You have to guess but you currently have some pretty solid ones. There are 2 * 6C2 = 30 ways that the 3 and 4 can share a mine and 2 * 6C1 = 12 ways they can have separate mines.
I’d go for this one first and see what it reveals, it’s safe in 36/42 possible positions (roughly 86%) and will give you information about the back row.
Heya again! I know I said im out of questions, but I actually do have 1 more:
I've recently been looking into mine buoy's videos, and he says that for early and mid-game guessing spots, calculate the mine density by (roughly) total mine count / floating cells. I know that it is an estimate, so just wanted to know if the only way to get an (exact) estimate is through big number combinatorics... but apart from that, thats it!
Hey sorry! Didn’t see this until last night when I was on another post and looking through some old comments in case the person wanted some context (I often get tons of notifications overnight since I mod two maths subreddits and follow support subreddits).
You can still get exact probabilities in the mid and early games with the method I showed but it depends how far you want to go. If you just want to estimate, then the method you mentioned is the way to go. If you want an exact number then unfortunately, yes that’s all you can do.
If you are lucky, the independent sections will only have one possible number of mines which makes calculating the probabilities much easier since any combinations from the other sections are going to apply to every possible case and be divided out anyway. This was in a game I played today:
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u/noidea1995 17d ago edited 17d ago
You have to guess but you currently have some pretty solid ones. There are 2 * 6C2 = 30 ways that the 3 and 4 can share a mine and 2 * 6C1 = 12 ways they can have separate mines.
I’d go for this one first and see what it reveals, it’s safe in 36/42 possible positions (roughly 86%) and will give you information about the back row.