Minesweeper Game Probability

Probability Level 1

In the image below, a minesweeper game on a 16 x 16 board has a total of 40 mines hidden in the cells at random, with 4 of them already open. The number in each of the 4 open cells represents the number of hidden mines in the 8 cells around that number. For example, the cell with 2 in it implies that there are mines in 2 of the 8 cells around it.

If a player must choose one cell (P, Q, R, S, or T) to click, which cell has the least probability of having a mine hidden in it?

P Q R S T

2 solutions

Victor Paes Plinio
Nov 27, 2017

The probability of having a mine in cell P is $\frac{2}{8}$ , because the number two means that we have 2 mines hidden in the 8 squares around the number;

The probability of having a mine in cell Q is $\frac{1}{8}$ ;

The probability of having a mine in cell S is $\frac{4}{8}$ ;

The probability of having a mine in cell T is $\frac{3}{8}$ ;

The probability of having a mine in cell R is giving by the number of mines left and the number of cells available to choose. Look the following image:

We don't need to consider the cells inside each red squares because we already find the probability for these cells. Also, we don't need to consider the mines we already know (the number of mines in the cells with numbers). Then we have $40-4-3-2-1=30$ mines to consider and $16\times16-9\times4=256-36=220$ cells to choose. Then the probability of having a mine in cell R is $\frac{30}{220}$ .

We can compare these fractions and find that $\frac{1}{8}$ is less than $\frac{30}{220}$ ( $\frac{55}{440}<\frac{60}{440}$ ), which means that the cell $\boxed{Q}$ is the cell with the least probability of having a mine in it.

Why do we subtract 9 x 4 from 256 ?

Firas Rajab - 3 years, 5 months ago

Because the 10 mines in those 36 squares are already accounted for.

Samir Betmouni - 3 years, 5 months ago

9 represents de open number + the 8 squares surrounding it. since it happens 4 times, he multiplied.

PS: I'd like you all to know that this question was on Brazil's National High School Exam (ENEM) that took place in 2017. Not even close to be the hardest one there, but in fact, was my favorite. Perhaps because I'm a minesweeper player, but it could be for its elegant style.

Whallyson Freire - 3 years, 5 months ago

I forgot to put the problem source hehe. Thank you for comment!

Victor Paes Plinio - 3 years, 5 months ago

For each of the four numbers shown, we have the information about the nine squares centered at the visible numbers. Hence 9 x 4. And we know exactly how many mines are in those 36 squares by adding up the four numbers: there are 10 there. The rest of the board consists of the original 256 less the 36 that we have information about. The only thing we know about the rest of the board is that there are a total of 30 mines in there.

Richard Desper - 3 years, 5 months ago

Shouldn't the remaining in number of mines be 26 (the four revealed, plus the count of mines in the surrounding squares)? This would leave a fraction of 26/220 or 11.8%, a figure lower than 1/8, making R the least probable.

Trea Dmill - 3 years, 5 months ago

I think it's just bad wording; he means that 4 of the cells are already opened, not the mines

Dylan Wedel - 3 years, 5 months ago

If you see a number in Minesweeper, there is no mine at that spot. The number indicates how many mines are in the neighboring spots.

Richard Desper - 3 years, 5 months ago

The statement of the problem is mathematically flawed. The statement that the mine locations are "random" can be understood as distributed with equal "a priori" probability. Then the "conditional" probability of a mine in a location would decrease when there are more mines in its vicinity. So the result would be the opposite of the given solution: S would have least probability to have a mine.

Samer Atasi - 3 years, 5 months ago

Think of all the ways you can arrange 4 mines in the eight squares around the "4". Half of those arrangements will have a mine in position S.

Samir Betmouni - 3 years, 5 months ago

You are overthinking it. You have the knowledge that S has 4 mines around it, since there are 8 spaces, it must be a 1/2 chance. And so on for the others. No matter what the arrangement of other mines may be, you can know for sure these probabilities because they are given to you.

Alex Li - 3 years, 5 months ago

Have to agree with Atasi. There is nothing magical about an arbitrary choice of eight closest neighbors of a cell as representing the statistics of the surrounded cell, when the statement of the problem provides an a priori probability of all each cell. If you are going to do a two-dimensional regression of the probability, to find the probability as a function of (x,y), then you must have an a priori reason to know that the probability is, indeed, a function of (x,y). But that is not included in the statement of the problem.

William Darr - 3 years, 5 months ago

sounds enigmatic.but,yes,there is some omission or commission in the logic given in the answer.what would a good question probably be

Sunil Nandella - 3 years, 5 months ago

Your answer for S, 4/8, is correct. But all other answers are wrong, because you have forgotten to take into account the cells adjacent to the red boxes, below the box. Those three cells provide a 3x(30/220) chance of having a bomb. Adding that quantity into the 1/8 for Q gives 0.534, considerably higher than R's probability of 30/220 = 0.136

Glen Goodale - 3 years, 5 months ago

I disagree on how the probabilities were obtained in the "official" solution. Take the point "P", for example. The probability of having a mine in "P" is indeed 1/4, but not because "2/8". The "2" indicates that in the 8 surrounding cells there are 2 mines. These 8 cells can be combined in C(8, 2) = 28 different ways, and 7 of them (cell P with each other of the remaining 7) contains the cell P. Thus, the probability of having a mine in P is 7/28 = 1/4. I didn´t calculate the others... Besides, we have no information about the external cells of each red square...

PAULO BOUHID - 3 years, 5 months ago

And 4 mine is already open. So available mine is 40-10-4=26.

মঈনুল শুভ - 3 years, 3 months ago

John Billingsley
Dec 31, 2017

Y(P)=2 mines out of 8 unchecked cells=2/8=>.25

Y(Q)=1 mine out of 8 unchecked cells=1/8=.125

Y(R)=40 total mines less the number of mines whose locations are known out of 64 total unchecked cells less unchecked cells whose values can be inferred=[40-(2+1+4+3)]/[256-(9+9+9+9)]=30/220=>.136

Y(S)=4 mines out of 8 unchecked cells=4/8=>.5

Y(T)=3 mines out of 8 unchecked cells=3/8=>.375

Minesweeper Game Probability

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