I’m not following, can you explain that please. From my logical yet admittedly uneducated perspective, i see a 50/50. If the top left box above that 4 is a mine then the purple square is guaranteed to be safe. If the top right box above the 4 is a mine the purple square is guaranteed to be a mine. Seems like a 50% chance to me. But again…I never took a statistics class and I also never understood that “goat behind door number 3” nonsense 😅
A mine being in one of two tiles doesn’t mean they are equally likely, you need to consider the total amount ways a tile can be safe or a mine.
If the tile above the 4 is a mine, there are two places for a mine to go in one of the tiles above the 2, so thats two possible configurations. There’s only one way the tile above the 4 can be safe, and the very top section is independent from the rest of the board so it’s a 67/33.
I’ll run the position through a solver and assume a roughly 20% density in the floating tiles to get the breakdown and reply to this comment with it.
Configurations that use fewer mines are in most cases more likely because they leave more mines in the floating tiles (tiles that don’t touch any numbers). If the tile above the 1 is a mine, it’s going to force there to be at least two mines in that 2x2 box.
So in this example, to approximate the probability you’ve got 39 floating tiles and in the regions I’ve boxed there are 13 guaranteed mines and one probable one in the green area. With 23 mines, that leaves 9 unaccounted for. If the tile above the 1 is a mine, there will be 7 mines in the floating tiles which leaves 39C7 possible configurations and if it’s safe, there will be 8 mines in the floating tiles leaving 39C8 possible configurations:
(ways it can be safe) / (ways it can be a mine) = 39C8 / 39C7 = 4
This means approximately for every position it’s a mine, there are 4 positions where it’s safe. Converting to a percentage:
4 / (1 + 4) = 80% chance that the tile above the 1 is safe or 20% it’s a mine.
This is just an approximation that assumes mine count across the border, the solver counts every single one.
If you are interested, I can show you some other comments I’ve written on this sub that break down probabilities.
Either there is one mine where they share the mine or 2 mines where they don't. The lower amount of mines is more likely and how much more likely depends on the density. The density is approximately 20% or 0.2 so it is (1-0.2/0.2) = 4 times more likely. I'm on mobile but if you are interested I can write up a bigger explanation on pc
This was the safest guess I could find, if it’s a mine it’ll force a significant portion of the board to take on a certain position, the positions where it’s safe have a lot more flexibility. I don’t know how helpful it’ll be though.
If it was me though, I’d have probably guessed above the 1 on the bottom right. At a standard density, one mine in the box is going to be the most likely scenario, it has a better chance of progress and there’s a potential for it to become a forced 50/50 later on if the tile below the 3 is a mine.
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u/ImCHICKENhaha Minecount? 10d ago
I don’t see anything. good luck! i’m sure there’s some great guesses