<Question>
A set of three points is chosen randomly from the grid shown. Each three-point set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

[asy]
size(50);
for (int i=0; i<3; ++i) {
for (int j=0; j<3; ++j) {
dot((i,j));};}
[/asy]
</Question>

<think>
<Plan_1>
Okay, so I need to find the probability that three randomly chosen points from a 3x3 grid all lie on the same straight line. Hmm, let me start by understanding the problem step by step.

First, the grid is 3x3, which means there are 9 points in total arranged in a square grid. The question is about selecting any three points from these nine, and we want to know the probability that these three points are colinear, meaning they lie on the same straight line.

<prompt_1.1> Calculate the total number of ways to choose three points from the nine available points in the grid. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
To find this probability, I remember that probability is generally the number of favorable outcomes divided by the total number of possible outcomes. So, in this case, the total number of ways to choose three points from nine, and the number of those combinations that lie on the same straight line.

First, let's compute the total number of ways to choose three points. Since the order of selection doesn't matter, this is a combination problem. The formula for combinations is C(n, k) = n! / (k! * (n - k)! ), where n is the total number of items, and k is the number of items to choose. Here, n = 9 and k = 3. So, the total number of possible sets is C(9, 3).

Calculating that: 9! / (3! * 6! ) = (9 * 8 * 7) / (3 * 2 * 1) = 84. So, there are 84 possible sets of three points.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, the next part is figuring out how many of these sets of three points lie on the same straight line. To do this, I need to identify all the possible straight lines in a 3x3 grid and then count how many sets of three points each of these lines contains.

<prompt_2.1> Identify all possible straight lines in a 3x3 grid and count how many sets of three points each line contains. </prompt_2.1>
-----

But wait, let me just make sure there isn't a different way people sometimes count lines in a grid. For example, sometimes people consider not just the obvious lines but also diagonals of different lengths.

<prompt_2.2> Re-evaluate the count of colinear triplets, considering all possible lines including shorter diagonals. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let me visualize the 3x3 grid. It's like a tic-tac-toe board with three horizontal rows, three vertical columns, and two main diagonals. But wait, are there any other lines that might have three points? For example, are there lines with different slopes that pass through three points in the grid?

Let me think. The main lines are the rows, columns, and the two main diagonals. Each of these has exactly three points. But is that all? Let's see. If I consider lines with different slopes, like 1/2 or 2, but in a 3x3 grid, the points are too close together for those slopes to pass through three points. For example, a line with slope 2 would go from (0,0) to (1,2), but the next point would be (2,4), which is outside the grid. Similarly, a line with slope 1/2 from (0,0) would go to (2,1), but there's no third point. So, maybe in a 3x3 grid, the only lines that contain three points are the rows, columns, and the two main diagonals.

Wait, but let me check again. Suppose we take a row: three horizontal points. Similarly, three vertical points. Then the two main diagonals. Are there any other lines? For example, if I take a diagonal that's not the main one, like starting from (0,1), but that's only two points. Let me list all possible lines.

Alternatively, maybe there are other lines with different slopes. Let's consider all possible lines that pass through at least three points in the grid. Let's check for each possible slope:

- Horizontal lines: slope 0. There are three rows, each with three points. So, three lines here.

- Vertical lines: undefined slope. There are three columns, each with three points. Three lines here.

- Diagonal lines with slope 1: the main diagonals. There are two of them, each with three points.

- Diagonal lines with slope -1: same as above.

What about other slopes? Let's see. For example, slope 2: starting at (0,0), next point would be (1,2), but that's the edge of the grid. Then (2,4), which is outside. So only two points. Similarly, slope 1/2: starting at (0,0), next point (2,1), but only two points. Similarly, slope -2: from (0,2) to (1,0), but only two points. Slope -1/2: from (0,2) to (2,1), but again only two points. So, in a 3x3 grid, any line with a slope other than 0, infinity, 1, or -1 can only pass through at most two points. Therefore, there are no lines with three points other than the rows, columns, and the two main diagonals.

Therefore, the total number of lines that have three colinear points is 3 (rows) + 3 (columns) + 2 (diagonals) = 8 lines.
</execution_2.1>

<execution_2.2>
But in a 3x3 grid, the only lines that have three points are the three rows, three columns, and the two main diagonals. Any other line would either be a diagonal with only two points or a line with a different slope that doesn't pass through three points. For instance, a line with a slope of 2 would pass through (0,0), (1,2), but that's only two points. Similarly, other slopes wouldn't give three points. So, I think the original count is correct.

Therefore, the number of favorable outcomes is 8, and the total number of possible outcomes is 84. Therefore, the probability is 8/84, which simplifies to 2/21.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Each of these lines has exactly three points, so each contributes one set of three colinear points.

Wait, but hold on. The problem says "a set of three points is chosen randomly".

<prompt_3.1> Verify the count of colinear triplets and calculate the probability. </prompt_3.1>
-----

Based on execution_2.2:
But let me just check once again to make sure I haven't missed any lines. Let me list all possible lines in the grid:

<prompt_3.2> Confirm the probability by listing all possible lines and counting the colinear triplets. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
So, each row is a set of three points in a line, each column is as well, and the two diagonals. So, 3 + 3 + 2 = 8. Therefore, there are 8 sets of three points that lie on a straight line. Therefore, the probability would be 8 divided by 84, which simplifies to 2/21. Wait, 8 divided by 84 is 2/21? Yes, because 8 divided by 84: divide numerator and denominator by 4, 2/21. So, the probability should be 2/21.
</execution_3.1>

<execution_3.2>
- Rows: 3 lines, each with three points.

