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.
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1Also, I have never seen an 8. Is that even possible? – Stephen Montgomery-Smith Jun 11 '16 at 21:05
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4Yes, I think it ought to be. – Paul Jun 11 '16 at 21:05
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1Interesting question. I have no clue how to solve it but it is interesting. One thing to take heed of though. Minesweeper may very well use premade constructs and not true randomness or an analogue of it. So the board might be built from four different "palletes" that were designed by the programmers for certain skill levels. In that case, the mathematical probability may depend on the programs code! – user64742 Jun 11 '16 at 21:12
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5@Steph Yes, can remember getting an 8 once. – 355durch113 Jun 11 '16 at 21:12
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@TheGreatDuck Lets assume it is truly random then this question is amazing. – Jun 11 '16 at 21:14
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2The good old days of Windows XP..... – bleh Jun 12 '16 at 02:48
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Minesweeper definitely will generate an 8 if you configure a custom game with an insane number of mines. If it doesn’t have a mechanism to prevent this from happening, then I would expect that it’s possible in the standard configurations. – Scott - Слава Україні Jun 12 '16 at 04:58
1 Answers
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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
$$\frac{392{471\choose91}}{480\choose99}\approx 0.0008219$$
Harsh Kumar
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5By applying Markov's Inequality, this shows that the probability of getting a $7$ on a randomly generated board is less than or equal to $0.02716\ldots$. – JimmyK4542 Jun 11 '16 at 22:26
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@JimmyK4542 The interesting question becomes is having one 7 highly correlated, or highly anti-correlated, with more 7s. If we presume they are uncorrelated, then 2.7% is going to be really close to the chance of getting a 7, as the second order effect will be fractions of a percent. – Yakk Jun 11 '16 at 22:38
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Most 7s will be uncorrelated, because their 9 squares won't overlap. So I think the approximation provided by @JimmyK4542 is extremely close to the correct answer. – Stephen Montgomery-Smith Jun 11 '16 at 23:18