Probability of getting a $7$ in Minesweeper

Solution 1:

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 $$\frac{392{471\choose91}}{480\choose99}\approx 0.0008219$$