6. Let $1 \leqslant r \leqslant n$. Then the arithmetic mean of the smallest numbers in all $r$-element subsets of the set $M=\{1,2, \cdots, n\}$ is $\qquad$
6. $\frac{n+1}{r+1}$.

The number of subsets of $M$ containing $r$ elements is $\mathrm{C}_{n}^{r}$, and the number of subsets of $M$ containing $r$ elements with the smallest number being the positive integer $k$ is $\mathrm{C}_{n-k}^{r-1}$. Therefore, the arithmetic mean of the smallest numbers in all subsets of $M$ containing $r$ elements is
$$
\begin{array}{l} 
\frac{1}{\mathrm{C}_{n}^{r}}\left[1 \times \mathrm{C}_{n-1}^{r-1}+2 \times \mathrm{C}_{n-2}^{r-1}+\cdots+(n-r+1) \times \mathrm{C}_{r-1}^{r-1}\right] \\
= \frac{1}{\mathrm{C}_{n}^{r}}\left[\left(\mathrm{C}_{n-1}^{r-1}+\mathrm{C}_{n-2}^{r-1}+\cdots+\mathrm{C}_{r-1}^{r-1}\right)+\right. \\
\left.\left(\mathrm{C}_{n-2}^{r-1}+\cdots+\mathrm{C}_{r-1}^{r-1}\right)+\cdots+\mathrm{C}_{r-1}^{r-1}\right] \\
= \frac{1}{\mathrm{C}_{n}^{r}}\left(\mathrm{C}_{n}^{r}+\mathrm{C}_{n-1}^{r}+\cdots+\mathrm{C}_{r}^{r}\right)=\frac{\mathrm{C}_{n+1}^{r+1}}{\mathrm{C}_{n}^{r}}=\frac{n+1}{r+1} .
\end{array}
$$