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So there are 480 squares and 99 mines on the advanced level of minesweeper. It got me thinking, what would be the chances of winning the game randomly clicking each square? So not being influenced by numbers and without it doing any multi openings, so you would need to click 381 boxes (I think) :)

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The number of mines is fixed: $99$. The probability of hitting a mine at first click is therefore $99/480$. Provided that you survived the first click, the probability of hitting a mine on the second click is $99/479$. The probability of surviving $n$ clicks, with $1\leq n\leq 381$, is therefore $1 - 99^n/\prod_{i=0}^{n-1}(480-i)$. – user12918723509187 Feb 18 '12 at 0:37
Hitting a mine at first click is impossible in the Minesweeper that comes with Windows. – user23211 Feb 18 '12 at 1:04
I was like "Oh so that's the... wait NOW WE MUST RECALCULATE IT ALL OVER! – user76894 May 8 '13 at 21:34

Using the information that the first click is never a mine, we need to do $380$ clicks out of $479$ squares successfully. The first is a chance of $\frac {380}{479}$ The second is $\frac {379}{478}$ and so on. The total is $\frac {380!\cdot 99!}{479!}\approx 2 \cdot 10^{-105}$. Don't hold your breath.
This presumes that neither you nor the computer are smart enough to click all the squares around a 0, but I think that is the spirit of the question.

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