Explanation:
Okay, so I need to find the greatest possible value of the greatest common divisor of $ n + 7 $ and $ 2n + 1 $, where $ n $ is a positive integer. Hmm, let me think about how to approach this.

First, I remember that the greatest common divisor (gcd) of two numbers also divides their linear combinations. Maybe I can use the Euclidean algorithm here? Let me recall: the Euclidean algorithm says that $ \gcd(a, b) = \gcd(b, a - qb) $ where $ q $ is some integer. So maybe I can apply that here.

Let me denote $ a = 2n + 1 $ and $ b = n + 7 $. Then, according to the Euclidean algorithm, $ \gcd(a, b) = \gcd(n + 7, (2n + 1) - 2(n + 7)) $. Let me compute that subtraction:

$ (2n + 1) - 2(n + 7) = 2n + 1 - 2n - 14 = -13 $.

So, $ \gcd(n + 7, -13) $. But since gcd is always non-negative, this simplifies to $ \gcd(n + 7, 13) $.

Therefore, the gcd of $ n + 7 $ and $ 2n + 1 $ is the same as the gcd of $ n + 7 $ and 13. So, the problem reduces to finding the maximum possible value of $ \gcd(n + 7, 13) $, where $ n $ is a positive integer.

Now, since 13 is a prime number, the possible values of the gcd are 1 and 13. Therefore, the maximum possible value is 13, provided that $ n + 7 $ is a multiple of 13.

So, if $ n + 7 = 13k $ for some integer $ k $, then $ \gcd(n + 7, 13) = 13 $. Therefore, the maximum possible value is 13.

But wait, let me check if this is possible. Let me take $ n = 13k - 7 $. Since $ n $ has to be a positive integer, $ 13k - 7 > 0 $, so $ k \geq 1 $, because if $ k = 1 $, then $ n = 13 - 7 = 6 $, which is positive.

Let me test with $ n = 6 $. Then $ n + 7 = 13 $, and $ 2n + 1 = 13 $. Therefore, $ \gcd(13, 13) = 13 $. So that works.

But let me check another value. Let me take $ k = 2 $, so $ n = 26 - 7 = 19 $. Then $ n + 7 = 26 $, and $ 2n + 1 = 39 $. Then $ \gcd(26, 39) $. Let me compute that. The divisors of 26 are 1, 2, 13, 26. Divisors of 39 are 1, 3, 13, 39. So the gcd is 13. So that also works.

Therefore, it seems that 13 is achievable. Therefore, the maximum possible value is 13.

But let me make sure there isn't a higher value. Since 13 is prime, the gcd can't be higher than 13. So 13 is the maximum.

Wait, but let me think again. Suppose there is some other way where the gcd is higher? But since we have reduced the problem to gcd(n + 7, 13), and since 13 is prime, the gcd can only be 1 or 13. Therefore, 13 is the maximum.

Therefore, the answer should be 13.

But let me check another example. Let me take n = 6, which gives gcd(13, 13) = 13. If I take n = 1, then gcd(8, 3) = 1. If n = 2, gcd(9, 5) = 1. n = 3, gcd(10, 7) = 1. n = 4, gcd(11, 9) = 1. n = 5, gcd(12, 11) = 1. n = 6, gcd(13, 13) = 13. n = 7, gcd(14, 15) = 1. n = 8, gcd(15, 17) = 1. n = 9, gcd(16, 19) = 1. n = 10, gcd(17, 21) = 1. n = 11, gcd(18, 23) = 1. n = 12, gcd(19, 25) = 1. n = 13, gcd(20, 27) = 1. n = 14, gcd(21, 29) = 1. n = 15, gcd(22, 31) = 1. n = 16, gcd(23, 33) = 1. n = 17, gcd(24, 35) = 1. n = 18, gcd(25, 37) = 1. n = 19, gcd(26, 39) = 13. So yes, when n is 6 or 19, the gcd is 13.

Therefore, the maximum possible value is indeed 13.

But let me think again if there's a case where the gcd could be higher. Suppose someone says, what if n + 7 and 2n + 1 have a common divisor d greater than 13? But since we have shown that d divides 13, which is prime, so d can't be greater than 13. Therefore, 13 is the maximum.

Alternatively, suppose we didn't use the Euclidean algorithm. Let me try another approach. Let me suppose that d divides both n + 7 and 2n + 1. Then, d divides any linear combination of these two numbers. For example, if d divides n + 7 and 2n + 1, then d divides 2(n + 7) - (2n + 1) = 2n + 14 - 2n - 1 = 13. Therefore, d divides 13. Therefore, the possible values of d are the divisors of 13, which are 1 and 13. Therefore, the maximum possible value is 13.

So this confirms the previous result. Therefore, regardless of the value of n, the gcd can't be more than 13, and it can be 13. Therefore, the answer is 13.

Therefore, the greatest possible value is 13.

**Final Answer**
The greatest possible value of $\gcd(n + 7, 2n + 1)$ is \boxed{13}.

To determine the greatest possible value of $ \gcd(n + 7, 2n + 1) $ for a positive integer $ n $, we begin by applying the **Euclidean algorithm**, which is a powerful tool for computing the greatest common divisor (gcd) of two integers.

---

### Step 1: Apply the Euclidean Algorithm

Let:
- $ a = 2n + 1 $
- $ b = n + 7 $

We compute:
$$
\gcd(2n + 1, n + 7) = \gcd(n + 7, (2n + 1) - 2(n + 7))
$$

Simplify the expression inside the gcd:
$$
(2n + 1) - 2(n + 7) = 2n + 1 - 2n - 14 = -13
$$

So:
$$
\gcd(2n + 1, n + 7) = \gcd(n + 7, -13)
$$

Since the gcd is always non-negative, we can write:
$$
\gcd(n + 7, 13)
$$

---

### Step 2: Analyze the Result

We now want to find the **maximum** value of $ \gcd(n + 7, 13) $, where $ n $ is a positive integer.

Note that 13 is a **prime number**, so the only possible values of the gcd are the **positive divisors of 13**, which are:
$$
1 \quad \text{and} \quad 13
$$

Thus, the **maximum possible value** of the gcd is **13**, and this occurs **if and only if** $ n + 7 $ is a multiple of 13.

---

### Step 3: Verify the Maximum is Achievable

Let’s find such an $ n $. Suppose:
$$
n + 7 = 13k \quad \text{for some integer } k
\Rightarrow n = 13k - 7
$$

We need $ n > 0 $, so:
$$
13k - 7 > 0 \Rightarrow k \geq 1
$$

Take $ k = 1 $, then $ n = 6 $. Check:
- $ n + 7 = 13 $
- $ 2n + 1 = 13 $

So:
$$
\gcd(13, 13) = 13
$$

This confirms that the maximum value of $ \gcd(n + 7, 2n + 1) $ is indeed **13**, and it is achievable.

---

### Final Conclusion

Since the gcd can only be 1 or 13 (as 13 is prime), and 13 is achievable, the **greatest possible value** of $ \gcd(n + 7, 2n + 1) $ is:

$$
\boxed{13}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest possible value of the greatest common divisor of $ n + 7 $ and $ 2n + 1 $, where $ n $ is a positive integer. Hmm, let me think about how to approach this.

First, I remember that the greatest common divisor (gcd) of two numbers also divides their linear combinations. Maybe I can use the Euclidean algorithm here? Let me recall: the Euclidean algorithm says that $ \gcd(a, b) = \gcd(b, a - qb) $ where $ q $ is some integer. So maybe I can apply that here.

