A chessboard has a dark bishop in the bottom left corner and a light bishop in the bottom right corner. Both bishops move up the board one square at a time, randomly moving left or right, until they reach the top row. What is the probability that the dark bishop reaches the top row to the right of the light bishop?
Let's first look at the dark pawn. The probability that it has been on the bottom-left square is 1, as it starts from there. From there, it can only move to the 2nd row, 2nd square, so the probability that it has been on the 2nd row, 2nd square is also 1:

$$
p_{(1,1)}=p_{(2,2)}=1
$$

From there, it can move in two directions, each with a probability of $\frac{1}{2}$:

$$
p_{(3,1)}=p_{(3,3)}=\frac{1}{2}
$$

From the 3rd row, 1st square, it will definitely move to the 4th row, 2nd square, and it can also move there from the 3rd row, 3rd square. Therefore, the probability for this is

$$
p_{(4,2)}=\frac{1}{2}+\frac{1}{2} \cdot \frac{1}{2}=\frac{3}{4}
$$

The probability for the 4th row, 4th square is

$$
p_{(4,4)}=\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}
$$

The other probabilities can be calculated similarly:

$$
\begin{array}{ll}
p_{(5,1)}=\frac{1}{2} \cdot \frac{3}{4}=\frac{3}{8}, & p_{(5,3)}=\frac{1}{2} \cdot \frac{3}{4}+\frac{1}{2} \cdot \frac{1}{4}=\frac{4}{8} \\
p_{(5,5)}=\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8} ; & p_{(6,2)}=\frac{3}{8}+\frac{1}{2} \cdot \frac{4}{8}=\frac{10}{16} \\
p_{(6,4)}=\frac{1}{2} \cdot \frac{4}{8}+\frac{1}{2} \cdot \frac{1}{8}=\frac{5}{16}, & p_{(6,6)}=\frac{1}{2} \cdot \frac{1}{8}=\frac{1}{16} ; \\
p_{(7,1)}=\frac{1}{2} \cdot \frac{10}{16}=\frac{10}{32}, & p_{(7,3)}=\frac{1}{2} \cdot \frac{10}{16}+\frac{1}{2} \cdot \frac{5}{16}=\frac{15}{32}, \\
p_{(7,5)}=\frac{1}{2} \cdot \frac{5}{16}+\frac{1}{2} \cdot \frac{1}{16}=\frac{6}{32}, & p_{(7,7)}=\frac{1}{2} \cdot \frac{1}{16}=\frac{1}{32} \\
p_{(8,2)}=\frac{10}{32}+\frac{1}{2} \cdot \frac{15}{32}=\frac{35}{64}, & p_{(8,4)}=\frac{1}{2} \cdot \frac{15}{32}+\frac{1}{2} \cdot \frac{6}{32}=\frac{21}{64} \\
p_{(8,6)}=\frac{1}{2} \cdot \frac{6}{32}+\frac{1}{2} \cdot \frac{1}{32}=\frac{7}{64}, & p_{(8,8)}=\frac{1}{2} \cdot \frac{1}{32}=\frac{1}{64} .
\end{array}
$$

|  | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 8 | 1 | 35 | 7 | 21 | $\underline{21}$ | 7 | 35 | 1 |
|  |  |  | 64 | 64 | 64 | 64 | 64 |  |
| 7 | 10 |  | $\underline{15}$ |  | $\underline{6}$ |  | $\frac{1}{n}$ |  |
|  |  | 10 |  |  |  |  |  |  |
| 6 |  | $\frac{10}{16}$ |  | $\frac{\circ}{16}$ |  | $\frac{\overline{1}}{16}$ |  |  |
|  | 3 |  | 4 |  | $\mathrm{r}_{1}$ |  |  |  |
| 5 | $\overline{8}$ |  | $\overline{8}$ |  | $\overline{8}$ |  |  |  |
| 4 |  | $\frac{3}{-}$ |  | $\underline{1}$ |  |  |  |  |
|  | 1 |  | 1 |  |  |  |  |  |
| 3 | $\frac{1}{2}$ |  | $\frac{1}{2}$ |  |  |  |  |  |
| 2 |  | 1 |  |  |  |  |  |  |
|  | 1 |  |  |  |  |  |  |  |

It can be seen that the sum of the probabilities of stepping on dark squares in each row is 1.

Due to symmetry, the probabilities of the light pawn landing in the last row can be easily given.

If the dark pawn is on the 8th square, the light pawn can be anywhere, always to its left. The probability of this is

$$
\frac{1}{64} \cdot 1=\frac{1}{64}
$$

If the dark pawn is on the 6th square, the light pawn can be on the 1st, 3rd, or 5th square, so the probability of this is

$$
\frac{7}{64} \cdot\left(\frac{1}{64}+\frac{7}{64}+\frac{21}{64}\right)=\frac{203}{64^{2}}
$$

If the dark pawn is on the 4th square, the light pawn can be on the 1st or 3rd square, so the probability of this is

$$
\frac{21}{64} \cdot\left(\frac{7}{64}+\frac{1}{64}\right)=\frac{168}{64^{2}}
$$

Finally, if the dark pawn is on the 2nd square, the light pawn can only be on the 1st square, and the probability of this is $\frac{35}{64} \cdot \frac{1}{64}=\frac{35}{64^{2}}$.

Thus,

$$
\frac{1}{64}+\frac{203}{64^{2}}+\frac{168}{64^{2}}+\frac{35}{64^{2}}=\frac{470}{64^{2}} \approx 0.115
$$

is the probability that the dark pawn arrives to the right of the light pawn.

## Ajtai Boglárka (Miskolc, Földes Ferenc Gimn., 11th grade)

Note. Many people calculated how many different paths each pawn could take to reach the squares in the top row of the chessboard, then divided this by the total number of possible paths, and thus arrived at the probabilities of reaching each square. However, according to the problem statement, the pawns did not choose randomly from among the paths, but at each step, they chose whether to move up to the left or up to the right (provided this did not result in stepping off the chessboard). On the edge of the chessboard, only one type of move was possible with a probability of 1, which resulted in the different paths not having the same probability. Thus, this solution is incorrect.