1 mine or 2 mines? what are the odds?

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1 mine or 2 mines? what are the odds?

Postby EWQMinesweeper » Fri May 03, 2013 4:33 pm

dayan showed me this pattern today and asked what the chances are, that the correct solution for it has 1 mine and respectively for 2 mines.

my first instinct was that it could be easily solved by just counting the mines. but let us ignore this case for the moment. assume that we are either playing to get the fastest time and have no time to count the mines or that we are not yet done with the rest of the board and can't obtain all neccessary information.

a primitive approach would be that assuming the chance for each unknown square to be a mine is the same (20.625% on expert, shortening it to 20 and 80 for a safe square here), we would get 80*80*20 for 1 mine and 20*20*80 for 2 mines. thus the pattern having only 1 mine comes at 80% chance.

but is this really correct? could we possibly argue that there are only 2 possible solutions and thus they have to be equally likely? what about similar situations? in density mode, does it matter how many mines are left to be identified and how many squares are remaining?
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Re: 1 mine or 2 mines? what are the odds?

Postby Dayan Marchezi » Fri May 03, 2013 5:16 pm

A simple way to see the solution [name removed]found is to admit just that one of the three squares has to be a mine and another one has to be safe. Only a third square is doubtful and thus the presence of a mine on it is at random, which means for us the own density of the board (20%), with 80% chance to be safe. Therefore, the pattern having only 1 mine comes at 80% chance.

I guess that is how speed players should see it. But I am not sure about precise odds.
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Re: 1 mine or 2 mines? what are the odds?

Postby CBright » Fri May 03, 2013 6:07 pm

It certainly does depend on the number of mines and squares left: if the board has a high density of mines on the remaining squares then it will be more likely to be a 2-mine pattern.

Formally, if the board has m mines and n unshown squares about which you know no information then there are (n choose (m−4)) ways of placing mines giving a 1-mine pattern and (n choose (m−5)) ways of placing mines giving a 2-mine pattern. So if this was from an expert board which is otherwise completely unsolved one gets a 0.79 chance a 1-mine pattern, but on a beginner board it would be 0.85.
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Re: 1 mine or 2 mines? what are the odds?

Postby wylx » Tue May 14, 2013 9:48 pm

If one's goal is to win as frequently as possible, one should assume that it's one mine, since sane minesweeper is almost exclusively played in mine densities less than one-half.

This is, however, minesweeper, and we're usually looking for having few specific games to go very well (won with greatest chance of a new record) rather than many games to go fairly well (won). If one's goal is to get a quick time, one should assume that it's two mines, since one click would occur faster than two clicks. I personally would go for the latter.
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Re: 1 mine or 2 mines? what are the odds?

Postby Tommy » Wed May 15, 2013 3:14 pm

wylx wrote:If one's goal is to win as frequently as possible, one should assume that it's one mine, since sane minesweeper is almost exclusively played in mine densities less than one-half.

This is, however, minesweeper, and we're usually looking for having few specific games to go very well (won with greatest chance of a new record) rather than many games to go fairly well (won). If one's goal is to get a quick time, one should assume that it's two mines, since one click would occur faster than two clicks. I personally would go for the latter.


Not on exp. Good boards are rare, and our sweeping time is limited. So, I agree close to the beginning of a game, but it would be a real shame to blast four times as often in a given situation at the end of a great board. The time advantage given by these awesome and rare boards is bigger than the cost of an additional click, IMO.

80-20 are really bad odds, especially for the sake of saving a single click. Given a certain number of games/time that we want to finish on average, it is _far_ better to take certain (even unforced) guesses that are less bad for completion and save multiple clicks - for example, uncertain deep flags next to ones can open a whole new area and have a survival chance of fairly close to 50%, while saving the time it takes to unravel a complex area (where you might have to end up guessing anyway).

Think of it this way: We are playing for speed, but we still want to win some games. Say that we are OK with winning only 1% of all games as long as that makes us faster. For every guessing strategy that is risky but makes us faster that we use, we finish less games, but those are faster. We want to add the most effective strategies first, until we get just above 1%. And this strategy is so ineffective (one click, no hesitation on the way, 80% fatality rate) that if it really makes the cut, our completion will be so low that we almost might as well be clicking randomly.

tl;dr: save your risks for more low-hanging fruit on expert if you want to finish an expert game at least once a year.
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Re: 1 mine or 2 mines? what are the odds?

Postby EWQMinesweeper » Wed May 15, 2013 5:17 pm

thought curtis' answer already solved it. thanks for the quick reply curtis.

but since there seems to be a new question i might try to answer it. there is a term called agressive guessing, coined by josh w. from australia in 2008 or 2009. in short you always go for the solution that requires the fewest clicks.

although it dramatically reduces your chances to complete a board, i wouldn't write it off as easily as tk does. once you have reached your maximum speed and can't improve any further it should be acceptable to use agressive guessing. on expert not even kamil has reached this point, but on intermediate some sweepers with a sub13 high score might consider this strategy. and on beginner everyone should use it when trying to beat their non-lucky record.
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