井字棋求和

给 $3\times3$ 网格的每个格子编号，并只看其中数字的奇偶性： 若某格中的整数为奇数，则在该格放一个 $1$；若为偶数，则放一个 $0$。 因为 1,3,5,7,9 一共有五个奇数，而 2,4,6,8 一共有四个偶数，所以我们需要一个恰好含有五个 $1$ 的 $3\times3$ 二元矩阵。

题目要求每一行和每一列的和都为奇数，也就是说每一行和每一列都必须含有奇数个 $1$。对于一行（或一列）中的 3 个位置来说，奇数个 $1$ 只能是 $1$ 个或 $3$ 个。由于总共有 5 个 $1$，所以每行的 $1$ 的个数（每列也一样）必须是 $(3,1,1)$ 的某种排列。

因此，恰好有一行全为 $1$，也恰好有一列全为 $1$。一旦选定哪一行（3 种选择）和哪一列（3 种选择）是“全 1 行”和“全 1 列”，这个矩阵就唯一确定了：把选定的那一整行和那一整列都填成 $1$。因此，一共得到 $$3\times 3 = 9$$ 个不同的奇偶性矩阵；每个矩阵都恰好有五个 $1$，并且每一行、每一列对模 2 都同余于 $1$。

对于每个这样的奇偶模式，可以把五个奇数填入这五个位置，共有 $5!$ 种方法；把四个偶数填入剩余四个位置，共有 $4!$ 种方法。因此，有利的排列数为 $$9\cdot 5!\cdot 4!.$$ 而总排列数是 $9!$，所以所求概率为 $$\frac{9\cdot 5!\cdot 4!}{9!} = \frac{9\cdot (5!\,4!)}{9!} = \frac{25920}{362880} = \boxed{\frac{1}{14}}.$$

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Original Explanation

Label the $3\times3$ grid cells and view each placed integer by its parity: put a $1$ in a cell if the integer there is odd and a $0$ if it is even. There are five odd integers (1,3,5,7,9) and four even integers (2,4,6,8), so we need a $3\times3$ binary matrix with exactly five $1$'s.

The requirement that each row and each column sum to an odd number means that each row and each column has an odd number of $1$'s. For a $3$-entry row (or column) the only possible odd counts are $1$ or $3$. Since the total number of $1$'s is $5$, the row-counts (and column-counts) must be a permutation of $(3,1,1)$.

Consequently exactly one row has three $1$'s and exactly one column has three $1$'s. Once we choose which row (3 choices) and which column (3 choices) are the "all-ones" row and column, the matrix is forced: put $1$'s in every cell of that chosen row and every cell of that chosen column. This yields exactly $$3\times 3 = 9$$ distinct parity matrices, each with exactly five $1$'s and with every row and column summing to $1\pmod 2$.

For each such parity pattern we can place the five odd numbers in the five cells in $5!$ ways and the four even numbers in the remaining four cells in $4!$ ways. Thus the number of favorable permutations of $1,\dots,9$ is $$9\cdot 5!\cdot 4!.$$ The total number of permutations is $9!$, so the desired probability is $$\frac{9\cdot 5!\cdot 4!}{9!} = \frac{9\cdot (5!\,4!)}{9!} = \frac{25920}{362880} = \boxed{\frac{1}{14}}.$$