What is the probability that there are two knights that can attack each other on a 3*3 chessboard...?

Question

What is the probability that there are two knights that can attack each other on a 3*3 chess board if each square contains a knight with 1/2 probability?

Answer 1

So the middle square is irrelevant.And we want to find the probability that two elements in the cyclic list , $A1-B3-C1-A2-C3-B1-A3-C2-A1$, is adjacent.

We could have any of $A_1,C_1,C_3,A_3$ appear together or any of $B_3,A_2,B_1,C_2$ appear together. Also the 3 coner pieces, then side rows and colums, or any two none of which are the middle square adjacent. That gives a total of $2^4+2^4-1+4+4+8=2^5+16=32+16=47$. So the probability is $\frac{209}{256}$.

Answer 2

HINT (in case this is homework)

As many have pointed out you need to count the number of ways knights can populate the $8$-square cycle $A1-B3-C1-A2-C3-B1-A3-C2-A1,$ without any adjacent positions being occupied. Note that each of the $2^8$ possibilities are equally likely.

I can't think of a one-shot way to count this, but we can certainly count by conditioning on the number of knights.

• If there is $0$ knight, that is $1$ case.

• If there is $1$ knight, then there are no adjacent pair of knights. The number of ways to do this is ${8 \choose 1}$.

• If there are $2$ knights, then the total number of ways is ${8 \choose 2}$, but you have to subtract all adjacent-pair cases, and there are $X$ of those.

• If there are $3$ knights, this is the hardest case. However, note that if the $3$ knights are not adjacent, then the gaps must be $2-2-1$ or $3-1-1$. Lets say there are $Y$ of the $2-2-1$ case and $Z$ of the $3-1-1$ case. Then just add them up.

• If there are $4$ knights, there are only $W$ ways to for them to be non-adjacent.

• If there are $5$ or more knights, there must be an adjacent pair, so these cases contribute nothing to the count.

The total number of ways for a set of knights to be non-adjacent $= 1 + 8 + ({8 \choose 2} - X) + Y + Z + W$.

Answer 3

"Knightwise" we have a circular graph with $8$ vertices, and on some of these vertices there is a knight. We have to compute the probability $p$ that two adjacent vertices are taken. There are $2^8=256$ equiprobable configurations, and we have to count the configurations containing at least one pair of taken adjacent vertices. These are the good configurations.

If there are two knights there are $8$ good configurations, if there are three knights there are $8\cdot 4+8=40$ good configurations, and if there are four knights all configurations except the two regular ones are good, making ${8\choose4}-2=68$ good configurations. All ${8\choose5}+\ldots+{8\choose8}=93$ configurations with $\geq5$ knights are good. Since there are $8+40+68+93=209$ good configurations we obtain $$p={209\over256}\ .$$