This situation arises all the time. What is the probability that the marked cell is a bomb?

This space is doted on by 2 probabilities: .5, .2

Does one probability dominate the other? Is it possible to combine or reconcile the two to find P(yellow cell= bomb)?

Do you know this is an unsolved problem? Solve it and win a million. Good luck!

http://www.claymath.org/Popular_Lectures/Minesweeper/

In principle, it's possible; you enumerate every possible configuration of mines for the whole board which is consistent with your information to date, and count up how many of em have a mine under the yellow dot.

We do have to be careful when we speak of probabilities here: the event which has probability 0.2 is "there is a mine on a cell chosen at random from the five hidden cells adjacent to the 3." That is NOT the same thing as P(under one designated cell)=0.2 at all.

In your particular case, the true answer to the yellow-dot probability might be close to 0.5 if there is a large unsolved area off to the left and up. Its exact value depends on everything we know about the unsolved portions of the board, and may be very different if the unexposed area is small.

To illustrate the method, let's suppose the 3x6 area you've shown is the entire board. Let me call the hidden squares abc in the first row, de in the second, fgh,ij in the third.

If there are only two unmarked mines, the possible arrangements for the remaining mines are hj, bi, ci, ei, gi, and P(yellow)=0.2.
If there are three unmarked mines, there are 15 possible arrangements (each of the above plus any of a,d,f for the last mine). Again p(yellow)=0.2.

In this case, where we have ONLY these two restrictions, the 3 is in control. On a real board, there will be some other exposed number in the 4th row up, which will constrain what can be placed in cells b and c, and will influence the probabilities of the bottom cells.

ahh. ive heard of p=np and the million dollar unsolved math problems. Had no idea this type of question was involved. Thanks.

Assuming that the upper right corner is the upper right corner of the playing area and labeling the top row from right to left as a1,b1,c1,... and the 2nd row from right to left labeled as a2,b2,c2,... etc.

Then the indicated continuation is clearing squares a4 and b4 and gaining the new information beneath hidden there.

As for the probability of d3 having a mine, that depends on the size of the entire grid and how many mines are left hidden. Having a mine at d3 tends to decrease the density of mines elsewhere, while not having a mine there tends to increase the density of mines elsewhere. In short, an accurate estimate is not possible with the information you have provided.

In any case, the question is irrelevant at this point since you have other moves you can make that gain information without risk.

This is true, but Im looking at the problem in a theoretical sense rather than a minesweper best strategy sense.

agreed. suppose this is the entire grid and we have 3 mines to place. What now?

Solving, I'm getting the probabilities for the hidden squares of the:
Top row: 1/6, 1/8, 1/8
Mid row: 1/6, 1/8
Bot row: 1/6, 1/8, 1/2, 1/2, 1/2

I find that hard to believe: one mine must be in the leftmost column (only one can be at bottom right, and only one can be around the "3")... so a probability of 1/6 each can't be right for those 3 cells...

I am getting 1/3, 1/5, 1/5; 1/3, 1/5; 1/3, 1/5, 1/5, 4/5, 1/5.

We have a mine located at d or e.
Assume a mine at e, then there must no mine in d, a mine at c, none in zone b and 1 in zone a, so

a = 1/3, b = 0, c = 1, d = 0, e = 1.

Checking with
3*a + 4*b + c + d + e =
3*(1/3) + 4*(0) + (1) + (0) + (1) =
3 , so this is consistent.

If instead there is a mine at d, then there is no mine at c or e, one in zone b and one in zone a, so

a = 1/3, b = 1/4, c = 0, d = 1, e = 0.

Checking with 3*a + 4*b + c + d + e =
3*(1/3) + 4*(1/4) + (0) + (1) + (0) =
3 , so this is consistent.

These 2 possibilities are equally likely, so

a = 1/3, b = 1/8, c = 1/2, d = 1/2, e = 1/2.

Checking with 3*a + 4*b + c + d + e =
3*(1/3) + 4*(1/8) + (1/2) + (1/2) + (1/2) =
3 , so this is consistent.

I erred with zone a. It's 1/3 and not 1/6.

As for your solution, it looks like you are lumping c together with zone b when clearly c cannot be treated in the same way as zone b.

Okay, this is interesting. I went over your approach a bit more carefully and I am now hard put to find the real reason why it is any less valid than my approach. I can settle this issue with a Monte Carlo program, but I am feeling a bit lazy about it right now. Maybe tomorrow.

After thinking about it some more, I now think Siegmund's approach is correct. Very interesting. No Monte Carlo program is necessary. I've learned something new here. Thanks.

Presumably where we differed before, and where we agree now, is that "mine in zone b" and "no mine in zone b" happen in different numbers of ways?

ok, i did it in my head in about 20 secs. 10%. where's my $1M?