Let me denote $ a = 2n + 1 $ and $ b = n + 7 $. Then, according to the Euclidean algorithm, $ \gcd(a, b) = \gcd(n + 7, (2n + 1) - 2(n + 7)) $. Let me compute that subtraction:

$ (2n + 1) - 2(n + 7) = 2n + 1 - 2n - 14 = -13 $.

So, $ \gcd(n + 7, -13) $. But since gcd is always non-negative, this simplifies to $ \gcd(n + 7, 13) $.

Therefore, the gcd of $ n + 7 $ and $ 2n + 1 $ is the same as the gcd of $ n + 7 $ and 13. So, the problem reduces to finding the maximum possible value of $ \gcd(n + 7, 13) $, where $ n $ is a positive integer.

Now, since 13 is a prime number, the possible values of the gcd are 1 and 13. Therefore, the maximum possible value is 13, provided that $ n + 7 $ is a multiple of 13.

So, if $ n + 7 = 13k $ for some integer $ k $, then $ \gcd(n + 7, 13) = 13 $. Therefore, the maximum possible value is 13.

But wait, let me check if this is possible. Let me take $ n = 13k - 7 $. Since $ n $ has to be a positive integer, $ 13k - 7 > 0 $, so $ k \geq 1 $, because if $ k = 1 $, then $ n = 13 - 7 = 6 $, which is positive.

Let me test with $ n = 6 $. Then $ n + 7 = 13 $, and $ 2n + 1 = 13 $. Therefore, $ \gcd(13, 13) = 13 $. So that works.

But let me check another value. Let me take $ k = 2 $, so $ n = 26 - 7 = 19 $. Then $ n + 7 = 26 $, and $ 2n + 1 = 39 $. Then $ \gcd(26, 39) $. Let me compute that. The divisors of 26 are 1, 2, 13, 26. Divisors of 39 are 1, 3, 13, 39. So the gcd is 13. So that also works.

Therefore, it seems that 13 is achievable. Therefore, the maximum possible value is 13.

But let me make sure there isn't a higher value. Since 13 is prime, the gcd can't be higher than 13. So 13 is the maximum.

Wait, but let me think again. Suppose there is some other way where the gcd is higher? But since we have reduced the problem to gcd(n + 7, 13), and since 13 is prime, the gcd can only be 1 or 13. Therefore, 13 is the maximum.

Therefore, the answer should be 13.

But let me check another example. Let me take n = 6, which gives gcd(13, 13) = 13. If I take n = 1, then gcd(8, 3) = 1. If n = 2, gcd(9, 5) = 1. n = 3, gcd(10, 7) = 1. n = 4, gcd(11, 9) = 1. n = 5, gcd(12, 11) = 1. n = 6, gcd(13, 13) = 13. n = 7, gcd(14, 15) = 1. n = 8, gcd(15, 17) = 1. n = 9, gcd(16, 19) = 1. n = 10, gcd(17, 21) = 1. n = 11, gcd(18, 23) = 1. n = 12, gcd(19, 25) = 1. n = 13, gcd(20, 27) = 1. n = 14, gcd(21, 29) = 1. n = 15, gcd(22, 31) = 1. n = 16, gcd(23, 33) = 1. n = 17, gcd(24, 35) = 1. n = 18, gcd(25, 37) = 1. n = 19, gcd(26, 39) = 13. So yes, when n is 6 or 19, the gcd is 13.

Therefore, the maximum possible value is indeed 13.

But let me think again if there's a case where the gcd could be higher. Suppose someone says, what if n + 7 and 2n + 1 have a common divisor d greater than 13? But since we have shown that d divides 13, which is prime, so d can't be greater than 13. Therefore, 13 is the maximum.

Alternatively, suppose we didn't use the Euclidean algorithm. Let me try another approach. Let me suppose that d divides both n + 7 and 2n + 1. Then, d divides any linear combination of these two numbers. For example, if d divides n + 7 and 2n + 1, then d divides 2(n + 7) - (2n + 1) = 2n + 14 - 2n - 1 = 13. Therefore, d divides 13. Therefore, the possible values of d are the divisors of 13, which are 1 and 13. Therefore, the maximum possible value is 13.

So this confirms the previous result. Therefore, regardless of the value of n, the gcd can't be more than 13, and it can be 13. Therefore, the answer is 13.

Therefore, the greatest possible value is 13.

**Final Answer**
The greatest possible value of $\gcd(n + 7, 2n + 1)$ is \boxed{13}.

To determine the greatest possible value of $ \gcd(n + 7, 2n + 1) $ for a positive integer $ n $, we begin by applying the **Euclidean algorithm**, which is a powerful tool for computing the greatest common divisor (gcd) of two integers.

---

### Step 1: Apply the Euclidean Algorithm

Let:
- $ a = 2n + 1 $
- $ b = n + 7 $

We compute:
$$
\gcd(2n + 1, n + 7) = \gcd(n + 7, (2n + 1) - 2(n + 7))
$$

Simplify the expression inside the gcd:
$$
(2n + 1) - 2(n + 7) = 2n + 1 - 2n - 14 = -13
$$

So:
$$
\gcd(2n + 1, n + 7) = \gcd(n + 7, -13)
$$

Since the gcd is always non-negative, we can write:
$$
\gcd(n + 7, 13)
$$

---

### Step 2: Analyze the Result

We now want to find the **maximum** value of $ \gcd(n + 7, 13) $, where $ n $ is a positive integer.

Note that 13 is a **prime number**, so the only possible values of the gcd are the **positive divisors of 13**, which are:
$$
1 \quad \text{and} \quad 13
$$

Thus, the **maximum possible value** of the gcd is **13**, and this occurs **if and only if** $ n + 7 $ is a multiple of 13.

---

### Step 3: Verify the Maximum is Achievable

Let’s find such an $ n $. Suppose:
$$
n + 7 = 13k \quad \text{for some integer } k
\Rightarrow n = 13k - 7
$$

We need $ n > 0 $, so:
$$
13k - 7 > 0 \Rightarrow k \geq 1
$$

Take $ k = 1 $, then $ n = 6 $. Check:
- $ n + 7 = 13 $
- $ 2n + 1 = 13 $

So:
$$
\gcd(13, 13) = 13
$$

This confirms that the maximum value of $ \gcd(n + 7, 2n + 1) $ is indeed **13**, and it is achievable.

---

### Final Conclusion

Since the gcd can only be 1 or 13 (as 13 is prime), and 13 is achievable, the **greatest possible value** of $ \gcd(n + 7, 2n + 1) $ is:

$$
\boxed{13}
$$

Solution:
Let $d = \gcd(n + 7, 2n + 1)$, so $d$ divides both $n + 7$ and $2n + 1$.  Then $d$ divides $2(n + 7) - (2n + 1) = 13$, so $d$ is at most 13.

If $n = 6$, then $\gcd(n + 7, 2n + 1) = \gcd(13,13) = 13$, which shows that the value of 13 is attainable.  Therefore, the greatest possible value of $\gcd(n + 7, 2n + 1)$ is $\boxed{13}$.


Explanation:
Okay, so I need to find the probability that three randomly chosen points from a 3x3 grid lie on the same straight line. Let me start by understanding the problem.

First, the grid is 3x3, which means there are 9 points in total. The question says a set of three points is chosen randomly, and each set has the same probability. So, I need to figure out two things: the total number of possible three-point sets, and the number of those sets that are colinear (lie on the same straight line). Then, the probability would be the number of colinear triplets divided by the total number of triplets.

