Let's see if Veeky Forums is better than /b/ at statistics

Let's see if Veeky Forums is better than /b/ at statistics.

I start a "Beginner" Minesweeper game. I click one box, it's a [1]. The odds of there being a mine in any of the 8 surrounding spaces is 1/8. Simple.

I click a second box two spaces away. It's also a [1]. Now:

1. What are the odds of there being a mine under one of the spaces in between the two [1]s (marked in green)?
2. What are the odds of there being a mine under one of the outside spaces (marked in orange)?

I'll give hints if people get too stumped, but it isn't a trick question, you have all the information you need to answer it.

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1, 0/1
2, 5/1

Well the second part is also poorly worded, because if one side of orange spaces has a mine the other side must necessarily have one as well.

Given that information, there are 28 possible outcomes, with 25 of them being combinations of a mine on one side in orange and another mine on the other side, and 3 of them being where the mine is shared and within the green space.

So 25/28 chance of mines in orange, and 3/28 chance of a single mine in green

Sorry if it was poorly worded. I suppose I should have said "odds of there being a mine under *two* of the outside spaces". You figured out what I meant though.

Anyway, your answers are wrong. 25/28 and 3/28 are on the right track, but are missing a crucial piece of information. Hint: Is each possible outcome really equally likely?

No they are not. I'm too lazy to fully solve this but I know the answer depends on the dimensions of the board and the number of total bombs. Good luck I hope a clear and concise solution comes along that I can understand.

A beginner minesweeper game has 10 mines and 81 spaces.

There are 3(66 choose 9) ways there can be one mine and 25(66 choose 8) ways there can be two mines in the highlighted space.

So the probability of there being one mine in the green space is

3(66 choose 9)/(3(66 choose 9)+25(66 choose 8))

= 58/133 ~ 0.44

And the probability of the opposite is

75/133 = 135/193 ~ 0.56

The answers to this seem to be totally determined by the first spot you clicked on, the mine is in the middle (3/8) or it's not (5/8), revealing the second mine doesn't change that
so if it's in the 3/8 on the first one, then it's gonna be between them, and if it's in the 5/8 on the first one, then there are going to be 2 mines in the orange, one on each side.
1. 3/8
2. 5/8

it does change, because there's a constraint on the size of the board and number of mines, and mines are uniformly distributed

we don't know anything about the game (size, # of mines), all we know is that there's a 1/8 chance there being a bomb in a particular spot. If OP wanted that kind of answer it should have been stated in the OP

oh, shit we do. my bad.

There are 13 spaces. If we place a mine randomly, 10/13 times it will land in orange manifold and imply another is in the other orange manifold. Thus 3/13 one mine in the green manifold.

One mine isn't being placed randomly, 10 are.

We're only looking at certain cases of random generation which require mines to be in certain places. What you said is like the random number generated in that XKCD comic.

...

You don't understand probability. The fact that we happened to find one or two mines in a certain group of spaces does not mean that mines were "required" to be there. This was the result of a random process, specifically the distribution of 10 mines randomly throughout the 80 spaces which were not clicked on when you made your first move (actually this is not strictly true, what happens is that if your first move is on a mine, that mine is moved to the upper-leftmost space which is not occupied by a mine, but we will ignore this for simplification's sake). In order to answer the question we must take into account this process. It means that the spaces we have information about are more likely to contain one mine instead of two mines than you think.

See for the true answer

Revealing the two positions puts us into a definite game state, so this is mostly a question of logic.

We know the left and right spaces both have to have exactly 1 bordering mine. Three spaces are shared.

So, we will deduce what must occur logically. There are 8 places for a mine to go around the left mine. If a mine is in one of the orange spaces, then we can't also have a mine in the green space, but we need a mine next to the right space, so there must be a mine in the other orange space.

If we place the mine in the green strip, then both spots are satisfied.

Thus, in this exact game state, the probability of a mine in the orange is 5/8, and 3/8 in the green.

Wrong. The states are not equally likely.

>You don't understand probability. The fact that we happened to find one or two mines in a certain group of spaces does not mean that mines were "required" to be there.

You don't understand the question. It's not asking about the probability of a game state in the space of all possible game states, it's asking about the probability of places within a definite game state. You don't need to know anything about the rest of the board to answer the question.

I was wrong because I overcounted orange spaces, so change the answer to 5/8 and 3/8.

>You don't understand the question. It's not asking about the probability of a game state in the space of all possible game states, it's asking about the probability of places within a definite game state.
I never said it asked about the probability among all game states. It's asking about the probability among all game states *that satisfy the information given by the two uncovered spots.* You have failed to do that, because you ignored everything outside of the spaces next to those spots.

If you had more than rudimentary knowledge in probability you would see that here >You don't need to know anything about the rest of the board to answer the question.
Of course you do. The number of states in which there are 9 mines randomly distributed throughout the rest of the board is larger than the number of states in which there are 8. This makes it more likely that there is only one mine in the colored spaces than you think.

>I was wrong because I overcounted orange spaces, so change the answer to 5/8 and 3/8.
No, that's still wrong.

Actually you're right, that's a neat problem. We're still all wrong because we're making false assumptions about Minesweeper's RNG, having all eight mines be clustered in a corner is impossible, but otherwise you're right.

I'm not clicking any boxes, you click the boxes

We know that the number of mines in that spot is a hypergeometric distribution, i.e. there are 79 spots total, there are 10 spots that are mines, and we are choosing from 13 (these numbers are because 2 were already revealed)
We know that in the 13 spaces there is either 1 mine or 2 mines

(these were calculated here: stattrek . com/online-calculator/hypergeometric.aspx )
Prob( exactly 1 mine in region ) = 0.33400
Prob( exactly 2 mines in region ) = 0.31096

since they are exclusive we can add them to Prob( exactly 1 or exactly 2 ) = 0.64496

so, probability of it being 1 given that its exactly 1 or exactly 2
Prob( 1 given exactly 1 or exactly 2 ) = Prob( exactly 1 and (exactly 1 or exactly 2)) / Prob( exactly 1 or exactly 2)
= .33400/.64496 = .51786

This is the chance that there's exactly 1 mine in that region, it would have to be in the green. It doesn't take into account like positioning at all, so this makes me think I didn't do it right, since there are only 3 spots for there to be a single mine. Of course, the chance of orange would be 1 - .51786.

I honestly don't think a provable method is easily done in a Veeky Forums thread because of how minesweeper generation works. During generation mines aren't simply placed down randomly, spots are weighted to increase the probability of mines being spread out.

OP here.

Correct!

You're missing the fact that within that 5/8, many scenarios would not result in the second box being a "1".

Well put.

Fair point, but for the purposes of the problem assume the Minesweeper RNG makes every board combination (that contains 10 mines) equally likely. I didn't expect people to inspect minesweeper's disassembly or anything.

You don't want the probability there's 1 or 2 mines in the region though, you want the probability that there's one mine specifically in the green section, and two mines specifically in the orange section (one on each side).