How rare is it to get a $8$ in minesweeper? (Bruh reputation requirments)

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I need help on this, ignore if its already answered.

Ok so today i was wondering, could you get a $8$ in minesweeper and how rare it is?

All i know is that it will be rare. Very rare indeed. I dont really know how to say it. Its so annoying to be honest (tbh). I also dont know what you'd need to answer, so its complicated, well, because of that. If your answering, it might be hard.

Oh, i probably dont know but heres a predicted formula of how rare it is

First, this variable.
p8 = How rare it is to get 8 mines forming a hole like below. X = empty / O = mine
O O O
O X O
O O O
Rc = How rare it is to randomly click inside a patch that has minesweepers all directions you look, using the same
example as the square mines forming a hole in the middle (This means there is no middle mine therefore)
And then, the predicted formula below. It isnt advanced so you could make a better formula. It would please me.
p8 ÷ Rc = 8r
Forgot to mention. 8r = formula result

So yeah. Not much to explain because im new to stack exchange. Anyways, the end of this probably.

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2 Answers

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It depends on the size of the grid and the number of mines. In the version of Minesweeper that comes with Windows, the options are:

Easy: $9\times9$ grid with $10$ mines ($\approx 12.3\%$ of the squares have mines)

Intermediate: $16\times16$ grid with $40$ mines ($\approx 15.6\%$ of the squares have mines)

Expert: $30\times16$ grid with $99$ mines ($\approx 20.6\%$ of the squares have mines)

In the easy grid, there are $\binom{81}{10}$, which is about $1.88$ trillion, ways to distribute the mines.

None of the border squares can be surrounded by $8$ mines, only squares in the middle $7\times7$ portion of the grid can possibly be an $8$.

Suppose the upper left square in the middle $7\times7$ grid is surrounded by $8$ mines. Then there are $81-9=72$ squares left to place the remaining $2$ mines. That can be done in $\binom{72}{2}=2556$ ways.

Thus each of the $7\times7=49$ non-border squares has $2556$ ways to be surrounded by $8$ mines. I don't think I double counted any arrangements, but maybe I am overlooking something.

We have $49\times2556=125244$ ways for an easy grid to have an $8$ somewhere. Out of the $1.88$ trillion total easy grids, this gives a probability of about $6\times10^{-8}$. So, very rare indeed!

Working out the probabilities for the bigger boards would be trickier due to the possibility of over-counting, though as the proportion of mines in the grid increases, the likeliness of an $8$ also increases. Also, people in the comments have pointed out other intricacies in the rules of the game on Microsoft that may change the true probability. Interesting question though!

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I've been playing the expert mode of this game for over 20 years. I don't remember getting an eight in either a lost game or a game I won. I got this one today and won the game.

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