Let me start with the total number of three-point sets. Since there are 9 points, the number of ways to choose 3 points is the combination of 9 taken 3 at a time. The formula for combinations is C(n, k) = n! / (k! * (n - k)!). So, plugging in the numbers:

C(9, 3) = 9! / (3! * 6!) = (9 × 8 × 7) / (3 × 2 × 1) = 84. So, there are 84 possible three-point sets. Got that.

Now, the tricky part is figuring out how many of these sets are colinear. Let me think about how lines can be formed on a 3x3 grid. Lines can be horizontal, vertical, or diagonal. Also, there might be other lines with different slopes that pass through three points. Wait, in a 3x3 grid, are there any other lines besides the obvious ones?

Let me visualize the grid. Let me imagine the grid with coordinates from (0,0) to (2,2). So, the points are (0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2).

First, horizontal lines: Each row is a horizontal line. There are 3 rows. Each row has 3 points, so each row contributes exactly one set of three colinear points. So, 3 horizontal lines.

Similarly, vertical lines: Each column is a vertical line. There are 3 columns, each with 3 points. So, another 3 vertical lines.

Then, the main diagonals: The diagonal from (0,0) to (1,1) to (2,2), and the diagonal from (0,2) to (1,1) to (2,0). So, that's 2 more lines.

But wait, are there any other lines that pass through three points? For example, lines with slopes like 2, 1/2, -1, etc. Let me check.

Let me think about lines with slope 1/2. For example, starting at (0,0), going up 1 and right 2 would be (0,0), (1, 0.5), (2, 1). But (1, 0.5) is not a grid point, so that line doesn't pass through three grid points. Similarly, starting at (0,1), going up 1 and right 2 would be (0,1), (1, 1.5), (2, 2). Again, (1, 1.5) is not a grid point. Similarly, slope 2: starting at (0,0), going up 2 and right 1 would be (0,0), (1,2), then (2,4), which is outside the grid. So, only two points. Similarly, starting at (0,1), going up 2 and right 1 would be (0,1), (1,3), which is outside. So, no.

What about slope -1/2? Starting at (0,2), going down 1 and right 2: (0,2), (1,1), (2,0). Wait, that's the main diagonal we already counted. Similarly, starting at (0,1), going down 1 and right 2: (0,1), (1,0), (2,-1). The last point is outside. So, only two points. Similarly, slope -2: starting at (0,2), going down 2 and right 1: (0,2), (1,0), then (2,-2). Again, only two points. So, no other lines with slopes other than 0, infinity, 1, -1 have three points.

Wait, but what about other diagonals? For example, in a 3x3 grid, are there any other diagonals with three points? Let me check. For example, starting at (0,0), going to (1,1), (2,2) is the main diagonal. Starting at (0,1), going to (1,2)... but that's only two points. Similarly, starting at (1,0), going to (2,1)... again two points. Similarly, starting at (0,2), going to (1,1), (2,0) is the other main diagonal. So, no other diagonals with three points.

Wait, but what about lines that aren't aligned with the grid or main diagonals? For example, maybe a line that goes through (0,1), (1,2), and... but (2,3) is outside. Similarly, (1,0), (2,1)... (3,2) is outside. So, no. Similarly, (0,0), (1,1), (2,2) is already counted. So, seems like only the horizontal, vertical, and two main diagonals have three colinear points.

Wait, but hold on. Let me think again. Are there any other lines? For example, in a 3x3 grid, there are also lines that go through three points with different slopes? Let me check for example, the line passing through (0,0), (1,2), and (2,1). Wait, does that line pass through three points? Let me check. The slope between (0,0) and (1,2) is 2. Then, from (1,2) to (2,1) is slope (1-2)/(2-1) = -1. So, different slopes, so not a straight line. So, that's not a straight line. How about (0,1), (1,0), (2,-1)? Again, the last point is outside.

Alternatively, maybe (0,0), (1,1), (2,2) is the main diagonal. What about (0,0), (1,0), (2,0)? That's horizontal. Similarly, (0,0), (0,1), (0,2) is vertical. So, seems like only those.

Wait, but maybe there are lines that are not aligned with the grid or main diagonals but still pass through three points? Let me think. For example, take points (0,1), (1,2), and (2,1). Wait, is there a straight line through these three? Let me check the slopes. Between (0,1) and (1,2): slope is (2-1)/(1-0) = 1. Between (1,2) and (2,1): slope is (1-2)/(2-1) = -1. Different slopes, so not colinear. How about (0,1), (1,1), (2,1)? That's horizontal, already counted.

How about (0,0), (1,1), (2,2)? Already counted. How about (0,2), (1,1), (2,0)? Also counted.

Wait, maybe there are lines with slope 2 or 1/2? Let me check. For example, starting at (0,0), going up 2 and right 1: (0,0), (1,2). Then, next would be (2,4), which is outside. So, only two points. Similarly, starting at (0,1), going up 2 and right 1: (0,1), (1,3) which is outside. Starting at (1,0), going up 2 and right 1: (1,0), (2,2). Then, (3,4) outside. So, only two points. Similarly, slope 1/2: starting at (0,0), going up 1 and right 2: (0,0), (2,1). Then, next would be (4,2), outside. So, only two points. Similarly, starting at (0,1), going up 1 and right 2: (0,1), (2,2). Then, (4,3) outside. So, again, only two points.

So, seems like there are no other lines with three points. Therefore, total number of colinear triplets is horizontal lines (3) + vertical lines (3) + main diagonals (2) = 8. Wait, but hold on. Wait, is that all?

Wait, but wait. Let me think again. Are there any other lines? For example, in some grids, there are more lines. Wait, for example, in a 3x3 grid, there are also lines that are not aligned with the main directions. Wait, but as we saw, they don't have three points.

Wait, but let me check another way. Let me count all possible lines that contain at least three points.

First, horizontal lines: 3 lines, each with 3 points.

Vertical lines: 3 lines, each with 3 points.

Diagonals with slope 1: 2 lines, each with 3 points.

Diagonals with slope -1: 2 lines, each with 3 points. Wait, no, slope -1 is already counted as the other diagonal. Wait, there are two diagonals with slope 1 and two with slope -1? Wait, no. Wait, in a 3x3 grid, there are two main diagonals: one from top-left to bottom-right (slope 1) and one from top-right to bottom-left (slope -1). Each of these diagonals has exactly three points. So, that's two diagonals.

But wait, are there other diagonals with three points? For example, if you consider diagonals that are shorter? But no, because they have to have three points. For example, a diagonal that starts at (0,1), goes to (1,2), but then there's no third point. Similarly, starting at (1,0), going to (2,1), no third point. Similarly, starting at (0,1), going to (1,0), then (2,-1), which is outside. So, no.

Therefore, total lines with three points: 3 horizontal + 3 vertical + 2 diagonals = 8 lines. Each of these lines has exactly three points, so each contributes exactly one set of three colinear points. Therefore, the number of colinear triplets is 8.

Wait, but hold on. Wait, is that correct? Because if there are 8 lines, each with exactly one triplet, then there are 8 colinear triplets. But I have a feeling that maybe there are more. Wait, let me think again.

Wait, for example, in the horizontal lines: each row is a line with three points. So, there are three rows, each contributing one triplet. Similarly, three columns. Then, two diagonals. So, 3 + 3 + 2 = 8. So, 8 triplets.

But wait, I remember that in some problems, there are more lines. Wait, maybe I need to check again. Let me think of all possible lines that can pass through three points.

Alternatively, maybe I can think of all possible lines and count how many points they have.