- Columns: 3 lines, each with three points.

- Diagonals: 2 lines (main diagonals), each with three points.

- Are there any other lines with three points? Let's check.

For example, consider the line passing through (0,1), (1,2). Wait, but that's only two points. Similarly, (1,0), (2,1) is another two-point line, but not part of the main diagonals. Wait, but the main diagonals are from corner to corner. Let me check if there are any other diagonals.

Wait, in a 3x3 grid, the only diagonals that have three points are the main diagonals. The other diagonals have only two points. So, for example, the diagonal starting at (0,1) going to (1,2) is part of the main diagonal that goes from (0,0) to (2,2), but that's already counted. So, no, there are no other diagonals with three points.

Similarly, vertical and horizontal lines are already counted. So, yes, only 8 lines.

Therefore, 8 favorable sets.

Therefore, the probability is 8/84, which simplifies to 2/21. So, the answer should be 2/21.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But let me verify this again. Because sometimes when you count lines, you might miss some.

<prompt_4.1> Verify the count of colinear triplets again to ensure no lines are missed. </prompt_4.1>
-----

Based on execution_3.2:
Therefore, the probability is 8/84, which reduces to 2/21. So, the probability is 2/21.

But let me just cross-check with another approach.

<prompt_4.2> Cross-verify the probability using a different approach. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
Wait, another way to think: how many triplets are colinear? For each line that has three points, there's exactly one triplet (since each line has exactly three points, and choosing all three gives one combination). So, if there are 8 lines with three points each, that gives 8 favorable combinations. Therefore, 8/84.

But wait, hold on. Wait, in the 3x3 grid, are there any other lines that pass through three points? Let me think again. For example, consider the vertical lines: each column has three points, so that's three lines. Similarly, horizontal lines: three lines. Diagonals: two lines. So 3 + 3 + 2 = 8. So, 8 lines, each contributing one set of three colinear points. Therefore, 8 favorable sets. So, 8/84 reduces to 2/21.
</execution_4.1>

<execution_4.2>
For example, sometimes people use coordinates to count. Let me assign coordinates to the grid points, with rows as x-coordinates from 0 to 2 and columns as y-coordinates from 0 to 2. So, each point is (x, y) where x and y are 0, 1, 2.

Then, a line in the grid can be defined by the equation ax + by + c = 0, where a, b, c are integers with no common divisor. For three points to be colinear, they must lie on such a line. But in the 3x3 grid, the only lines that pass through three points are the ones we already considered: rows, columns, and the two main diagonals. So, again, 8 lines.

Alternatively, suppose we fix a point and count how many lines pass through it with two other points. But that might complicate things. Alternatively, perhaps use the concept of slopes.

For any three points to be colinear, they must have the same slope between each pair. So, for three points A, B, C, the slope between A and B must equal the slope between B and C. Let's see how many such triplets exist.

But this approach might be more involved. Let me see.

Alternatively, perhaps I can list all possible lines:

Rows:

1. (0,0), (0,1), (0,2)

2. (1,0), (1,1), (1,2)

3. (2,0), (2,1), (2,2)

Columns:

4. (0,0), (1,0), (2,0)

5. (0,1), (1,1), (2,1)

6. (0,2), (1,2), (2,2)

Diagonals:

7. (0,0), (1,1), (2,2)

8. (0,2), (1,1), (2,0)

So, that's 8 lines. Each of these lines has exactly three points, so each contributes exactly one combination of three colinear points. Therefore, 8 favorable combinations.

Therefore, the probability is 8 / C(9,3) = 8 / 84 = 2 / 21. So, that seems correct.

But let me check once more. Let's think about how many lines there are with three points. In a 3x3 grid, as mentioned, the only lines with three points are the rows, columns, and the two main diagonals. So, 3 + 3 + 2 = 8. So, 8 lines, each with exactly one combination of three points. Therefore, 8 favorable combinations.

Total combinations: C(9,3) = 84. So, probability is 8/84 = 2/21 ≈ 0.0952, or 9.52%.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.2:
But wait, just to ensure that there isn't a different interpretation of the problem. For example, does the problem consider the grid as 3x3 dots, which form a 2x2 grid of squares?

<prompt_5.1> Ensure there is no misinterpretation of the grid and confirm the probability. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Wait, the Asymptote code given is:

size(50);
for (int i=0; i<3; ++i) {
for (int j=0; j<3; ++j) {
dot((i,j));}
}

So, this code dots the points from (0,0) to (2,2), which is a 3x3 grid of points, arranged in a square. So, it's a 3x3 grid of points, which forms a 2x2 grid of squares. But in terms of colinear points, it's still a 3x3 grid. So, the analysis remains the same.

Alternatively, sometimes people get confused between 3x3 grid of points and 3x3 grid of squares, but in this case, the problem says a set of three points is chosen from the grid, with the Asymptote code showing dots at (i,j) for i and j from 0 to 2. So, it's a 3x3 grid of points. So, the original analysis holds.

Therefore, I think the answer is 2/21.
</execution_5.1>
</Execution_5>

<Final_answer>
Therefore, the probability that three randomly chosen points from this grid lie on the same straight line is $\boxed{\frac{2}{21}}$.
</Final_answer>
</think>