Can we have a small probability theory discussion involving Minesweeper? I will assume you understand the rules of the game.
I just came across this particular scenario which raises an interesting question. In this case, based on the number of mines remaining, there are 3 possible board configurations. In 2 of the 3 possibilities, ? is a mine.
However, if you were just looking at the bottom side (or the far right side for that matter), these elements indicate a 50/50 configuration where the mine that satisfies the 4's vaue is an unrelated element.
How do you reconcile two conflicting sets of cases to determine the probability that ? is a mine?
Hunter Price
>How do you reconcile two conflicting sets of cases to determine the probability that ? is a mine?
you already did it in your post. what are you even asking?
Alexander Howard
>what are you even asking? I'm not convinced the probability is 2/3
Gavin Evans
you're overthinking it.
Austin Reed
I think those 33s should be 25s, and again, that's only one way of seeing it
Luis Allen
I don't understand what there is to "reconcile". You can't look at a section of a board in isolation and complain that configurations change when you tack new squares onto the side.
Eli Allen
This is a corner. What's happening beyond the corner would make this way too complicated, so let's just assume you solve the rest of the board and end up in this corner every time
Wyatt Murphy
I agree with this.
James Taylor
I'm not referring to the fact that there's a board outside this corner, I'm referring to >where the mine that satisfies the 4's vaue is an unrelated element. This is false, since the location of the last mine on 4 affects whether one of the possibilities for ? is legal or not.
Ethan Russell
So I assume you think it's 66%?
Noah Cox
If this is what you meant, where does the 1/3 chance come from?
Connor Torres
>In 2 of the 3 possibilities, ? is a mine. 2/3 = 66% yes.
The fact that there are only 2 squares doesn't mean that the probabilities are split 50/50.
Lincoln Edwards
You have to take into account ALL possibilities. You can't just ignore unknown spaces.
Xavier Evans
...
Andrew Ramirez
the chance is in relation to the 4. the 3 squares in its vicinity share probability of having a mine equally, however the probability of the bottom right square in relation to the 4 is overridden by it's relation to the 2, as squares do not split probabilities between relations; they have independent probabilities for each relation.
Levi Parker
>you can't just ignore unknown spaces I DON'T KNOW
But if you were to evaluate each one as contingent on the next (like you'd normally solve a minesweeper grid), seems like a correct assessment.
What you're suggesting is pic related, but that's only possible if you solve the whole board at once and not a series of contingencies
John Adams
...
Jacob Butler
Then help us math wizard, surely this applies to this situation too
Ian Hughes
OK, show me the rest of the board...
Christopher Diaz
No
James Cruz
Assuming the rest of the board is solved the answer is obviously 2/3. Boring.
Jace Wood
This didn't sit right with me and this is why. You just can't solve the entire board at once. Minesweeper is a system of contingencies. By betting that it's more likely to be in the top corner, you're betting it's less likely to be in the left, which is paradoxical.
Parker Gutierrez
Those events are not mutually exclusive, so adding their probabilities is fallacious. A mine being in the topmost position does not preclude a mine from being in the leftmost position.
>You just can't solve the entire board at once. I'm not. I'm counting all random configurations of 10 mines which adhere to the information given by the revealed tiles. 5/8 of those configurations result in a mine being in the topmost position. 5/8 result in a mine being in the leftmost position (but not all the same configurations as the latter 5/8, obviously). So 5/8 of the time you are playing, and you get a board that looks exactly like that, there will be a mine in the topmost corner. Understand?
Now, why is the answer 5/8 and not 6/7? Because it's a little more likely that randomly distributing the mines will have resulted in 2 mines being in the 9 mines that we have some information about than 3 mines. When you distribute 10 mines among 81 spaces, the most likely amount of mines in 9 spaces is whatever is closest to 10/81, which is 2.
Sebastian Cook
>5/8 of those configurations result in a mine being in the topmost position. 5/8 result in a mine being in the leftmost position (but not all the same configurations as the latter 5/8, obviously). I understand this much. My problem is that, if you were to look at those sets, where they overlap or in other words, the times when they're both mines is obviously 5(67 choose 7)/(2(67 choose 8)+5(67 choose 7) or 1/4. In 3/4 cases, only one is a mine
If you're guessing that top is a mine, you're by extension guessing that left isn't, but why would that be the case when they have a symmetrical configuration
Daniel Wood
>My problem is that, if you were to look at those sets, where they overlap or in other words, the times when they're both mines is obviously 5(67 choose 7)/(2(67 choose 8)+5(67 choose 7) or 1/4. In 3/4 cases, only one is a mine Correct.
>If you're guessing that top is a mine, you're by extension guessing that left isn't No, you're not. They're not mutually exclusive. Both can be mines. You JUST said that they are both mines 1/4 of the time
Mason Flores
>No, you're not. They're not mutually exclusive. Both can be mines. They can be, but that's the least likely outcome
Imagine it like you're looking at them not from the perspective of an individual square, but from possible arrangements.
I think I'm making my mistake by changing the question itself but that's the problem with minesweeper and in playing the game, you're doing so from a certain perspective. You can do so from the perspective of which square is safe (in which I agree the inner squares are safer), or you can do it from the perspective of which arrangement is most likely, which the "top not left" and vice versa configurations are much more likely than both
I thought it was paradoxical but maybe it's not, at any rate I'm done thinking about it
Jacob Sanchez
>They can be, but that's the least likely outcome Yeah, so what? I didn't say both being mines was likely.
If you want to use the probabilities to actually determine optimal strategy, then you should click on one of the mines to the bottom right. If you really want to get into it, then you can also figure out the probabilities of what information you can get from each action and minimizing the chance of having to make random choices.
Caleb Bell
For the hell of it I ran these numbers for an expert grid and I got 69%. Did I do something wrong? ((466 Choose 97) + 5(466 Choose 96))/(2(466 Choose 97) + 5(466 Choose 96)) Or was I wrong in assuming it would be the same probability
Dominic Richardson
That's correct. The answer should be closer to 6/7 because on an expert board there is a higher density of mines, meaning 3 mines being in 9 spaces on an expert board is more likely than 3 mines being in 9 spaces on a beginner board. The density of mines changing will always change the answer.
Asher Allen
But the answer wouldn't change if the density remained the same, regardless of grid size?
Jeremiah Lee
Yes, just adjust the amount of mines on an expert board to have the same density as a beginner board and you'll see.
Carter Bennett
Thank you, this will all be very useful information
Thomas Jones
Brute forced it. This is the objectively correct answer.