But maybe there's another way. Let me think of all possible lines.

First, horizontal lines: y = 0, y = 1, y = 2. Each has three points.

Vertical lines: x = 0, x = 1, x = 2. Each has three points.

Diagonals with slope 1: y = x, y = x - 1, y = x - 2? Wait, let me check.

For slope 1, the lines are:

- Starting at (0,0): y = x. Points (0,0), (1,1), (2,2).

- Starting at (0,1): y = x + 1. Points (0,1), (1,2). But only two points.

- Starting at (0,2): y = x + 2. Points (0,2), (1,3) which is outside. So, only one point.

Similarly, starting from the other side:

- Starting at (1,0): y = x - 1. Points (1,0), (2,1). Only two points.

- Starting at (2,0): y = x - 2. Points (2,0). Only one point.

So, only one diagonal with slope 1 that has three points. Similarly for slope -1:

Lines with slope -1:

- Starting at (0,2): y = -x + 2. Points (0,2), (1,1), (2,0).

- Starting at (0,1): y = -x + 1. Points (0,1), (1,0). Only two points.

- Starting at (0,0): y = -x. Points (0,0), (1,-1) which is outside. So, only one point.

Similarly, starting from the other side:

- Starting at (1,2): y = -x + 3. Points (1,2), (2,1). Only two points.

- Starting at (2,2): y = -x + 4. Points (2,2). Only one point.

So, only one diagonal with slope -1 that has three points. Therefore, total diagonals with three points: 2.

So, total lines with three points: horizontal (3) + vertical (3) + diagonal (2) = 8. Therefore, 8 lines, each contributing one triplet. Therefore, 8 colinear triplets.

But wait, I think I remember that in some cases, there are more lines. Wait, let me check again. For example, is there a line that goes through (0,0), (1,2), and (2,1)? Wait, earlier I thought that wasn't a straight line. Let me check coordinates.

Let me calculate the slope between (0,0) and (1,2): slope is (2 - 0)/(1 - 0) = 2. Then between (1,2) and (2,1): slope is (1 - 2)/(2 - 1) = -1. Different slopes, so not colinear.

How about (0,1), (1,2), (2,3)? But (2,3) is outside the grid.

How about (1,0), (1,1), (1,2)? That's vertical, already counted.

How about (0,0), (1,1), (2,2)? Already counted.

How about (0,2), (1,1), (2,0)? Already counted.

How about (0,0), (2,1)? That's a line with slope 1/2, but only two points.

How about (0,1), (1,0), (2,-1)? Again, only two points inside the grid.

So, seems like there are no other lines with three points. Therefore, total colinear triplets are 8.

Wait, but hold on. Wait, I think I might be missing something here. Let me think again.

Wait, in the 3x3 grid, there are also lines that are not aligned with the grid or the main diagonals but still pass through three points. Wait, for example, take the points (0,0), (1,1), (2,2) – that's the main diagonal. What about (0,1), (1,1), (2,1)? That's horizontal.

Wait, but maybe there are other lines. For example, take points (0,0), (1,2), and (2,1). Wait, we saw that they are not colinear. How about (0,0), (2,1), and (something else)? Let me check.

Wait, if I take (0,0), (1, 0.5), (2,1). But (1, 0.5) is not a grid point.

Alternatively, maybe there are lines with three points that are not aligned with the grid? For example, in a 3x3 grid, is there a line that goes through three points with different slopes?

Wait, let me think of all possible triplets and see if any of them are colinear but not in the 8 lines I mentioned.

For example, take the points (0,0), (1,1), (2,2). That's already counted.

How about (0,0), (1,0), (2,0)? Horizontal, counted.

How about (0,0), (0,1), (0,2)? Vertical, counted.

How about (0,1), (1,1), (2,1)? Horizontal, counted.

How about (1,0), (1,1), (1,2)? Vertical, counted.

How about (0,2), (1,1), (2,0)? Diagonal, counted.

How about (0,0), (1,2), (2,1)? Not colinear.

How about (0,1), (1,2), (2,1)? Not colinear.

How about (0,2), (1,0), (2,1)? Let me check if they are colinear.

Slope between (0,2) and (1,0): (0 - 2)/(1 - 0) = -2.

Slope between (1,0) and (2,1): (1 - 0)/(2 - 1) = 1. Different slopes. Not colinear.

How about (0,0), (2,1), (something)? Let me check if there's a third point.

The line from (0,0) to (2,1) has slope 1/2. The equation is y = (1/2)x. The other points on this line would be (0,0), (2,1), and (4,2), which is outside. So, only two points.

How about (0,1), (1,2), (2,3)? Again, (2,3) is outside.

How about (1,0), (2,1), (3,2)? Outside.

How about (0,0), (1,1), (2,2)? Already counted.

How about (0,2), (1,1), (2,0)? Already counted.

How about (0,0), (0,1), (0,2)? Vertical, counted.

How about (1,0), (1,1), (1,2)? Vertical, counted.

How about (2,0), (2,1), (2,2)? Vertical, counted.

How about (0,0), (1,0), (2,0)? Horizontal, counted.

How about (0,1), (1,1), (2,1)? Horizontal, counted.

How about (0,2), (1,2), (2,2)? Horizontal, counted.

So, seems like all possible triplets that are colinear are already counted in the 8 lines. Therefore, there are 8 colinear triplets.

Wait, but hold on. Wait, I think I might be missing some lines. Let me think again.

Wait, in the 3x3 grid, there are also lines that are not axis-aligned or diagonals but still pass through three points. Wait, but as we saw, they don't.

Wait, but let me think of another approach. Let me count all possible lines that can be formed with three points.

Each line is determined by two points. So, maybe I can count all lines that have at least three points.

But since there are only 9 points, maybe it's manageable.

First, horizontal lines: 3 lines, each with 3 points.

Vertical lines: 3 lines, each with 3 points.

Diagonals with slope 1: 1 line with 3 points.

Diagonals with slope -1: 1 line with 3 points.

Are there any other lines?

For example, lines with slope 2: Let me check if there are any lines with slope 2 that pass through three points.

Take points (0,0) and (1,2). The line is y = 2x. Next point would be (2,4), which is outside. So, only two points.

Similarly, points (0,1) and (1,3): (1,3) is outside.

Points (1,0) and (2,2): y = 2x - 2. Next point would be (3,4), outside.

Similarly, slope 1/2: points (0,0) and (2,1). Line y = (1/2)x. Next point would be (4,2), outside.

Points (0,1) and (2,2): y = (1/2)x + 1. Next point would be (4,3), outside.

Slope -2: points (0,2) and (1,0). Line y = -2x + 2. Next point would be (2, -2), outside.

Slope -1/2: points (0,2) and (2,1). Line y = (-1/2)x + 2. Next point would be (4, 0), outside.

So, no lines with slopes 2, -2, 1/2, -1/2 have three points.

What about other slopes? For example, slope 3? Points (0,0) and (1,3): (1,3) is outside.

Slope -3? Similarly, points (0,2) and (1,-1): outside.

Slope 1/3? Points (0,0) and (3,1): outside.

So, no.

How about slope 3/2? Points (0,0) and (2,3): (2,3) is outside.

Slope 2/3? Points (0,0) and (3,2): outside.

So, no.

Therefore, there are no other lines with three points.

Therefore, total number of colinear triplets is 8.

Wait, but hold on. Wait, I think I remember that in some problems, there are more lines. Wait, let me check again.

