Probability question: What is the chance of A being a mine? What

exactly 50%. not meming

>>8720550
50% that A is a mine. 50% that the spot above A is a mine. The other 2 are 0%

>>8720550
There is only one mine in all 4 squares. The sum of the probability of a mine in each square must be 1. If A is .5, then the square above A is also .5, same for B and the square to its right. That sums to a probability of 2.

You're either memeing or you're bad at math.

>>8720551
You do realize the two 1's could be set off by a mine off screen right?

>>8720555
>"so the rest of the board cannot provide useful information"

>>8720558
Sorry, I didn't mean that the rest of the board is invisible to the squares we see, I meant that it is invisible to us.

Can anything interesting be done with minesweeper? My first thought is quantum minesweeper or cellular automata minesweeper, and I have no idea how either of those would work.

>>8720547
Just find all possible permutations and divide the number of permutations there is a mine in that spot by the total number of permutations.

Your second assumption is stupid, the board size and number of mines is always known and can give you a significant edge. Generally you should assume mines will not be close to each other.

>>8720566
http://web.mat.bham.ac.uk/R.W.Kaye/minesw/ordmsw.htm

File: mines.png (33KB, 772x707px) Image search: [Google]

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>>8720547
While waiting to see if someone can do this with math, heres a brute force solution so we can check the answers

Pic shows all possible configurations, with blue off screen tiles that can influence the numbers.

Assuming all configuration are equally probable, there are 33, of which only 1 has a mine on A, and 7 have a mine on B

Probability of A a mine is 1/33
Probability of B a mine is 7/33

>>8721176
those 33s should be 32s, the last one has no mine and i forgot to delete it.

Either A or the square above A must be a mine, both with a 50% chance. B and the square to the right of B cannot be mines.

But note that this does not (always) translate to an actual game, as the rest of the board will have an effect on those eight squares.

>>8720573
>Generally you should assume mines will not be close to each other.
Do you even play the game?

File: Puzzle for Beginners.png (7KB, 383x216px) Image search: [Google]

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>>8721510
Yes. Do you understand probability?

Here, try this.

>>8721545
0.5 right?

My minesweeper intuition tells me the black dot is not a mine and the square below it is but I'm not sure why.

>>8721545
1/2

>>8721557
never mind I think the bottom-right 1 makes it a little lower.

File: Beginner.png (23KB, 652x634px) Image search: [Google]

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>>8721557
>>8721562
>>8721564
Nope. You didn't take into account the information given by the board size and number of total mines. It's more likely than not that there is a mine in that space because that would mean the mines are less densely packed. If 10 mines are distributed randomly among 81 spaces, it's less likely that three of them will be found in the 16 spaces in the top left corner. 2/16 is closer to 10/81 than 3/16. Understand?

File: minesweepter.png (3MB, 1568x845px) Image search: [Google]

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>>8721545
>Do you understand probability?
obviously you do not

you do know that selecting random points on a plane tends to give you clumps that are not evenly distributed right?
http://home.iscte-iul.pt/~jaats/myweb/Efficient%20methods%20for%20solving%20SDEs.html
look at the pics, the random one has many points next to each other.

Pic related. a real minesweeper game showing how many times mines are next to each other.

>>8721571
>you do know that selecting random points on a plane tends to give you clumps that are not evenly distributed right?
Yes, and those clumps are less likely than an evenly distributed area. See >>8721567

I'm not saying clumps, can't occur, I'm saying they are less likely to occur. This is a general rule of thumb that can help you greatly in situations in which you need to guess where the mine is. Since you don't have time to calculate the actual probability, this rule of thumb is the next best thing. If you did not use it then the problem I gave would seem like a 50/50 guess when it is in fact not.

What's so hard about this question? The mine cannot be anywhere else than in the two dark gray squares on the left (A and the one above it), with an equal chance of being in either of them.

>>8721627
>The mine cannot be anywhere else than in the two dark gray squares on the left (A and the one above it), with an equal chance of being in either of them.
It could be in neither.

>>8720547
There's no way a mine can be in B
A is 50/50 with the tile above it

>>8721776
Wrong.

Does increasing the detection radius of squares, necessarily increase the difficulty of the game?

File: msp.png (35KB, 768x395px) Image search: [Google]

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>>8721849
The "1"s are implying that there's something right of them and since the "2" only needs 1 additional mine, we know where the last mine is. That mask B and the tile right of it clear.
When cleared the B will show a "2", the other block will be "1"
If that number would be "3" the B tile would gain a 50% chance to be a mine.

>>8721933
You're ignoring the rest of the board not shown in the picture.

>>8721567
You are a retard. Assuming the mines are randomly distributed, any possible combination is equally probable.

Like seriously, kill your self, brainlet.

>>8720547

a has a 1/2 chance of being a mine.

b has a 1/2 chance of being a mine.

>>8721958
>Assuming the mines are randomly distributed, any possible combination is equally probable.
Yes, which means you have to take into account ALL permutations, not just treat the permutations within the spaces you have information about as all the permutations. The number of mines outside of those spaces matters, because the number of ways 8 mines can be distributed in the rest of the space is different from the number of ways 7 mines can be distributed in the rest of the space. Once you learn probability you will understand this. Until then, you ironically call others retards.

>>8721968
Wrong. See >>8721176

>>8720547
A is 42.9%
B is 14.3%

A is in a zone of 3 numbers
B is in zone of 1 number
Each number covers 7 tiles. Chance that a mine is in a tile affected by a single number is 14.3%

A has a 50% chance to be a mine.

B cannot be a mine, assuming the flag is correct.

>>8720547
Exactly what probability distribution on infinite minesweeper boards are we using?

>>8722019
Wrong. You're ignoring the rest of the board.

>>8722011
Totally wrong. The probability is dependent on the number of possible combinations, the chance of being around a number is not necessarily equally distributed.

>>8721176
Im retarded and forgot some combinations, heres the correct number (I think):
The possible combinations are:

Possible places a mine can be that's around Top1, and no other number: 3
Possible places a mine can be that's around Bottom1, and no other number: 2
Possible places a mine can be that's around 2, and no other number: 4

Possible places a mine can be that's around Both 1s, and not 2: 2
Possible places a mine can be that's around Top1 and 2, and not Bottom1: 0
Possible places a mine can be that's around Bottom1 and 2, and not Top1: 1

Possible places a mine can be that's all 3: 2

Sums to the 14 possible place a mine can be. The above is necessary and sufficient to solve this.

Now enumerate all possibilities (you can probably generalize this step, but ill write all combinations):
(all 3) = 2
(Bottom1 and 2) * (Top1) = 1*3 = 3
(Top1 and 2) * (Bottom1) = 0
(Both 1s) * (2) = 2 * 4 = 8
(Top1) * (Bottom1) * (2) = 3 * 2 * 4 = 24

All possibilities = 24 + 8 + 3 + 2 = 37

A falls in the category 'all 3', and since there are 2 tiles in it its 1/2 the probability of all 3, which is 2. Thus 1/2*2/37 = 1/37
B falls in the category 'just 2', and since there are 4 tiles in it its 1/4 the probability of just 2, which is (8 + 24). Thus its 1/4*(8+24)/37 = 8/37

This should be easily generalizable to any number of visible numbers and mines.

File: solution.png (21KB, 805x616px) Image search: [Google]

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>>8722277
Yeah that's what I got.

>>8720547
>Probability
>chance
Consistency much?