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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:

  1. A "row" being the same color
  2. A "column" being the same color
  3. A "diagonal" line being the same color

But how do we calculate that?

Especially Lime
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Cyh1368
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  • You can find the probability of a red three, of a green three, and of a yellow three, and add these up. But there is the possibility of more than $1$ three, so that sum is over-counting. So you need to take account of the probability of there being exactly $2$ threes $(0)$ and of there being $3$ threes and adust your answer to be the sum of the probability of a red three and no others, of a green three and no others, of a yellow three and no others, and of $3$ threes. – Henry Oct 31 '21 at 11:17
  • Consistent with the comment of @Henry, I suggest using Inclusion Exclusion, as described in the first part (only) of this answer (ignore the stars and bars part of the answer). Here, I would let $A_1, A_2, A_3$ denote the set of 3 consecutive red, 3 consecutive green, and 3 consecutive yellow, respectively. So, the problem reduces to enumerating $|A_1 \cup A_2 \cup A_3|.$ – user2661923 Oct 31 '21 at 11:21
  • See also this Inclusion Exclusion article. – user2661923 Oct 31 '21 at 11:22
  • Following the approach in my comments, the problem is complicated. For example, when exploring $A_1$, you have to distinguish 3 separate cases, depending on whether there are $3$ reds in a row, or in a column, or on one of the main diagonals. One shortcut: If for example there are 3 reds on a main diagonal, then you can not have 3 consecutive yellows or 3 consecutive greens, anywhere. Alternatively, if for example, there are 3 reds in a row, the only way of having 3 consecutive yellows is if the yellows are also in a row, which implies that the 3 greens would also be in a row. – user2661923 Oct 31 '21 at 11:27
  • Another shortcut : consider the Addendum in the answer that I linked to. Because of symmetry considerations, you have (for example) that $|A_1| = |A_2|$ and that $|A_1 \cap A_2| = |A_1 \cap A_3|$. – user2661923 Oct 31 '21 at 11:31
  • I am not allowed to expand my comments into a clear cut solution, because your posted question doesn't compare well with the mathSE protocol expressed in this article. Advice: although it may be possible to divide the problem into the 3 cases, based on a bingo being vertical, horizontal or diagonal, I advise against that approach. Instead, I advise using $A_1, A_2, A_3$, as I have defined them. – user2661923 Nov 02 '21 at 12:06

1 Answers1

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There are five different ways this can happen:

  1. Every row is a single colour.
  2. Exactly one row is a single colour.
  3. Every column is a single colour.
  4. Exactly one column is a single colour.
  5. 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.

  1. There are $3!=6$ ways to do this - we can arrange the three colours to the three rows in any order.
  2. 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.
  3. Is the same as 1. rotated.
  4. is the same as 2. rotated.
  5. 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 $\frac{6+162+6+162+120}{1680}=\frac{19}{70}$.

Especially Lime
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