Wait, for example, in the 3x3 grid, there are also lines that go through the center and two other points. Wait, but those are already counted in the diagonals or the horizontal/vertical.

For example, the center point (1,1) is on the two main diagonals.

But if I take the line through (0,1), (1,1), (2,1), that's horizontal.

Similarly, (1,0), (1,1), (1,2) is vertical.

So, all lines through the center are already counted.

Wait, but maybe there are other lines? For example, take points (0,0), (1,1), (2,2) – that's the main diagonal.

How about points (0,0), (1,0.5), (2,1)? Not on grid.

No.

Alternatively, maybe there are lines that pass through three points not aligned with the grid or diagonals?

Wait, for example, take points (0,0), (1,1), (2,2) – already counted.

How about points (0,1), (1,2), (2,3)? (2,3) is outside.

No.

How about points (1,0), (1,1), (1,2)? Vertical, counted.

How about points (0,0), (0,1), (0,2)? Vertical, counted.

So, I think I have to accept that there are only 8 lines with three points. Therefore, 8 colinear triplets.

But wait, hold on. Wait, I think I made a mistake here. Because in the 3x3 grid, there are more lines with three points. Wait, for example, the line that goes through (0,0), (1,1), (2,2) is one. The line through (0,2), (1,1), (2,0) is another. Then, horizontal lines: 3, vertical lines: 3. So, 3 + 3 + 2 = 8.

But I think that's correct. Wait, but let me check with another approach.

Total number of lines in a 3x3 grid. Wait, but maybe I can calculate the number of lines that contain exactly three points.

But maybe I can think of all possible lines.

Alternatively, maybe I can think of all possible triplets and see how many are colinear.

But that would be tedious, but maybe manageable.

Total number of triplets is 84.

Number of colinear triplets: ?

If I can find that, then I can compute the probability.

But if I think that there are 8 lines, each with exactly one triplet, then 8 colinear triplets.

But I have a feeling that maybe there are more. Wait, for example, in the horizontal lines, each line has three points, so there is only one triplet per line. Similarly for vertical and diagonals.

But maybe there are other lines with three points?

Wait, let me think of the line that goes through (0,0), (1,2), and (2,1). Wait, earlier I thought they are not colinear. Let me check again.

Let me calculate the equation of the line passing through (0,0) and (1,2). The slope is 2, so the equation is y = 2x. Now, does (2,1) lie on this line? Plugging x=2, y=2*2=4. But (2,1) has y=1 ≠ 4. So, no.

How about the line through (0,1) and (1,2). Slope is 1. Equation is y - 1 = 1*(x - 0) => y = x + 1. Does (2,1) lie on this line? Plug x=2: y=2 + 1 = 3 ≠ 1. No.

How about the line through (0,0) and (2,1). Slope is 1/2. Equation is y = (1/2)x. Does (1, 0.5) lie on this line? Yes, but (1, 0.5) is not a grid point.

How about the line through (0,1) and (2,0). Slope is (0 - 1)/(2 - 0) = -1/2. Equation is y - 1 = -1/2 x => y = -1/2 x + 1. Does (1, 0.5) lie on this line? Yes, but again, not a grid point.

So, no.

Wait, maybe there are lines with three points that are not aligned with the grid or diagonals?

Wait, for example, take points (0,0), (1,1), (2,2). That's the main diagonal.

How about points (0,0), (1,0), (2,0). Horizontal.

How about points (0,0), (0,1), (0,2). Vertical.

How about points (0,1), (1,1), (2,1). Horizontal.

How about points (1,0), (1,1), (1,2). Vertical.

How about points (0,2), (1,1), (2,0). Diagonal.

How about points (0,0), (1,1), (2,2). Diagonal.

How about points (0,2), (1,2), (2,2). Horizontal.

How about points (0,0), (0,1), (0,2). Vertical.

Wait, so all the lines with three points are already accounted for.

Therefore, total number of colinear triplets is 8.

But wait, I think I remember that in some references, the number is higher. Wait, let me check with another approach.

Let me think of all possible lines.

First, horizontal lines: 3 lines, each with 3 points.

Vertical lines: 3 lines, each with 3 points.

Diagonals with slope 1: 1 line with 3 points.

Diagonals with slope -1: 1 line with 3 points.

But wait, are there other diagonals? For example, in a 3x3 grid, there are also diagonals of length 2. But they don't have three points.

Wait, but maybe there are other lines with three points?

Wait, for example, take the line that goes through (0,0), (1,1), (2,2). That's slope 1.

Another line with slope 1: if you start at (0,1), then the next point would be (1,2), but then the third point would be (2,3), which is outside.

Similarly, starting at (1,0), next point (2,1), then (3,2) outside.

So, only one line with slope 1 that has three points.

Same with slope -1.

So, total lines with three points: 3 + 3 + 1 + 1 = 8.

Therefore, 8 colinear triplets.

But then, if that's the case, the probability would be 8 / 84 = 2 / 21 ≈ 0.095.

But I have a feeling that the answer is different. Wait, let me check with an example.

Wait, let me think of all possible lines.

Wait, for example, in the 3x3 grid, there are also lines that pass through three points if you consider different slopes. Wait, but we saw that there are none.

Alternatively, maybe I made a mistake in counting the number of lines.

Wait, let me think of all possible lines that can be formed with three points.

Each line is uniquely determined by two points. So, maybe I can count the number of lines that have exactly three points.

But how?

Alternatively, think of all possible pairs of points and see which ones are colinear with a third point.

But that might take time, but let me try.

Total number of pairs: C(9, 2) = 36.

For each pair, check if there is a third point on the line.

But this is tedious, but maybe manageable.

Alternatively, think of how many lines have exactly three points.

We know that there are 8 lines with three points.

But maybe there are more lines with exactly two points?

But the question is about lines that have at least three points.

But since we already considered that there are only 8 lines with three points, then the number of colinear triplets is 8.

But let me check with an example.

Take the line y = x. It has three points: (0,0), (1,1), (2,2).

Similarly, y = -x + 2: (0,2), (1,1), (2,0).

Horizontal lines: y = 0, y = 1, y = 2.

Vertical lines: x = 0, x = 1, x = 2.

So, that's 3 + 3 + 2 = 8 lines.

Therefore, 8 lines, each contributing one triplet.

Therefore, 8 colinear triplets.

Therefore, probability is 8 / 84.

Simplify that: divide numerator and denominator by 4: 2 / 21.

But wait, I think I remember that the answer is different. Wait, let me check with another approach.

Wait, maybe I missed some lines. For example, in the 3x3 grid, there are also lines that go through the center and two other points. Wait, but those are already counted in the diagonals or horizontal/vertical.

Wait, for example, the line through (0,1), (1,1), (2,1) is horizontal.

The line through (1,0), (1,1), (1,2) is vertical.

The line through (0,0), (1,1), (2,2) is diagonal.

The line through (0,2), (1,1), (2,0) is diagonal.

So, all lines through the center are already counted.

Wait, but what about lines that pass through the center and two other points not aligned with the grid? For example, (0,1), (1,1), (2,1) is horizontal.

(1,0), (1,1), (1,2) is vertical.

(0,0), (1,1), (2,2) is diagonal.

(0,2), (1,1), (2,0) is diagonal.

So, no other lines.

Therefore, I think my original count of 8 is correct.

But let me check online... Wait, no, I can't. But maybe I can think of another way.

Wait, suppose I think of all possible lines.

In a 3x3 grid, the number of lines that contain exactly three points is 8.

But maybe there are more lines with three points?

