Probability of Bingo on a 3x3 grid Consider a $3\times 3$ grid
We paint 3 grids yellow, 3 grids green and 3 grids red
What is the probability of that there exists three consecutive grids that are the same color (aka "bingo"), including diagonals?
Note: "Bingo" in this question has no relations to American BINGO
My work:
The total number of ways to paint the 3x3 grid is:$$\frac{9!}{3!3!3!} = 1680$$
Now, I suppose we should seperate the problem into three parts:

*

*A "row" being the same color

*A "column" being the same color

*A "diagonal" line being the same color

But how do we calculate that?
 A: There are five different ways this can happen:

*

*Every row is a single colour.

*Exactly one row is a single colour.

*Every column is a single colour.

*Exactly one column is a single colour.

*Exactly one diagonal is a single colour.

These cover all possibilities with no overlap, since if e.g. one row is a single colour, then every column or diagonal can't be all a different colour (since they intersect), or all the same colour (since there are only three tiles of that colour).
We can therefore count the possibilities separately and add.

*

*There are $3!=6$ ways to do this - we can arrange the three colours to the three rows in any order.

*There are $3$ ways to choose which row, and $3$ ways to choose the colour. Then we need to assign a second colour to three of the remaining six squares, without creating another one-colour row. There are $\binom 63$ ways to assign the second colour to three of the remaining squares, $2$ of which would create another solid row, so there are $3\times 3\times(\binom 63-2)=162$ ways.

*Is the same as 1. rotated.

*is the same as 2. rotated.

*There are $3\times2$ ways to choose a colour and diagonal, and then the second colour can be allocated to three other squares in $\binom 63$ ways, all of which are ok. So there are $3\times 2\times\binom63=120$ ways.

Thus the overall probability is $\frac{6+162+6+162+120}{1680}=\frac{19}{70}$.
