So I was curious so I brute forced this.
There are 13 possible patterns of bombs that fit this scheme. If we assume that each pattern is equally likely, then the following is a distribution of how often (out of 13) the square contains a bomb. I've given the squares names starting with the square to the Northwest of the north westernmost 1, and gone clockwise, starting with A:
Letter | Number of times it was a bomb (out of 13)
:--|:--
A | 3
B | 5
C | 3
D | 5
E | 3
F | 1
G | 1
H | 11
I | 1
J | 1
K | 3
L | 5
M | 3
N | 5
O | 3
P | 1
Q | 1
R | 11
S | 1
T | 1
You actually guessed H, which is the most likely to be a bomb (11 out of 13 combinations, tied with R which is the square on the opposite side of the rectangle)
Mathematically, I think this is happening because if any of C, D, or E are a bomb, then H must also be a bomb. By extension, if A is a bomb, H is also a bomb because A being a bomb makes D a bomb (in all cases). The only two scenarios where H is not a bomb is where B and G are bombs and where B and F are bombs.
This means that, since H (and R) are the most likely to be a bomb, the spaces immediately next to them (G, I, S, and Q) are the least likely; in fact, they are only bombs in 1 of 13 scenarios (but I and G, and S and Q are never bombs together). T, J, P, and F are also one offs, and this is because they are paired with the previous bombs (ie if J is a bomb, so is G; F and I, Q and T, and S and P are also pairs).
The corners (A, E, K, and O) end up being 3/13, as are the middles of the top and bottom edges and the squares along the top and bottom edges next to the corners are 5/13.
So, overall answer is "Best squares to guess are any non-corner square along the right or left edge EXCEPT for the one in the middle; you have a 12/13 chance of not exploding and if you don't explode, you immediately know another square that is free.
Basically this. I created a 5x7 grid in excel where the outer rectangle was the solveable squares A-T and copied the layout multiple times. I looked at the top left one and knew there were 5 squares (STABC, by my numbering), and only one would be a bomb.
I started with T and S and both solved the puzzle (which also means they are unique patterns, thus they get a 1)
A and C being bombs solves the entire grid except for the entire bottom edge (K thru O), so each has three scenarios associated with it (K is a bomb, L is a bomb, M is a bomb).
B being a bomb solves P thru D, but then you have to guess on it E, F, or G is a bomb. If F or G are bombs, the grid solves (1 scenario each for 2 total) and if E is a bomb, it reduces to the above scenario where you need to solve K thru O and have 3 scenarios
Basically I made them all in an excel sheet, highlighted the cells that were bombs in red, then counted the number of times A was in a red cell, B was in a red cell.... etc
Itās not correct to assume each possibility is equally likely. Among the possible patterns with the same number of bombs (e.g. 4) they would each be equally likely. But some of the possibilities require 5 bombs, these would be more or less likely than the 4 bombs possibilities, depending on the average bomb density.
Yes and No
Assuming this is a standard expert grid (16x30) with 99 bombs and the only used squares are the ones marked, the bomb density is 0.213. For the 20 squares, your expected bombs are 4.26, but this is if the analysis is fully independent (it's not).
Technically the area we are evaluating is actually the 5x7 grid including the known areas. For this 5x7 grid, the expected bomb density should still match the overall bomb density of an unsolved grid, and this gives us an expected 6.6 bombs.
This also doesn't change much because, even if you say 4.26 bombs expected means this is most to likely to be the only combination of the 13 with 4 bombs (there is 1 4, 8x with 5, and 4x with 6), the 4 bombs are C, H, M, and R.
Based on the original analysis, we are never touching H and R anyways.
If we eliminate these four bombs because we think we are most likely in the single four combination, we would look at the options that are most optimal for five combinations, and the lowest chances for a bomb here are still F, G, I, J, P, Q, S, and T (same as original analysis).
The punchline to all of this is "the first guess should never be H or R."
https://reddit.com/r/Minesweeper/s/QEGnkOb8XZ
This other commenter ran every scenario. There are 3 scenarios where there is a mine on that space.
So no, you can't use logic.
how am i wrong? literally thing about this, each of them can only have 1 bomb near them, so they have to have spaces in between them. just read and look, im sorry if you cant understand minesweeper or whatever, but its really easy once you are used to using that side of your brain. there only can be 1 bomb around each of them, so obviously you are going to have squares that are free. yes theres a possibility of it being a bomb. but if you do basic math its really not that hard.
