Arnold, Bendegúz, Cézár, Dorottya, Egon and Ferdinánd are preparing for a role-playing game. They haven't decided who will be the game master yet. (In this game, one of them will be the game master, and the others will form the party.) They want to decide on the game master by drawing lots, using a 10-sided die that is used in the game, as follows:

In a predetermined order, everyone rolls the die once. The one who rolls a 1 or a 2 first will be the game master. If no one has become the game master by the end of the round, they start over in the same order, rolling until someone rolls a 1 or a 2.

Dorottya has never run a game in this group before and really wants to this time. The boys are polite and let her choose her turn to roll. What would you recommend to Dorottya?
I. solution. The probability that the first roll will be a 1 or a 2 is $P_{1}=\frac{2}{10}=\frac{1}{5}$. The probability that the first roll is not, but the second is, is $P_{2}=\frac{4}{5} \cdot \frac{1}{5}$. Similarly, we get $P_{3}=\left(\frac{4}{5}\right)^{2} \cdot \frac{1}{5}, \ldots, P_{k}=\left(\frac{4}{5}\right)^{k-1} \cdot \frac{1}{5}$.

Therefore, the probability that the first player to roll will become the game master is

$$
p_{1}=\sum_{i=0}^{\infty} P_{6 i+1}=\sum_{i=0}^{\infty}\left(\frac{4}{5}\right)^{6 i} \cdot \frac{1}{5}
$$

Similarly,

$$
\begin{gathered}
p_{2}=\sum_{i=0}^{\infty} P_{6 i+2}=\sum_{i=0}^{\infty}\left(\frac{4}{5}\right)^{6 i+1} \cdot \frac{1}{5}=\sum_{i=0}^{\infty} \frac{4}{5} \cdot\left(\frac{4}{5}\right)^{6 i} \cdot \frac{1}{5}=\frac{4}{5} p_{1}p_{4}>p_{5}>p_{6}$. Thus, I would advise Dorottya to roll first.

Zsófia Szalay (Budapest, Szent István Gymnasium, 11th grade)

Note. The probabilities in question can be calculated precisely:

$$
p_{1}=\sum_{i=0}^{\infty}\left(\frac{4}{5}\right)^{6 i} \cdot \frac{1}{5}=\frac{1}{5}\left(1+\left(\frac{4}{5}\right)^{6}+\left(\left(\frac{4}{5}\right)^{6}\right)^{2}+\ldots\right)=\frac{1}{5} \cdot \frac{1}{1-\left(\frac{4}{5}\right)^{6}}=\frac{5^{5}}{5^{6}-4^{6}}=\frac{3125}{11529}
$$

and so on. Their sum, as a check, is indeed 1:

$$
p_{1}+p_{2}+\cdots+p_{6}=\left(1+\frac{4}{5}+\cdots+\left(\frac{4}{5}\right)^{5}\right) \cdot \frac{1}{5} \cdot \frac{1}{1-\left(\frac{4}{5}\right)^{6}}=\frac{\left(\frac{4}{5}\right)^{6}-1}{\frac{4}{5}-1} \cdot \frac{1}{5} \cdot \frac{1}{1-\left(\frac{4}{5}\right)^{6}}=1
$$

II. solution. Let's examine the probability that the lottery will be decided on the $k$-th roll, when a 1 or 2 is rolled for the first time: as already seen, $P_{k}=\left(\frac{4}{5}\right)^{k-1} \cdot \frac{1}{5}$, so the $P_{k}$ probabilities strictly decrease:

$$
\underbrace{P_{1}>P_{2}>P_{3}>P_{4}>P_{5}>P_{6}}_{1 . \text { round }}>\underbrace{P_{7}>P_{8}>\ldots}_{2 . \text { round }}
$$

It can be seen that in every round, the player who rolls first has the highest chance of becoming the game master. Therefore, if Dorottya wants to be the game master, she should roll first.

Péter Lang (Győr, Révai M. Gymnasium, 10th grade)

Note. Many wrote that since $P_{1}>P_{2}>\ldots$, the probability of rolling a 1 or 2 decreases with increasing $n$. Therefore, for $k=1$, this probability is the highest, so Dorottya should roll first. This is a less complete version of the above solution.