We have a $10 \mathrm{Ft}$ coin and several (at least four) $20 \mathrm{Ft}$ coins in our pocket. We take out one coin at a time randomly until the amount we have taken out is not a positive integer multiple of $30 \mathrm{Ft}$. On average, how many draws are needed for this?
Solution. Let's examine what the outcomes of the draws can be:

1.) If we draw a 10 first, then we are sure to draw a 20 second.

2.) If we draw a 20 first, then a 10 second.

3.) If we draw a 20 first, then another 20, and finally a 10. At this point, only 20s are left, so we draw one more 20 twice.

4.) The first three draws are all 20s.

There are no more possibilities.

Let $n$ be the number of 20s. Then the probabilities of the cases are respectively:

$$
\begin{aligned}
& p_{1}=\frac{1}{n+1} \cdot 1=\frac{1}{n+1} \\
& p_{2}=\frac{n}{n+1} \cdot \frac{1}{n}=\frac{1}{n+1} \\
& p_{3}=\frac{n}{n+1} \cdot \frac{n-1}{n} \cdot \frac{1}{n-1} \cdot 1 \cdot 1=\frac{1}{n+1} \\
& p_{4}=\frac{n}{n+1} \cdot \frac{n-1}{n} \cdot \frac{n-2}{n-1}=\frac{n-2}{n+1}
\end{aligned}
$$

(The sum of the four probabilities is indeed 1.)

On average, therefore,

$$
2 \cdot \frac{1}{n+1}+2 \cdot \frac{1}{n+1}+5 \cdot \frac{1}{n+1}+3 \cdot \frac{n-2}{n+1}=3
$$

draws are needed.