Probability of Forming a Line on a 3x3 Grid with Random Selections

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The discussion focuses on calculating the probability of forming at least one line of three squares on a 3x3 grid when randomly selecting five squares. There are six possible combinations for forming a line, and the total number of ways to choose five squares from nine is 126. Participants explore the challenge of determining how many of these combinations result in at least one line of three squares. One user has identified 49 valid arrangements, considering rotations and reflections, and seeks advice on whether to continue this method or find a simpler approach. The conversation emphasizes the importance of systematically analyzing the arrangements to determine the probability accurately.
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Consider a 3-by-3 square grid. Suppose you pick 5 of the squares at random. What is the probability that at least 1 line of 3 squares is formed? (3 diagonal squares is NOT a line).

I do know that there are 6 combinations of lines that are possible. After that, I'm not sure how to follow along with this.
 
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number of possible lines/number of possible ways you can select 3 out of 9
 
I considered that, but you are not choosing 3 squares at a time... but instead, you are choosing a group of 5 squares. And from these 5 squares, what is the probability that there is a "line of 3 squares".
 
The fact that _9C_5=126 shows that there are a total of 126 possible ways a group of five squares can be selected at random. Now how many of these 126 arrangements of squares contain at least one "line of 3 squares"? Do I have to draw all the possibilities (and consider some of the possibilities can be rotated or reflected to get another possibility)?

So far I have found 49 of them (a total of 10 diagram arrangements which can occur multiple times with rotations or reflections). Should I continue this way, or is there a simpler way?
 
If the line goes up a side, top or bottom it's probably always a rotation of a form where it goes up the right

note you can only have one line. So you put the line up the right, you have 6 spaces for 2 squares, multiply by 4 for rotation. Then you have to consider if the line goes up the middle
 
Thank you very much for the insight Office_Shredder. Now I can make only the possible arrangements, and the remaining arrangements of the 126 will be the ones that don't create a "line of 3 squares".
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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