Wait, for example, take the line that goes through (0,0), (1,2), and (2,1). Wait, we saw that they are not colinear.

How about (0,0), (2,1), and (something)?

No.

Alternatively, think of the line that goes through (0,1), (1,2), and (2,3). But (2,3) is outside.

No.

Alternatively, think of the line that goes through (1,0), (1,1), (1,2). Vertical, already counted.

So, no.

Therefore, I think my count is correct.

Therefore, the probability is 8 / 84.

Simplify that: divide numerator and denominator by 4: 2 / 21.

But wait, let me check with another approach.

Suppose I think of all possible triplets.

Total triplets: 84.

Number of colinear triplets: ?

If I can think of all possible lines with three points, which are 8, each contributing one triplet, then 8.

But maybe there are more?

Wait, for example, in the horizontal lines, each line has three points, so there is only one triplet per line. Similarly for vertical and diagonals.

But maybe there are lines with more than three points? No, because the grid is 3x3, so maximum three points on a line.

Therefore, each line with three points contributes exactly one triplet.

Therefore, 8 triplets.

Therefore, probability is 8 / 84 = 2 / 21.

But I have a doubt because I remember that in some similar problems, the answer is different. Wait, for example, in a 3x3 grid, the number of lines is more. Wait, let me think again.

Wait, here's a different approach.

In a 3x3 grid, how many lines are there?

Well, the number of lines can be calculated as follows:

- Horizontal lines: 3

- Vertical lines: 3

- Diagonals with slope 1: 2 (the main diagonal and the ones that are shorter? Wait, no. Wait, in a 3x3 grid, there are two diagonals with three points: the main ones.

Wait, but actually, there are more diagonals with two points. But they don't contribute to triplets.

But if I think of all possible lines, regardless of how many points they have, but the question is only about lines that have at least three points.

So, only the 8 lines mentioned before.

Therefore, 8 lines with three points.

Therefore, 8 triplets.

Therefore, probability is 8 / 84 = 2 / 21.

But let me check with an example.

Suppose I list all the colinear triplets.

Horizontal lines:

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

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

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

Vertical lines:

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)

Yes, that's 8 triplets.

Therefore, the number of favorable outcomes is 8.

Total number of possible triplets is 84.

Therefore, probability is 8 / 84.

Simplify: divide numerator and denominator by 4: 2 / 21.

So, the probability is 2/21.

But wait, I think I remember that the answer is different. Wait, let me check again.

Wait, maybe I missed some lines. For example, are there lines with three points that are not horizontal, vertical, or diagonal?

Wait, for example, take the points (0,0), (1,2), and (2,1). Wait, we saw they are not colinear.

How about (0,1), (1,0), (2, -1)? No.

How about (0,1), (1,2), (2,3)? No.

How about (1,0), (2,1), (3,2)? No.

How about (1,1), (1,2), (1,3)? No.

So, no.

Alternatively, maybe there are lines with three points that are not aligned with the grid or diagonals.

Wait, for example, take points (0,0), (1,1), (2,2). That's diagonal.

How about points (0,0), (1,0), (2,0). Horizontal.

How about points (0,0), (0,1), (0,2). Vertical.

So, no.

Therefore, I think my count is correct.

Therefore, the probability is 8 / 84 = 2 / 21.

But let me check with another method.

Suppose I calculate the number of lines with three points.

As we said, 8 lines.

Each line has exactly one triplet.

Therefore, 8 triplets.

Alternatively, maybe there are more triplets?

Wait, for example, suppose there is a line with three points, but we have counted it once.

But no, each line is unique.

Therefore, 8.

Alternatively, maybe I can think of the number of lines with three points as follows:

In a 3x3 grid, the number of lines is calculated as follows:

For each possible slope, count the number of lines with that slope that contain at least three points.

Slope 0 (horizontal): 3 lines.

Slope infinity (vertical): 3 lines.

Slope 1: 1 line.

Slope -1: 1 line.

Other slopes: 0 lines.

Therefore, total lines: 3 + 3 + 1 + 1 = 8.

Therefore, 8 lines.

Therefore, 8 triplets.

Therefore, probability is 8 / 84 = 2 / 21.

Therefore, I think that is the answer.

But to be thorough, let me check if there are any other lines.

For example, take the line that goes through (0,0), (1,1), (2,2).

Another line through (0,1), (1,2), but only two points.

Another line through (1,0), (2,1), only two points.

Similarly for others.

Therefore, no.

Therefore, I think my answer is correct.

Therefore, the probability is 2/21.

But let me check with another approach.

Suppose I calculate the total number of colinear triplets.

For horizontal lines: 3 lines, each with C(3,3) = 1 triplet. So, 3 * 1 = 3.

Vertical lines: same, 3 * 1 = 3.

Diagonals: 2 lines, each with 1 triplet. So, 2 * 1 = 2.

Total: 3 + 3 + 2 = 8.

Yes, same as before.

Therefore, 8.

Therefore, probability is 8 / 84 = 2 / 21.

Therefore, the answer is 2/21.

But wait, I just found a resource in my mind that says that in a 3x3 grid, there are 8 lines with three points, so the probability is 8 / 84 = 2 / 21.

But I also recall that sometimes people consider more lines, but in this case, I think it's correct.

Therefore, I think the answer is 2/21.

But to be 100% sure, let me think of all possible triplets.

Total triplets: 84.

Number of colinear triplets: 8.

Therefore, probability is 8/84 = 2/21.

Yes.

Therefore, the probability is 2/21.

**Final Answer**
The probability is \boxed{\dfrac{2}{21}}.

To determine the probability that three randomly chosen points from a 3×3 grid lie on the same straight line, we proceed in two main steps: **counting the total number of possible three-point sets**, and **counting the number of those sets that are colinear**.

---

### **Step 1: Total Number of Three-Point Sets**

The grid consists of 9 points. The number of ways to choose any 3 points from these 9 is given by the combination:

$$
\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
$$

So, there are **84** possible sets of three points.

---

### **Step 2: Counting Colinear Triplets**

We now count how many of these 84 sets lie on the same straight line. A line can contain at most 3 points in a 3×3 grid, and we need to find all such lines that contain exactly 3 points.

#### **Horizontal Lines**
There are 3 rows, each containing 3 points. Each row forms a horizontal line with exactly 3 points. So, we have:

- 3 horizontal lines → 3 colinear triplets

#### **Vertical Lines**
Similarly, there are 3 columns, each with 3 points. Each column forms a vertical line with exactly 3 points. So, we have:

- 3 vertical lines → 3 colinear triplets

#### **Diagonal Lines**
There are two main diagonals that pass through all 3 points:

- Diagonal from (0,0) to (1,1) to (2,2)
- Diagonal from (0,2) to (1,1) to (2,0)

So, we have:

- 2 diagonal lines → 2 colinear triplets

#### **Other Lines**
We now check if there are any other lines that pass through exactly 3 points. Consider lines with slopes like 2, 1/2, -1/2, or -2. For example:

- A line with slope 2 through (0,0) would pass through (1,2), but the next point (2,4) is outside the grid.
- A line with slope 1/2 through (0,0) would pass through (2,1), but the next point (4,2) is outside the grid.

Similarly, no other lines with different slopes pass through three grid points. Therefore, **no additional lines** with three points exist.

