Two-Click Mine Sweeper

Let's take a look at this problem through three cases: The first square Diana clicked is either 1) in a corner, 2) in an edge, or 3) anywhere else.

1) First click in a corner:

The number may be either 2 or 3. If it is 2, the mineless square is next to the one Diana clicked. The chance for the number 2 is $\frac{3}{n^2-1}$ as there are 3 squares next to the clicked square and the total number of non-clicked squares is $n^2-1$ . Then, Diana can guess for any adjacent square, so she hits with the chance of $\frac{1}{3}$ . If the number is 3, the mineless square can be anything but adjacent to the one Diana clicked. The chance for the number three is $\frac{n^2-4}{n^2-1}$ as there are $n^2-4$ squares that are non-clicked and not next to the clicked one. Now Diana, avoiding the squares adjacent to the one she clicked first, hits the mineless square with the chance of $\frac{1}{n^2-4}$ .

Thus, the total chance of succeeding is

$\frac{3}{n^2-1} \times \frac{1}{3}+\frac{n^2-4}{n^2-1} \times \frac{1}{n^2-4} =\boxed {\frac{2}{n^2-1}}$ .

2) First click in an edge:

The number may be either 4 or 5. If it is 4, the mineless square is next to the one Diana clicked. The chance for the number 4 is $\frac{5}{n^2-1}$ as there are 5 squares next to the clicked square. Then, Diana can guess for any adjacent square, so she hits with the chance of $\frac{1}{5}$ . If the number is 5, the mineless square can be anything but adjacent to the one Diana clicked. The chance for the number three is $\frac{n^2-6}{n^2-1}$ as there are $n^2-6$ squares that are non-clicked and not next to the clicked one. Now Diana, avoiding the squares adjacent to the one she clicked first, hits the mineless square with the chance of $\frac{1}{n^2-6}$ .

Thus, the total chance of succeeding is

$\frac{5}{n^2-1} \times \frac{1}{5}+\frac{n^2-6}{n^2-1} \times \frac{1}{n^2-6} =\boxed {\frac{2}{n^2-1}}$ .

3) First click neither in a corner nor in an edge:

The number may be either 7 or 8. If it is 7, the mineless square is next to the one Diana clicked. The chance for the number 7 is $\frac{8}{n^2-1}$ as there are 8 squares next to the clicked square. Then, Diana can guess for any adjacent square, so she hits with the chance of $\frac{1}{8}$ . If the number is 8, the mineless square can be anything but adjacent to the one Diana clicked. The chance for the number three is $\frac{n^2-9}{n^2-1}$ as there are $n^2-9$ squares that are non-clicked and not next to the clicked one. Now Diana, avoiding the squares adjacent to the one she clicked first, hits the mineless square with the chance of $\frac{1}{n^2-9}$ .

Thus, the total chance of succeeding is

$\frac{8}{n^2-1} \times \frac{1}{8}+\frac{n^2-9}{n^2-1} \times \frac{1}{n^2-9} =\boxed {\frac{2}{n^2-1}}$ .

As the chance of hitting the mineless square with the second click is the same in all cases, the final answer is $\boxed {\frac{2}{n^2-1}}$ .

Split the game into three parts and define the events $A:=$ "The mineless field is adjacent to the starting point", $B:=$ "Diana wins":

1. Diana chooses a starting point with $m$ adjacent squares
2. The game chooses a mineless field from the remaining $n^2-1$ squares and displays the number $N\in\{m,\:m-1\}$
3. Diana chooses a second square and either wins or looses

If $N=m-1$ , the mineless field is adjacent to the starting block and and if $N=m$ , it lies elsewhere. As Diana plays the best she can, she must choose her second square accordingly. We calculate her probability to win with Bayes' Theorem: $P(B)=P(B|A)P(A)+P(B|A^c)P(A^c)=\frac{1}{\cancel{m}}\cdot\frac{\cancel{m}}{n^2-1}+\frac{1}{\cancel{n^2-m-1}}\cdot\frac{\cancel{n^2-m-1}}{n^2-1}=\boxed{\frac{2}{n^2-1}}$