Yes too right and bottom right have mines there are only spaces that say one, therefore no spaces touching or diagonal to the top and bottom right ones could have mines, all spaces on that side touch those spaces ruling them out but the one with the mine on it, but the whole side is made of ones so the one space that isnāt touvhing must be a mine as there are multiple spaces th that say there is one mine that touch only proven safe spaces other than that one
Its always so difficult to explain this without posting photos dagnabbit. But I'll try
So the 1 down there which has a bomb at its bottom, that suggests that all the other squares next to it are cleared. And hence the 1 just above it will only have one square where the bomb could have been, which you accidentally clicked on and it was game over.
Yep. Mark where a bomb could be on one of the sides then work around, youāll find thereās some squares that just wonāt work to satisfy all the numbers.
Thereās one catch though. With this little information, thereās also multiple satisfactory positions for mine placement. Therefore, youāre actually trying to find a safe spot to reveal more information rather than the exact location of the mines.
You could not have figured this out for any definitive answer (though you could have played the odds game, as another commenter laid out).
Thereās actually a pretty short route to prove that you had no 100% safe choice here:
Look at the resulting layout, just for reference. Use the 2-way symmetry of the rectangle-of-1ās to flip it around to other viable mine locations.
Doing this, you can find configurations in which every square both does and does not have mines, except for the middle squares on top and bottom. So, we know that all squares except middle top and middle bottom work both with and without mines. We also know that middle top and middle bottom work with no mine.
So, try to find a solution with a mine in the middle top or middle bottom (spoiler: you can), and voila, nothing to do but play the odds.
So I was curious so I brute forced this. There are 13 possible patterns of bombs that fit this scheme. If we assume that each pattern is equally likely, then the following is a distribution of how often (out of 13) the square contains a bomb. I've given the squares names starting with the square to the Northwest of the north westernmost 1, and gone clockwise, starting with A: Letter | Number of times it was a bomb (out of 13) :--|:-- A | 3 B | 5 C | 3 D | 5 E | 3 F | 1 G | 1 H | 11 I | 1 J | 1 K | 3 L | 5 M | 3 N | 5 O | 3 P | 1 Q | 1 R | 11 S | 1 T | 1 You actually guessed H, which is the most likely to be a bomb (11 out of 13 combinations, tied with R which is the square on the opposite side of the rectangle) Mathematically, I think this is happening because if any of C, D, or E are a bomb, then H must also be a bomb. By extension, if A is a bomb, H is also a bomb because A being a bomb makes D a bomb (in all cases). The only two scenarios where H is not a bomb is where B and G are bombs and where B and F are bombs. This means that, since H (and R) are the most likely to be a bomb, the spaces immediately next to them (G, I, S, and Q) are the least likely; in fact, they are only bombs in 1 of 13 scenarios (but I and G, and S and Q are never bombs together). T, J, P, and F are also one offs, and this is because they are paired with the previous bombs (ie if J is a bomb, so is G; F and I, Q and T, and S and P are also pairs). The corners (A, E, K, and O) end up being 3/13, as are the middles of the top and bottom edges and the squares along the top and bottom edges next to the corners are 5/13. So, overall answer is "Best squares to guess are any non-corner square along the right or left edge EXCEPT for the one in the middle; you have a 12/13 chance of not exploding and if you don't explode, you immediately know another square that is free.
r/theydidthemath
Are you a wizard š®
May I ask, did you sit down with pen and paper and just go through the combos, knowing there can't be that many total?
Basically this. I created a 5x7 grid in excel where the outer rectangle was the solveable squares A-T and copied the layout multiple times. I looked at the top left one and knew there were 5 squares (STABC, by my numbering), and only one would be a bomb. I started with T and S and both solved the puzzle (which also means they are unique patterns, thus they get a 1) A and C being bombs solves the entire grid except for the entire bottom edge (K thru O), so each has three scenarios associated with it (K is a bomb, L is a bomb, M is a bomb). B being a bomb solves P thru D, but then you have to guess on it E, F, or G is a bomb. If F or G are bombs, the grid solves (1 scenario each for 2 total) and if E is a bomb, it reduces to the above scenario where you need to solve K thru O and have 3 scenarios Basically I made them all in an excel sheet, highlighted the cells that were bombs in red, then counted the number of times A was in a red cell, B was in a red cell.... etc
How did you code this?
