Gombóc Artúr lives at 1 Édes Street, while the chocolate shop is at the other end of the street, at number $n$. Every day, Artúr follows this fitness routine: he starts in front of the house numbered 2. If he stands in front of the house numbered $k$ (where $1<k<n$), he flips an expired but fair chocolate coin. If it lands heads, he moves to the house numbered ( $k-1$ ), and if it lands tails, he moves to the house numbered ( $k+1$ ). If he reaches the chocolate shop, he goes in and drinks a chocolate milkshake, then moves to the house numbered ( $n-1$ ). When he gets home, his workout is over. On average, how many chocolate milkshakes does Artúr drink each day?
On average, he rolls down 1 chocolate ball per day. We call an $n$-training the exercise performed on an $n$-length street, and let $f(n)$ be the average number of chocolate balls eaten. We prove by complete induction that $f(n)=1$. The statement is clearly true for $n=2$. Assuming it is true for $n=k \geq 2$, we examine the case for $n=k+1$. At the beginning of the $(k+1)$-training, with $\frac{1}{2}$ probability, he performs a $k$-training and returns to house 2, and with $\frac{1}{2}$ probability, he returns to house 1 with 0 chocolate balls. If he performed a $k$-training, afterwards he again has $\frac{1}{2}$ probability to perform another $k$-training, and $\frac{1}{2}$ probability to return to house 1, etc. Using the fact that

$$
\frac{1}{2^{l}}+\frac{1}{2^{l+1}}+\ldots=\frac{\left(\frac{1}{2}\right)^{l}}{1-\frac{1}{2}}=\left(\frac{1}{2}\right)^{l-1}
$$

the following can be written:

$$
\begin{aligned}
f(k+1) & =\frac{1}{2} \cdot 0+\frac{1}{4} f(k)+\frac{1}{8} \cdot 2 f(k)+\frac{1}{16} \cdot 3 f(k)+\ldots= \\
& =f(k)\left(\frac{1}{4}+\frac{1}{8}+\ldots\right)+f(k)\left(\frac{1}{8}+\frac{1}{16}+\ldots\right)+\ldots= \\
& =f(k)\left(\frac{1}{2}+\frac{1}{4}+\ldots\right)=f(k)
\end{aligned}
$$

This proves the statement for $n=k+1$ as well.

Therefore, Arthur rolls down on average 1 chocolate ball every day.