What's the probability that the ? is a mine?

1/13

>>7727587
1/13

Why?

1/2, either it is or it isn't.

>>7727587
With that much info 1/10

>>7727597
>>7727598
you are forgetting the possibility that there are two mines - one to the left and one to the right.

>>7727611
or 1/7 actually

>>7727597
>>7727598
Yeah. Yeah

Is it though?

I didn't tell you that the board has a finite size (16x30) and that there's 99 mines on the board. So that changes it, right?

>>7727612
This. And I think we need to know how many mines there are.

>>7727608
Ummm no.

What about a total probability of
1) there is one common mine for those two open cells
2) there are two mines (one for each cell)

>>7727618
There are 99 mines and 480 total squares, 2 of which are revealed and do not contain a mine.

>>7727635
Yeah I feel like this is one of those combinatorics problems where you divide by common outcomes

>>7727587
.2071

>>7727635
there's 9 ways there could be 2 mines, and 4 ways there could be 1. So for 1) 4/13, for 2) 9/13, and for OP's its 1/13.

>>7727587
Just enumerate all possible states, and fine
[math] P = \frac{number of states where ? is a mine}{all possible states} [/math]

>>7727587
you don't have enough information to tell if there are one or two mines among the squares next to the ones.

See
>>7727612

>>7727597
>>7727598
>>7727616

Only these three posts make sense. Please stop being retards.

Yes. That changes it and makes it a more complicated problem.

File: phdinstats.png (28KB, 491x170px) Image search: [Google]

28KB, 491x170px

two ease

>>7727662
how does it change it? Even if there were 1 million tiles, We know for a fact there is at most 2 around those 10 tiles. The two 1s tell us how many mines there are, or at lease give us an upper bound. If it was only those 12 tiles as the entire board, the chance would still be 1/13.

File: 11a.png (3KB, 704x581px) Image search: [Google]

3KB, 704x581px

>>7727677
Well imagine you have a 4x4 board with 2 bombs. Now you see how the "outside" corners affect the number of total states.

is 1/7, seen as 1/14+1/14. In fact, if the four central squares (which are seen by both the 1) have 1/7 (1/14+1/14) each one, and the external ones have 1/14 (they're seen only by one "1"), and you sum up every possibility in every square, you get 1 (as it MUST be). [that's true if there's only one mine; i've assumed that]

>>7727587
Case 1: there is mine in left column. Probability is 3/8.
In that case probability of that ? is a mine equal to zero
Case 2: there is no mine in left column with probability 5/8, and therefore no mine in right column. This means mine is in one of the four cells of intersecting areas with probability 1/4 be in each of them.
Total 3/8 * 0 + 5/8 * 1/4 = 5/32

>>7727769
Oops, a little mistake:
3/7 and 4/7, so it will be 1/7.

>>7727587
3/13 - 1/13 - 1/13 - 3/13
3/13 - - - - - - - - - - 3/13
3/13 - 1/13 - 1/13 - 3/13

>>7727693
>changes the problem and then limits the boundary conditions until it actually matters

>thinks he has then made his point

should probably just skip these logic threads, anon...

File: answer.png (28KB, 612x468px) Image search: [Google]

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Here are all possibilities.

>>7727818
Not interested in how the problem changes?

>>7727822
Yes. What if we know that there are 2 mines and there are X outside squares besides the 12 squares we were already working with?

>>7727612
you are wrong

3*3 for the left-right placement + 4 for one mine top or bottom

there are no other configurations of mines to produce those two ones

>>7727834
outside squares don't matter because we have to meet the revealed condition

conditional probability

>>7727834
Huh? The external mines have no effect on the tiles in the picture.

>>7727856
Whatever you're saying is covered by this picture so stop posting.
>>7727822

>>7727854
>>7727853
sure they do, look at:

>>7727693

>>7727869
Yes they have an effect if the two resolved tiles are diagonal like that.
They do NOT have an effect in OP's situation no matter what.

>>7727853
>>7727854
You didn't read his post. Lets say you have a 4x4 grid (add an extra row to the top or bottom of >>7727822) and in this 4x4 grid, we know there are 2 mines. Now we need to consider all the different variations where the original has only 1 mine. Those original 4 ways becomes 16 ways, so the chance for it being in the mystery spot is 1/16 instead of 1/13. This is not a different problem, minesweeper is a game where you have a given number of tiles and mines, so each combination must be taken into account.

>>7727876
If there are outside tiles and you know the number of mines is more than one, then they certainly have an effect. The outside tiles attract cases of mines toward them and make cases of 2 mines less likely.

>>7727879
>becomes 1/16
excuse me, that isn't correct. There are 4 different ways it could be in the mystery position, and a total of 25 ways. So the probability is 4/25 instead of 1/13.

>>7727587
What if we place bombs at random until condition is met?

So we get 1/10 for each of the center squares and the outcome is invalid when a bomb is put at either end.

>>7727889
4/25 for the 4 of that type, right? Same 4/25 for the 4 on the bottom. And then 3/25 for each of the 6 on the two sides. For a totals of

(4)4/25 + (4)4/25 + 6(3/25) = 50/25 bingo

File: autism.png (7KB, 571x571px) Image search: [Google]

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Sup minesweeper dudes.
Here are all possible layouts for a 4x4 grid with 2 mines.

>>7727974
They're not grouped by symmetry

there are 13 possible mine placements in this situation, that spot holding the mine is a 1/13 chance

>>7727974
How many orbits are there with respect to D4 symmetry?

1/13

File: Untitled.png (45KB, 214x108px) Image search: [Google]

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Help me out with this one, /sci/. Is it 1/2 or 1/6?

>>7728208
If that is the entire board the probability is 0

>>7728208
Gladly.

2/12 - 2/12 - 6/12
2/12 ---------- 6/12
2/12 - 2/12 - 2/12 ---------- 10/12 - 2/12

>>7727974
This is quite possibly the rarest pepe i have ever gazed upon

>>7727974

Did you write a script to do that or do you have actual autism and drew it?

>>7727587
It is 2/7

>>7727597
If you think about it for more than 5 seconds its clear that the answer isn't 1/13

>>7728281

ha

>>7727587
1/7

>>7727834
Hurr, what if we know there are a total of 40 mines and we already found 39 mines outside the 12 square we were already working with??

Then the problem changes because we know there is only one mine in there!!!!!


What if someone randomly added another mine into one of the 12 fields after we found the first mine!! What then though??

1/2.

File: 1450384170848.png (14KB, 1005x681px) Image search: [Google]

14KB, 1005x681px

>>7727587
>>7728294

1/7 + 1/7 = 2/7

>>7728239
shiaaat, how did you calculate this?

>>7727587
1/4

>>7729469
brute forced it
with so few tiles it's easy