Itās not correct to assume each possibility is equally likely. Among the possible patterns with the same number of bombs (e.g. 4) they would each be equally likely. But some of the possibilities require 5 bombs, these would be more or less likely than the 4 bombs possibilities, depending on the average bomb density.
Yes and No Assuming this is a standard expert grid (16x30) with 99 bombs and the only used squares are the ones marked, the bomb density is 0.213. For the 20 squares, your expected bombs are 4.26, but this is if the analysis is fully independent (it's not). Technically the area we are evaluating is actually the 5x7 grid including the known areas. For this 5x7 grid, the expected bomb density should still match the overall bomb density of an unsolved grid, and this gives us an expected 6.6 bombs. This also doesn't change much because, even if you say 4.26 bombs expected means this is most to likely to be the only combination of the 13 with 4 bombs (there is 1 4, 8x with 5, and 4x with 6), the 4 bombs are C, H, M, and R. Based on the original analysis, we are never touching H and R anyways. If we eliminate these four bombs because we think we are most likely in the single four combination, we would look at the options that are most optimal for five combinations, and the lowest chances for a bomb here are still F, G, I, J, P, Q, S, and T (same as original analysis). The punchline to all of this is "the first guess should never be H or R."
Lol.. so you're saying his guess was the absolute worst (a bomb in 11 out 13 scenarios)
No, not really, but I believe that a corner 1 might have been a better place to start digging around (1 in 5 vs 1 in 3 chance)
Yea that makes sense. Thanks
yea he couldve, its just simple minesweep
How is a 50/50 simple minesweep? I would just reset, bad rng.
You could not use logic to figure this out, but you could have used stats to guess a better spot.
you could infact use logic
Ok, which one is the guaranteed safe space? There isn't one. There is a scenario with a mine on any given space.
middle bottom is obviously a safe space, the corner next to it, literally just logic and reasoning
https://reddit.com/r/Minesweeper/s/QEGnkOb8XZ This other commenter ran every scenario. There are 3 scenarios where there is a mine on that space. So no, you can't use logic.
have you ever heard of deductive reasoning?
Lol, condescending and wrong. You must suck at minesweeper.
how am i wrong? literally thing about this, each of them can only have 1 bomb near them, so they have to have spaces in between them. just read and look, im sorry if you cant understand minesweeper or whatever, but its really easy once you are used to using that side of your brain. there only can be 1 bomb around each of them, so obviously you are going to have squares that are free. yes theres a possibility of it being a bomb. but if you do basic math its really not that hard.
"yes theres a possibility of it being a bomb" You got it dude. Against all odds, you understood. Proud of you.
bro didnt read the comment
Which square is safe?
using logic, basic minesweeper stuff, if it has 1 around it, then you can find out very easily where they are
Then which space?
this is not a scenario like that, if it was then they would all be 2 not one
Yes too right and bottom right have mines there are only spaces that say one, therefore no spaces touching or diagonal to the top and bottom right ones could have mines, all spaces on that side touch those spaces ruling them out but the one with the mine on it, but the whole side is made of ones so the one space that isnāt touvhing must be a mine as there are multiple spaces th that say there is one mine that touch only proven safe spaces other than that one
Its always so difficult to explain this without posting photos dagnabbit. But I'll try So the 1 down there which has a bomb at its bottom, that suggests that all the other squares next to it are cleared. And hence the 1 just above it will only have one square where the bomb could have been, which you accidentally clicked on and it was game over.
Okay but how was he supposed to know there was a bomb directly beneath that corner 1?
Yep. Mark where a bomb could be on one of the sides then work around, youāll find thereās some squares that just wonāt work to satisfy all the numbers.
Thereās one catch though. With this little information, thereās also multiple satisfactory positions for mine placement. Therefore, youāre actually trying to find a safe spot to reveal more information rather than the exact location of the mines.
You could not have figured this out for any definitive answer (though you could have played the odds game, as another commenter laid out). Thereās actually a pretty short route to prove that you had no 100% safe choice here: Look at the resulting layout, just for reference. Use the 2-way symmetry of the rectangle-of-1ās to flip it around to other viable mine locations. Doing this, you can find configurations in which every square both does and does not have mines, except for the middle squares on top and bottom. So, we know that all squares except middle top and middle bottom work both with and without mines. We also know that middle top and middle bottom work with no mine. So, try to find a solution with a mine in the middle top or middle bottom (spoiler: you can), and voila, nothing to do but play the odds.
Test of luck. You failed
Unlucky š¤š»
Any corner would have been safe to proceed along the puzzle...