We colored the fields of a chessboard with three colors as shown in the diagram. We randomly place a knight on the board, then move it one step randomly (but legally). What is the probability that the knight lands on a field of the same color as the starting field?

![](https://cdn.mathpix.com/cropped/2024_05_02_83d829d1e3ac2391f3cbg-1.jpg?height=371&width=374&top_left_y=297&top_left_x=865)
I. Solution. Let's label the chessboard squares as shown in the diagram. Since we place the knight randomly on the board, the probability of selecting any square is $\frac{1}{64}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_cde5ff29be62bc3c526fg-1.jpg?height=404&width=409&top_left_y=286&top_left_x=847)

For the following, we calculate the probability of moving to a square of the same color in one move for each square. Multiplying the sum of these probabilities by $\frac{1}{64}$ gives us the desired probability. The colored board is symmetric about the H1-A8 diagonal, so we only calculate the sum of probabilities for the squares bounded by the thick line, then add the probabilities for the squares on the H1-A8 diagonal to twice the result.

A "valid" move in every case means moving to a square of the same color as the starting square. I will detail the first four diagonal symmetries, and for the other cases, I will only provide the probabilities, with the numerator of the fraction representing the valid moves and the denominator representing the possible moves.

1.) A1: 2 possible and 0 valid moves, so the desired probability is 0.

2.) B1-A2: For both B1 and A2, there are 3 possible and 1 valid move, so the desired probability is $\frac{1}{3}$. The sum of probabilities for the squares on the B1-A2 "diagonal" is $\frac{2}{3}$.

3.) C1-A3:

For all three squares, there are 4 possible and 2 valid moves, so the desired probability is $\frac{2}{4}=\frac{1}{2}$. The sum of probabilities for the squares on the C1-A3 "diagonal" is $\frac{3}{2}$.

4.) D1-A4:

For D1 and A4, there are 4 possible and 2 valid moves, so the desired probability is $\frac{2}{4}=\frac{1}{2}$. For C2 and B3, there are 6 possible and 3 valid moves, so the desired probability is $\frac{3}{6}=\frac{1}{2}$. The sum of probabilities for the squares on the D1-A4 "diagonal" is $\frac{4}{2}=2$.

5.) The sum of probabilities for the squares on the E1-A5 diagonal:

$$
\frac{2}{4}+\frac{3}{6}+\frac{4}{8}+\frac{3}{6}+\frac{2}{4}=\frac{5}{2}
$$

6.) The sum of probabilities for the squares on the F1-A6 diagonal:

$$
\frac{2}{4}+\frac{3}{6}+\frac{4}{8}+\frac{4}{8}+\frac{3}{6}+\frac{2}{4}=3
$$

7.) The sum of probabilities for the squares on the G1-A7 diagonal:

$$
\frac{2}{3}+\frac{3}{6}+\frac{4}{8}+\frac{4}{8}+\frac{4}{8}+\frac{3}{6}+\frac{2}{3}=\frac{23}{6}
$$

8.) The sum of probabilities for the squares on the H1-A8 diagonal:

$$
\frac{2}{2}+\frac{2}{4}+\frac{4}{8}+\frac{4}{8}+\frac{4}{8}+\frac{4}{8}+\frac{2}{4}+\frac{2}{2}=5
$$

Thus, the sum of probabilities for the squares:

$$
2 \cdot\left(0+\frac{2}{3}+\frac{3}{2}+2+\frac{5}{2}+3+\frac{23}{6}\right)+5=2 \cdot \frac{27}{2}+5=32
$$

Therefore, the probability that the randomly placed knight moves to a square of the same color as the starting square is $\frac{1}{64} \cdot 32=\frac{1}{2}$.

II. Solution (outline). The result of $\frac{1}{2}$ is too beautiful not to try to find a more elegant solution. This (since the probability of selecting any square is $\frac{1}{64}$) suggests a symmetry such that every valid move can be uniquely paired with an invalid move. The pairing is as follows: for each move, we assign the move obtained by reflecting it over the perpendicular bisector of the first row of the chessboard.

Remarks. 1. Those who solved the problem well thought similarly to the first solution. No one worked similarly to the second solution.

2. The most common mistake was assuming that every move is equally probable, and thus the desired probability is the ratio of favorable moves to all possible moves. However, this assumption is clearly not true, as the moves from squares with fewer possible knight moves are more likely. Additionally, calculation errors were very common (for example, miscalculating the probability of a favorable move for a particular square).