---

### **Total Number of Colinear Triplets**

Adding up all the lines that contain exactly 3 points:

$$
3 \text{ (horizontal)} + 3 \text{ (vertical)} + 2 \text{ (diagonal)} = 8 \text{ colinear triplets}
$$

---

### **Final Probability Calculation**

The probability that a randomly chosen set of three points lies on the same straight line is:

$$
\frac{\text{Number of colinear triplets}}{\text{Total number of triplets}} = \frac{8}{84} = \frac{2}{21}
$$

---

### **Final Answer**

$$
\boxed{\dfrac{2}{21}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that three randomly chosen points from a 3x3 grid lie on the same straight line. Let me start by understanding the problem.

First, the grid is 3x3, which means there are 9 points in total. The question says a set of three points is chosen randomly, and each set has the same probability. So, I need to figure out two things: the total number of possible three-point sets, and the number of those sets that are colinear (lie on the same straight line). Then, the probability would be the number of colinear triplets divided by the total number of triplets.

Let me start with the total number of three-point sets. Since there are 9 points, the number of ways to choose 3 points is the combination of 9 taken 3 at a time. The formula for combinations is C(n, k) = n! / (k! * (n - k)!). So, plugging in the numbers:

C(9, 3) = 9! / (3! * 6!) = (9 × 8 × 7) / (3 × 2 × 1) = 84. So, there are 84 possible three-point sets. Got that.

Now, the tricky part is figuring out how many of these sets are colinear. Let me think about how lines can be formed on a 3x3 grid. Lines can be horizontal, vertical, or diagonal. Also, there might be other lines with different slopes that pass through three points. Wait, in a 3x3 grid, are there any other lines besides the obvious ones?

Let me visualize the grid. Let me imagine the grid with coordinates from (0,0) to (2,2). So, the points are (0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2).

First, horizontal lines: Each row is a horizontal line. There are 3 rows. Each row has 3 points, so each row contributes exactly one set of three colinear points. So, 3 horizontal lines.

Similarly, vertical lines: Each column is a vertical line. There are 3 columns, each with 3 points. So, another 3 vertical lines.

Then, the main diagonals: The diagonal from (0,0) to (1,1) to (2,2), and the diagonal from (0,2) to (1,1) to (2,0). So, that's 2 more lines.

But wait, are there any other lines that pass through three points? For example, lines with slopes like 2, 1/2, -1, etc. Let me check.

Let me think about lines with slope 1/2. For example, starting at (0,0), going up 1 and right 2 would be (0,0), (1, 0.5), (2, 1). But (1, 0.5) is not a grid point, so that line doesn't pass through three grid points. Similarly, starting at (0,1), going up 1 and right 2 would be (0,1), (1, 1.5), (2, 2). Again, (1, 1.5) is not a grid point. Similarly, slope 2: starting at (0,0), going up 2 and right 1 would be (0,0), (1,2), then (2,4), which is outside the grid. So, only two points. Similarly, starting at (0,1), going up 2 and right 1 would be (0,1), (1,3), which is outside. So, no.

What about slope -1/2? Starting at (0,2), going down 1 and right 2: (0,2), (1,1), (2,0). Wait, that's the main diagonal we already counted. Similarly, starting at (0,1), going down 1 and right 2: (0,1), (1,0), (2,-1). The last point is outside. So, only two points. Similarly, slope -2: starting at (0,2), going down 2 and right 1: (0,2), (1,0), then (2,-2). Again, only two points. So, no other lines with slopes other than 0, infinity, 1, -1 have three points.

Wait, but what about other diagonals? For example, in a 3x3 grid, are there any other diagonals with three points? Let me check. For example, starting at (0,0), going to (1,1), (2,2) is the main diagonal. Starting at (0,1), going to (1,2)... but that's only two points. Similarly, starting at (1,0), going to (2,1)... again two points. Similarly, starting at (0,2), going to (1,1), (2,0) is the other main diagonal. So, no other diagonals with three points.

Wait, but what about lines that aren't aligned with the grid or main diagonals? For example, maybe a line that goes through (0,1), (1,2), and... but (2,3) is outside. Similarly, (1,0), (2,1)... (3,2) is outside. So, no. Similarly, (0,0), (1,1), (2,2) is already counted. So, seems like only the horizontal, vertical, and two main diagonals have three colinear points.

Wait, but hold on. Let me think again. Are there any other lines? For example, in a 3x3 grid, there are also lines that go through three points with different slopes? Let me check for example, the line passing through (0,0), (1,2), and (2,1). Wait, does that line pass through three points? Let me check. The slope between (0,0) and (1,2) is 2. Then, from (1,2) to (2,1) is slope (1-2)/(2-1) = -1. So, different slopes, so not a straight line. So, that's not a straight line. How about (0,1), (1,0), (2,-1)? Again, the last point is outside.

Alternatively, maybe (0,0), (1,1), (2,2) is the main diagonal. What about (0,0), (1,0), (2,0)? That's horizontal. Similarly, (0,0), (0,1), (0,2) is vertical. So, seems like only those.

Wait, but maybe there are lines that are not aligned with the grid or main diagonals but still pass through three points? Let me think. For example, take points (0,1), (1,2), and (2,1). Wait, is there a straight line through these three? Let me check the slopes. Between (0,1) and (1,2): slope is (2-1)/(1-0) = 1. Between (1,2) and (2,1): slope is (1-2)/(2-1) = -1. Different slopes, so not colinear. How about (0,1), (1,1), (2,1)? That's horizontal, already counted.

How about (0,0), (1,1), (2,2)? Already counted. How about (0,2), (1,1), (2,0)? Also counted.

Wait, maybe there are lines with slope 2 or 1/2? Let me check. For example, starting at (0,0), going up 2 and right 1: (0,0), (1,2). Then, next would be (2,4), which is outside. So, only two points. Similarly, starting at (0,1), going up 2 and right 1: (0,1), (1,3) which is outside. Starting at (1,0), going up 2 and right 1: (1,0), (2,2). Then, (3,4) outside. So, only two points. Similarly, slope 1/2: starting at (0,0), going up 1 and right 2: (0,0), (2,1). Then, next would be (4,2), outside. So, only two points. Similarly, starting at (0,1), going up 1 and right 2: (0,1), (2,2). Then, (4,3) outside. So, again, only two points.

So, seems like there are no other lines with three points. Therefore, total number of colinear triplets is horizontal lines (3) + vertical lines (3) + main diagonals (2) = 8. Wait, but hold on. Wait, is that all?

Wait, but wait. Let me think again. Are there any other lines? For example, in some grids, there are more lines. Wait, for example, in a 3x3 grid, there are also lines that are not aligned with the main directions. Wait, but as we saw, they don't have three points.

Wait, but let me check another way. Let me count all possible lines that contain at least three points.

First, horizontal lines: 3 lines, each with 3 points.

Vertical lines: 3 lines, each with 3 points.

Diagonals with slope 1: 2 lines, each with 3 points.

Diagonals with slope -1: 2 lines, each with 3 points. Wait, no, slope -1 is already counted as the other diagonal. Wait, there are two diagonals with slope 1 and two with slope -1? Wait, no. Wait, in a 3x3 grid, there are two main diagonals: one from top-left to bottom-right (slope 1) and one from top-right to bottom-left (slope -1). Each of these diagonals has exactly three points. So, that's two diagonals.

But wait, are there other diagonals with three points? For example, if you consider diagonals that are shorter? But no, because they have to have three points. For example, a diagonal that starts at (0,1), goes to (1,2), but then there's no third point. Similarly, starting at (1,0), going to (2,1), no third point. Similarly, starting at (0,1), going to (1,0), then (2,-1), which is outside. So, no.

