Ten people want to play a board game, and they want to draw the starting player. The host is entrusted to put some red and some white balls into a hat. They draw cyclically without replacement, and the one who first draws a red ball starts. The first drawer is the host. How should the host fill the hat if they don't really want to start?
Let there be $n$ white and $k$ red balls in the hat, and let $p$ and $q$ denote the probability that the first and the second person, respectively, draw a red ball first. It is easy to see that if $k+n>1$, then

$$
p-q=\frac{k}{k+n}-\frac{n}{k+n} \frac{k}{k+n-1}=p \cdot \frac{k-1}{k+n-1}
$$

If therefore $k>0$, then $p \geq q$. Equality can only occur if $k=1$. Thus, in the procedure, everyone has at least as high a probability of drawing a red ball as the next person (this is also true in the cases $k+n=1, k=0$), and the two probabilities are only equal if there are 0 or 1 red balls in the hat before the draw (if $k=0$, obviously $p=q=0$). Thus, the host can only achieve that he draws a red ball with the same probability as everyone else. For this, on the one hand, he must put 1 red ball into the hat, and on the other hand, he must ensure that everyone draws the same number of times, so he must put $10 h-1$ white balls next to it, where $h$ is any integer.

Kiss Emil (Budapest, Fazekas M. Gyak. Gimn.)