the points rolled reaches 2010.
Solution. Let $X_{n}$ be the number of rolls required to reach a sum of $n$. Introduce the random variable

$$
I_{k}=\left\{\begin{array}{l}
1, \text { if the first roll gives } k \text { points, } \\
0 \text { otherwise. }
\end{array}\right.
$$

Obviously, $\mathrm{E} I_{k}=0 \cdot \frac{5}{5}+1 \cdot \frac{1}{6}=\frac{1}{6}$.

If we denote $X_{n-k}$ as the number of rolls, starting from the second, required to reach a sum of $n-k$, then

$$
X_{n}=X_{n-1} I_{1}+X_{n-2} I_{2}+\ldots+X_{n-6} I_{6}+1
$$

The quantity $X_{n-k}$ is independent of the result of the first roll. Therefore, the quantities $X_{n-k}$ and $I_{k}$ are independent. Transitioning to expectations, we get:

$$
\mathrm{E} X_{n}=\sum_{k=1}^{6} \mathrm{E} X_{n-k} \mathrm{E} I_{k}=\frac{1}{6} \sum_{k=1}^{6} \mathrm{E} X_{n-k}+1
$$

Denoting $\mathrm{E} X_{k}=e_{k}$ for brevity, we write the recurrence formula:

$$
e_{n}=\frac{1}{6} \sum_{k=1}^{6} e_{n-k}+1
$$

In the reasoning, the positivity of the numbers $n-2, n-3$, etc., was nowhere used. Therefore, the obtained formula is valid for all $n \geq 1$, if we set $\mathrm{E} X_{k}=0$ for all $k \leq 0$.

From (1), another formula can be derived. Write the equality (1) for $e_{n-1}$:

$$
e_{n-1}=\frac{1}{6} \sum_{k=1}^{6} e_{n-k-1}+1
$$

Now, subtracting the second equality from the first term by term, we get:

$$
e_{n}-e_{n-1}=\frac{e_{n-1}-e_{n-7}}{6} \text {, hence } e_{n}=\frac{7}{6} e_{n-1}-\frac{1}{6} e_{n-7}
$$

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

Using formulas (1) and (2), we sequentially compute:

$$
\begin{array}{ll}
e_{1}=1+\frac{0+0+0+0+0+0}{6}=1 ; & e_{2}=\frac{7}{6} \cdot 1-\frac{1}{6} \cdot 0=\frac{7}{6} \\
e_{3}=\frac{7}{6} \cdot \frac{7}{6}-\frac{1}{6} \cdot 0=\frac{49}{36} ; & e_{4}=\frac{7}{6} \cdot \frac{49}{36}-\frac{1}{6} \cdot 0=\frac{343}{216} ; \ldots
\end{array}
$$

Remark. It seems that the expectations form a geometric progression with a common ratio of $\frac{7}{6}$. This does not align with the intuitively obvious fact that the number of rolls should grow approximately as an arithmetic progression with a difference of $\frac{2}{7}$ - exactly 2 rolls on average are needed to increase the sum by 7. Of course, this impression is false - the geometric growth is replaced by a slower growth after the sum of 6 (as soon as $e_{n-7}$ becomes a positive number), and everything converges to the expected values.

Naturally, to approximate $e_{2010}$, one can use the found recurrence formula and some computational tool. For example, using Excel, we quickly find: $e_{2010}=574.761904$.

If we assume that the desired quantity grows approximately as an arithmetic progression with a difference of $\frac{2}{7}$, we can find $e_{2010} \approx 2010 \cdot \frac{2}{7} \approx 574.286$. As we see, the error is small.