A fair die is rolled many times. It is known that at some point the sum of the points became exactly 2010.

Find the expected number of rolls made by this point.

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The first method. Let the event $A_{n}$ be "the sum of points is $n$", and $X_{n}$ be the number of rolls made to achieve this. We consider the event $A_{n}$ to have already occurred, and we are interested in the expectation $\mathrm{E} X_{n}$. Let $I_{k}$ be 1 if the first roll gives $k$ points, and 0 otherwise.

Events $B_{k}$ "the first roll gives $k$ points" and $A_{n-k}$ (for the remaining rolls) relate to different rolls and are therefore independent. Consequently, the conditional probability of the event ($I_{k}=1$) is

$\mathrm{P}\left(B_{k} \mid A_{n}\right)=\frac{\mathrm{P}\left(B_{k} \cap A_{n}\right)}{\mathrm{P}\left(A_{n}\right)}=\frac{\mathrm{P}\left(B_{k} \cap A_{n-k}\right)}{\mathrm{P}\left(A_{n}\right)}=\frac{\mathrm{P}\left(A_{n-k}\right)}{6 \mathrm{P}\left(A_{n}\right)}$.

It is clear that $X_{n}=X_{n-1} I_{1}+X_{n-2} I_{2}+\ldots+X_{n-6} I_{6}+1$.

Since the random variables $X_{n-k}$ and $I_{k}$ are independent, as they relate to different series of rolls, we have

$\mathrm{E}\left(X_{n}\right)=\sum_{k=1}^{6} \mathrm{E} X_{n-k} \mathrm{E} I_{k}+1=\sum_{k=1}^{6} \frac{\mathrm{E} X_{n-k} \mathrm{P}\left(A_{n-k}\right)}{6 \mathrm{P}\left(A_{n}\right)}+1$

Introducing the notation $\mathrm{E} X_{k}=e_{k}$ and $\mathrm{P}\left(A_{k}\right)=p_{k}$, we get: $e_{n}=\sum_{k=1}^{6} \frac{e_{n-k} p_{n-k}}{6 p_{n}}+1$.

It remains to use the fact that $6 p_{n}=p_{n-1}+\ldots+p_{n-6}$. This follows, for example, from the fact that the sum of the probabilities $P\left(I_{k}=1\right)$ is one. Therefore, $e_{n}=\frac{e_{n-1} p_{n-1}+\ldots+e_{n-6} p_{n-6}}{p_{n-1}+\ldots+p_{n-6}}+1$.

Initial values: $p_{-5}=p_{-4}=\ldots=p_{-1}=0 ; p_{0}=1 ; e_{-5}=e_{-4}=\ldots=e_{0}=0$. Sequential calculation gives $e_{1}=1, p_{1}=1 / 6$

$e_{2}=\frac{1 \cdot \frac{1}{6}+0 \cdot 1+0 \cdot 0+0 \cdot 0+0 \cdot 0+0 \cdot 0}{\frac{1}{6}+1+0+0+0+0}+1=\frac{8}{7}, \quad p_{2}=\frac{\frac{1}{6}+1}{6}=\frac{7}{36}$

$e_{3}=\frac{\frac{8}{7} \cdot \frac{7}{36}+1 \cdot \frac{1}{6}+0 \cdot 1+0 \cdot 0+0 \cdot 0+0 \cdot 0}{\frac{7}{36}+\frac{1}{6}+1+0+0+0}+1=\frac{9}{7} ; \ldots$

$e_{2010}=574.5238095 \ldots$

The second method. It is known that the sum of points equals $n$. Let the event $A_{k}$ be "at some point the sum is $k$". Then the subsequent rolls must lead to the event $A_{n-k}$. Therefore,

$\mathrm{P}\left(A_{k} \mid A_{n}\right)=\frac{\mathrm{P}\left(A_{k} \cap A_{n}\right)}{\mathrm{P}\left(A_{n}\right)}=\frac{\mathrm{P}\left(A_{k} \cap A_{n-k}\right)}{\mathrm{P}\left(A_{n}\right)}=\frac{\mathrm{P}\left(A_{k}\right) \mathrm{P}\left(A_{n-k}\right)}{\mathrm{P}\left(A_{n}\right)}$ (events $A_{n-k}$ and $A_{k}$ relate to different series of rolls and are therefore independent).

Consider the quantity $J_{k}$, which is 1 if the sum at some point equals $k$, and 0 otherwise.

Obviously, the random variable "number of obtained sums" $X_{n}=J_{1}+J_{2}+\ldots+J_{n}$ equals the number of rolls made.

Introducing the short notations $\mathrm{E} X_{k}=e_{k}, \mathrm{P}\left(A_{k}\right)=p_{k}$. The expectation $\mathrm{E} J_{k}$ is

$0 \cdot\left(1-\mathrm{P}\left(A_{k} \mid A_{n}\right)\right)+1 \cdot \mathrm{P}\left(A_{k} \mid A_{n}\right)=\mathrm{P}\left(A_{k} \mid A_{n}\right)=\frac{p_{k} p_{n-k}}{p_{n}}$, therefore,
$e_{n}=\mathrm{E} X_{n}=\sum_{k=1}^{n} \mathrm{E} J_{k}=\sum_{k=1}^{n} \frac{p_{k} p_{n-k}}{p_{n}}=\sum_{k=1}^{n-1} \frac{p_{k} p_{n-k}}{p_{n}}+1$

Using the relation $p_{n-k}=1 / 6\left(p_{n-k-1}+\ldots+p_{n-k-6}\right)$, we can rewrite the expectation as follows:

$$
\begin{aligned}
& e_{n}=\frac{1}{6 p_{n}} \sum_{k=1}^{n-1} \sum_{i=1}^{6} p_{k} p_{n-k-i}+1=\frac{1}{6 p_{n}} \sum_{i=1}^{6} \sum_{k=1}^{n-1} p_{k} p_{n-k-i}+1= \\
& \frac{1}{6 p_{n}} \sum_{i=1}^{6} p_{n-i} \sum_{k=1}^{n-1} \frac{p_{k} p_{n-k-i}}{p_{n-i}}+1=\sum_{i=1}^{6} \frac{e_{n-i} p_{n-i}}{6 p_{n}}
\end{aligned}
$$

## Answer