10. Given a positive integer $n \geqslant 2$, Betya and Vasya are playing a game. Betya selects $2 n$ (possibly equal) non-negative numbers $x_{1}$, $x_{2}, \cdots, x_{2 n}$, whose sum equals 1. Vasya then places them in some order around a circle, after which he calculates the product of each pair of adjacent numbers and writes the largest of these $2 n$ products on the board. Betya wants the number written on the board to be as large as possible, while Vasya wants it to be as small as possible. What will the number written on the board be under the correct strategy?
10. $\frac{1}{8(n-1)}$.

If Betya chooses the array
$$
0, \frac{1}{2}, \frac{1}{4(n-1)}, \frac{1}{4(n-1)}, \cdots, \frac{1}{4(n-1)},
$$

then $\frac{1}{2}$ will be adjacent to some $\frac{1}{4(n-1)}$, indicating that there will be a product of $\frac{1}{8(n-1)}$, and the rest of the products will not be larger. Thus, $\frac{1}{8(n-1)}$ will appear on the board.

Next, we show that for any array, Vasya can ensure that the number appearing on the board does not exceed $\frac{1}{8(n-1)}$.
Let the numbers chosen by Betya be arranged in non-increasing order as
$$
x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{2 n} \text {. }
$$

Place $x_{1}$ on the circumference, starting from it, place $x_{2}, x_{3}, \cdots, x_{n}$ in a clockwise direction, then place $x_{2 n}$ between $x_{n}$ and $x_{1}$, $x_{2 n-1}$ between $x_{1}$ and $x_{2}$, $x_{2 n-2}$ between $x_{2}$ and $x_{3}$, and so on, placing $x_{n+1}$ between $x_{n-1}$ and $x_{n}$. Thus, the products of adjacent pairs are
$$
\begin{array}{l}
x_{n} x_{2 n}, x_{1} x_{2 n}, x_{2} x_{2 n-1}, x_{3} x_{2 n-2}, \cdots, \\
x_{k} x_{2 n-k+1}, \cdots, x_{n} x_{n+1},
\end{array}
$$

and $x_{1} x_{2 n-1}, x_{2} x_{2 n-2}, x_{3} x_{2 n-3}, \cdots, x_{k} x_{2 n-k}, \cdots, x_{n-1} x_{n+1}$.
Since $x_{k} x_{2 n-k+1} \leqslant x_{k} x_{2 n-k}(k=1,2, \cdots, n-1)$ and $x_{n} x_{n+1} \leqslant x_{n-1} x_{n+1}, x_{n} x_{2 n} \leqslant x_{n-1} x_{n+1}$, the largest product can only appear in the second row.
$$
\begin{array}{l}
\text { For } 1 \leqslant k \leqslant n-1 \text {, let } a=x_{k} x_{2 n-k} \text {. } \\
\text { By } x_{k} \leqslant x_{k-1} \leqslant \cdots \leqslant x_{1} \text {, we have } \\
k x_{k} \leqslant x_{1}+x_{2}+\cdots+x_{k} \text {. } \\
\text { Then } k a=k x_{k} x_{2 n-k} \leqslant\left(x_{1}+x_{2}+\cdots+x_{k}\right) x_{2 n-k} \text {. } \\
\text { By } x_{2 n-k} \leqslant x_{2 n-k-1} \leqslant x_{2 n-k-2} \leqslant \cdots \leqslant x_{k+1} \text {, we have } \\
(2 n-2 k) x_{2 n-k} \\
\leqslant x_{2 n-k}+x_{2 n-k-1}+\cdots+x_{k+1} \\
\leqslant x_{k+1}+x_{k+2}+\cdots+x_{2 n} \\
=1-x_{1}-x_{2}-\cdots-x_{k}, \\
2 k(n-k) a \\
\leqslant\left(x_{1}+x_{2}+\cdots+x_{k}\right)\left(1-x_{1}-x_{2}-\cdots-x_{k}\right) \\
=x(1-x),
\end{array}
$$

where $x=x_{1}+x_{2}+\cdots+x_{k}$.
By the AM-GM inequality, we have
$$
x(1-x) \leqslant\left(\frac{x+(1-x)}{2}\right)^{2}=\frac{1}{4} \text {. }
$$

Noting that, for $k \leqslant n-1$,
$$
\begin{array}{l}
(k-1)(n-k-1) \geqslant 0 \\
\Rightarrow k(n-k) \geqslant n-1 .
\end{array}
$$

Thus, $x_{k} x_{2 n-k}=a$
$$
\leqslant \frac{1}{8 k(n-k)} \leqslant \frac{1}{8(n-1)} .
$$