TITLE: Probability of getting a $7$ in Minesweeper
QUESTION [19 upvotes]: Let's say you have a $30$ by $16$ grid and $99$ mines. What is the probability of having at least one empty block surrounded by exactly $7$ mines?

For the sake of this question, assume that the mines are generated randomly.

REPLY [12 votes]: The average number of $7's$, which is slightly different.  
There are $14\times28 =392$ places to put a $7$.
There are eight places to put the non-mine.
There are $9$ squares involved with the $7$, so $480-9=471$ other squares.
These other squares contain the $92$ other mines.  So the number of grids with a $7$ at a particular spot is $$8\times {471\choose 92}.$$  That is out of a total of $(480$ choose $99)$ different grids.
The chance of a $7$ in any one of those is 
$$\frac{{8\choose1}{480-9\choose 92}}{480\choose99}\approx 0.00006928$$
so the average number of $7$s is $392$ times that, or approximately 
$$0.02716$$
The average number of $8s$ would be 
$$\frac{392{471\choose91}}{480\choose99}\approx 0.0008219$$