Anna and Zsófi take turns rolling a die, and they always add the rolled number to the sum of the numbers rolled so far. The one who, after their roll, makes the sum first divisible by 4 wins. If Anna starts, what is the probability that she wins?
Solution. Regardless of the sum of the numbers rolled up to a given moment - unless it is divisible by 4, in which case the game ends - the next player can roll a number that ensures a win: if the sum is of the form $4k+1$, they can roll a 3; if it is of the form $4k+2$, they can roll a 2 or a 6; and if it is of the form $4k+3$, they can roll a 1 or a 5.

Thus, with each roll, there is at least a $\frac{1}{6}$ chance that the game will end with the next roll. The probability that the game has not ended after $n$ rolls is at most $\left(\frac{5}{6}\right)^{n}$, and this value approaches zero. Therefore, the game ends with probability 1.

Suppose that at a given state of the game, the sum of the rolled numbers is of the form $4k+r$ $(0 \leq r \leq 3)$. If $P_{r}$ denotes the probability that the next player - who could be either Anna or Zsófi, let's call them $X$ - eventually wins, then this probability clearly depends only on the remainder $r$. On the other hand, since the game ends, the other player, who follows $X$, wins with probability $1-P_{r}$. Here, it is clear that $P_{0}=0$, since $X$ has already lost. If $X$'s roll results in a sum of the form $4k+r'$, then now it is not $X$ who follows, so according to the above, $1-P_{r'}$ is the probability that $X$ wins. In the following, we use this simple observation when writing down the probabilities $P_{r}$.

Let's examine the non-zero remainders; let $r=1$ first, and see what the probability is that $X$ wins the game in the situation that arises after their possible rolls. If they roll a 3 - the probability of which is $\frac{1}{6}$ - then they win with probability $\left(1-P_{0}\right)=1$. If they roll a 1 or a 5 - the probability of which is $\frac{1}{3}$ - then, arriving at a number of the form $4k+2$, they win with probability $\left(1-P_{2}\right)$. Similarly, if they roll a 2 or a 6 - the probability of which is also $\frac{1}{3}$ - then they win with probability $\left(1-P_{3}\right)$, and if they roll a 4, they win with probability $\left(1-P_{1}\right)$. Therefore:

$$
P_{1}=\frac{1}{6}\left(1-P_{0}\right)+\frac{1}{3}\left(1-P_{2}\right)+\frac{1}{3}\left(1-P_{3}\right)+\frac{1}{6}\left(1-P_{1}\right)
$$

Similarly, the probabilities $P_{2}$ and $P_{3}$ can be written as:

$$
\begin{aligned}
& P_{2}=\frac{1}{3}\left(1-P_{0}\right)+\frac{1}{3}\left(1-P_{3}\right)+\frac{1}{6}\left(1-P_{1}\right)+\frac{1}{6}\left(1-P_{2}\right), \quad \text { and } \\
& P_{3}=\frac{1}{3}\left(1-P_{0}\right)+\frac{1}{3}\left(1-P_{1}\right)+\frac{1}{6}\left(1-P_{2}\right)+\frac{1}{6}\left(1-P_{3}\right)
\end{aligned}
$$

Rearranging the three equations, we get the following system of equations:

$$
7 P_{1}+2 P_{2}+2 P_{3}=6
$$

From (2):

$P_{1}+7 P_{2}+2 P_{3}=6$

From (3):

$2 P_{1}+P_{2}+7 P_{3}=6$.

Solving the system of equations, we find that $P_{1}=\frac{50}{99}, P_{2}=\frac{60}{99}, P_{3}=\frac{62}{99}$.

Now let's turn to the game between the two players, Anna and Zsófi. If Anna rolls a number of the form $4k+r$ on her first turn, then according to the above, she wins with probability $1-P_{0}=1$ if $r=0$; and with probability $1-P_{r}$ if $r=1,2,3$. Since she rolls a 0 or 3 remainder with probability $\frac{1}{6}$ and a 1 (1 and 5) or 2 (2 and 6) with probability $\frac{1}{3}$, the probability that Anna wins as the starting player is:

$$
\frac{1}{6}\left(1-P_{0}\right)+\frac{1}{3}\left(1-P_{1}\right)+\frac{1}{3}\left(1-P_{2}\right)+\frac{1}{6}\left(1-P_{3}\right)=\frac{1}{6}+\frac{1}{3} \cdot \frac{49}{99}+\frac{1}{3} \cdot \frac{39}{99}+\frac{1}{6} \cdot \frac{37}{99}=\frac{52}{99}
$$

which is slightly greater than $\frac{1}{2}$.

Remark. Many omitted the proof that the game ends with probability 1, thus the above reasoning is incomplete.