Hello all, I've been playing Nono for about 3 days now, loving it (not that I'm all that good... yet! )

Anyways, I've been working out various chances of certain things on the Beginner board:

The total number of possible beginner boards is 36C9 = 94143280
(for those unfamiliar with this notation, nCr means the number of ways of choosing r objects from a set of n. The formula is n! ÷ r!(n-r)!)

Therefore the chances of just choosing a particular board and getting it is 1 in 94143280.

You could expect a 50% probability that you will get this in the first 50% of 94143280 games, i.e. 47071640.

At 1 game every 1/2 second, you would need to play beginner for 23535820s = 272 days non-stop to reach this total. And then you have half a chance of getting the board you want

To make things a bit more likely, say you wanted a board with a blank row or column. In this case, there are 12 possible rows or columns the blanks could occupy, so the remaining 3 blanks have to be in the remaining 30 squares:
12 × 30C3 = 48720

Therefore the chances of getting a board with a blank row or column is 48720 / 94143280 which is about 1 in 1932 (attainable!)

Another good way to get a nice beginner level is to have lots of 1-1-2 or 1-2-1 or 2-1-1. Ideally, we want 3 of these and 3 of 1-4, 2-3, 3-2 or 4-1 along the rows (by my beginner tactics, and I think I saw that someone else used this as well).
The 1-1-2s could be in any of 6C3 places, and there are 3 options for the order. Then there are 4 options for each of the remaining 3 rows:
3 × 6C3 × 4³ = 3840
The chances of getting one of these boards is about 1 in 24516, and by the half-second half-chance timing estimate, you'd need to play for about 1h42m to get a 50% probability of getting one.

Happy sweeping!

Hehe, thanks Qetuth

As for beginner boards that can be solved in a single click, somehow we need to find a method of quickly counting all possible sets of 9 squares connected by edges. Hmm... ???

I guess this is essentially a Graph Theory problem, so take the squares as vertices and edges between squares as edges joining two vertices? Eek, this is hard. I think some of the critical problems could be if you try and build from a starting point, you could cross over yourself, or making sure you don't count things twice.

Maybe I'll just write a program to test every possibility...

Just saw another!

I'd figure out the probability myself but I don't exactly know how to factor the number 180  without getting a serious head ache  

A specific empty row and empty column on expert means you're basically putting 90 mines into a 14x11 area, or, 154c90 = 154c64 = 154! / 90!*64! = 1.6x10^44.

There are 180c90 boards for expert: 9.1x10^52

So, chance of a board with a specific row and column empty is 1 in 555 million.

But, there are 15 columns and 12 rows, so 180 possible choices of specific row and column, so if you don't care WHICH row and column, its a mere 1 in 3 million (That's not taking into account the rather negligible number of boards which have more than 2 blanks: only 2 per million boards with a blank row and column also have at least one other blank column. So even though we're counting those boards twice, my rounding has more effect than they do.)

@Q: EVERY board is solvable using only one click   It´s a bit tricky though:
1) you can leave the grid to get into it at another place
2) you can move the cursor exactly at the line between cells - it doesn´t cause cells to behave as being clicked

OK, this was just for fun  , good job using all the math stuff  8) looking forward to see how many one click genius boards exist  

<Edit> @nikolaj: gah!! Oh well, I'll keep going, I guess, since that's sort of cheating. I dont imagine a single click which goes out of the board is any faster than 2 clicks. Actually, many of these aren't actually fast to solve, but anyway.... </Edit>

we've covered everything with 5 in a row. So, category 3: boards within a 4x4 square:

You cant have 2 blank rows plus a blank column in that square. A blank row and column leaves two blanks remaining:
two sides give: 4x(5c2+4) = 56.
Side and middle: 8x(5c2+2) = 96
two middles: 4x(8c2+2) = 120
So subtotal for that section: ~272

Now, with a blank top row, you have 5 remaining, and we've counted anything that uses a blank column as well. So:
Blank top row, 5 in 2nd and 3rd rows: 56-12=44
Blank top row, one in bottom row, no complete columns: 4+8+8+4=24
Blank top row, two in bottom row: 4
total with blank row: 72

Blank 2nd row, 5 in 3rd and 4th: 44
blank 2nd row, 1 above, 4 below: 28+16+16+28=88
blank 2nd row, 2 above, 9+12+9+20=50
blank 2nd row, 3 above, 11x4=44
blank 2nd row, 4 above is counted
so, blank second row gives 226

total with a blank row or a column but not both is then 298x4= ~1192

So now we only need to find the possible sets of 9 with no blank row or column, in a 4x4 area.

With a 3x2 section (can go in 6 places), placing 3 more around it without completing a row or column, gives 4x4+2x13=42. so with a 2x3 section must also be 42. 12 of these have both 2x3 and 3x2 so we've counted them twice. 42+42-12=72

With a 3 row, then an adjacent but displaced 3 row, but without creating any of the above: 4 for top row, so 16 for top and bottom, mirrored.13 for middle row, so 26 for mirrored: 16+26=42

We can now try and list all the shapes which have none of the above formations.

With a ring: 8
With a part-ring (missing a side): 8+8=16
With a part ring (missing a corner): 8+40+24=72
ring missing side and corner: 4+24=28
Claw: 4
Cup: 8
b3t3b2b1: 4
b2b3t3t1: 2

Thats all I can find, so 142+42+72=~256

If that's right, 272+1192+256=1720 formations within a 4x4 square, of which there are 9 in a beginner board: ~~15480~~

So, our running total is 26108, with a fairly small but very hard to find set remaining to count.

I wrote a program to find single-click boards, and there are only apparently 17955... but of course I could be wrong

Doing it a case-by-case way, it's very easy to count ones twice.