Therefore, total lines with three points: 3 horizontal + 3 vertical + 2 diagonals = 8 lines. Each of these lines has exactly three points, so each contributes exactly one set of three colinear points. Therefore, the number of colinear triplets is 8.

Wait, but hold on. Wait, is that correct? Because if there are 8 lines, each with exactly one triplet, then there are 8 colinear triplets. But I have a feeling that maybe there are more. Wait, let me think again.

Wait, for example, in the horizontal lines: each row is a line with three points. So, there are three rows, each contributing one triplet. Similarly, three columns. Then, two diagonals. So, 3 + 3 + 2 = 8. So, 8 triplets.

But wait, I remember that in some problems, there are more lines. Wait, maybe I need to check again. Let me think of all possible lines that can pass through three points.

Alternatively, maybe I can think of all possible lines and count how many points they have.

But maybe there's another way. Let me think of all possible lines.

First, horizontal lines: y = 0, y = 1, y = 2. Each has three points.

Vertical lines: x = 0, x = 1, x = 2. Each has three points.

Diagonals with slope 1: y = x, y = x - 1, y = x - 2? Wait, let me check.

For slope 1, the lines are:

- Starting at (0,0): y = x. Points (0,0), (1,1), (2,2).

- Starting at (0,1): y = x + 1. Points (0,1), (1,2). But only two points.

- Starting at (0,2): y = x + 2. Points (0,2), (1,3) which is outside. So, only one point.

Similarly, starting from the other side:

- Starting at (1,0): y = x - 1. Points (1,0), (2,1). Only two points.

- Starting at (2,0): y = x - 2. Points (2,0). Only one point.

So, only one diagonal with slope 1 that has three points. Similarly for slope -1:

Lines with slope -1:

- Starting at (0,2): y = -x + 2. Points (0,2), (1,1), (2,0).

- Starting at (0,1): y = -x + 1. Points (0,1), (1,0). Only two points.

- Starting at (0,0): y = -x. Points (0,0), (1,-1) which is outside. So, only one point.

Similarly, starting from the other side:

- Starting at (1,2): y = -x + 3. Points (1,2), (2,1). Only two points.

- Starting at (2,2): y = -x + 4. Points (2,2). Only one point.

So, only one diagonal with slope -1 that has three points. Therefore, total diagonals with three points: 2.

So, total lines with three points: horizontal (3) + vertical (3) + diagonal (2) = 8. Therefore, 8 lines, each contributing one triplet. Therefore, 8 colinear triplets.

But wait, I think I remember that in some cases, there are more lines. Wait, let me check again. For example, is there a line that goes through (0,0), (1,2), and (2,1)? Wait, earlier I thought that wasn't a straight line. Let me check coordinates.

Let me calculate the slope between (0,0) and (1,2): slope is (2 - 0)/(1 - 0) = 2. Then between (1,2) and (2,1): slope is (1 - 2)/(2 - 1) = -1. Different slopes, so not colinear.

How about (0,1), (1,2), (2,3)? But (2,3) is outside the grid.

How about (1,0), (1,1), (1,2)? That's vertical, already counted.

How about (0,0), (1,1), (2,2)? Already counted.

How about (0,2), (1,1), (2,0)? Already counted.

How about (0,0), (2,1)? That's a line with slope 1/2, but only two points.

How about (0,1), (1,0), (2,-1)? Again, only two points inside the grid.

So, seems like there are no other lines with three points. Therefore, total colinear triplets are 8.

Wait, but hold on. Wait, I think I might be missing something here. Let me think again.

Wait, in the 3x3 grid, there are also lines that are not aligned with the grid or the main diagonals but still pass through three points. Wait, for example, take the points (0,0), (1,1), (2,2) – that's the main diagonal. What about (0,1), (1,1), (2,1)? That's horizontal.

Wait, but maybe there are other lines. For example, take points (0,0), (1,2), and (2,1). Wait, we saw that they are not colinear. How about (0,0), (2,1), and (something else)? Let me check.

Wait, if I take (0,0), (1, 0.5), (2,1). But (1, 0.5) is not a grid point.

Alternatively, maybe there are lines with three points that are not aligned with the grid? For example, in a 3x3 grid, is there a line that goes through three points with different slopes?

Wait, let me think of all possible triplets and see if any of them are colinear but not in the 8 lines I mentioned.

For example, take the points (0,0), (1,1), (2,2). That's already counted.

How about (0,0), (1,0), (2,0)? Horizontal, counted.

How about (0,0), (0,1), (0,2)? Vertical, counted.

How about (0,1), (1,1), (2,1)? Horizontal, counted.

How about (1,0), (1,1), (1,2)? Vertical, counted.

How about (0,2), (1,1), (2,0)? Diagonal, counted.

How about (0,0), (1,2), (2,1)? Not colinear.

How about (0,1), (1,2), (2,1)? Not colinear.

How about (0,2), (1,0), (2,1)? Let me check if they are colinear.

Slope between (0,2) and (1,0): (0 - 2)/(1 - 0) = -2.

Slope between (1,0) and (2,1): (1 - 0)/(2 - 1) = 1. Different slopes. Not colinear.

How about (0,0), (2,1), (something)? Let me check if there's a third point.

The line from (0,0) to (2,1) has slope 1/2. The equation is y = (1/2)x. The other points on this line would be (0,0), (2,1), and (4,2), which is outside. So, only two points.

How about (0,1), (1,2), (2,3)? Again, (2,3) is outside.

How about (1,0), (2,1), (3,2)? Outside.

How about (0,0), (1,1), (2,2)? Already counted.

How about (0,2), (1,1), (2,0)? Already counted.

How about (0,0), (0,1), (0,2)? Vertical, counted.

How about (1,0), (1,1), (1,2)? Vertical, counted.

How about (2,0), (2,1), (2,2)? Vertical, counted.

How about (0,0), (1,0), (2,0)? Horizontal, counted.

How about (0,1), (1,1), (2,1)? Horizontal, counted.

How about (0,2), (1,2), (2,2)? Horizontal, counted.

So, seems like all possible triplets that are colinear are already counted in the 8 lines. Therefore, there are 8 colinear triplets.

Wait, but hold on. Wait, I think I might be missing some lines. Let me think again.

Wait, in the 3x3 grid, there are also lines that are not axis-aligned or diagonals but still pass through three points. Wait, but as we saw, they don't.

Wait, but let me think of another approach. Let me count all possible lines that can be formed with three points.

Each line is determined by two points. So, maybe I can count all lines that have at least three points.

But since there are only 9 points, maybe it's manageable.

First, horizontal lines: 3 lines, each with 3 points.

Vertical lines: 3 lines, each with 3 points.

Diagonals with slope 1: 1 line with 3 points.

Diagonals with slope -1: 1 line with 3 points.

Are there any other lines?

For example, lines with slope 2: Let me check if there are any lines with slope 2 that pass through three points.

Take points (0,0) and (1,2). The line is y = 2x. Next point would be (2,4), which is outside. So, only two points.

Similarly, points (0,1) and (1,3): (1,3) is outside.

Points (1,0) and (2,2): y = 2x - 2. Next point would be (3,4), outside.

Similarly, slope 1/2: points (0,0) and (2,1). Line y = (1/2)x. Next point would be (4,2), outside.

Points (0,1) and (2,2): y = (1/2)x + 1. Next point would be (4,3), outside.

Slope -2: points (0,2) and (1,0). Line y = -2x + 2. Next point would be (2, -2), outside.

Slope -1/2: points (0,2) and (2,1). Line y = (-1/2)x + 2. Next point would be (4, 0), outside.

So, no lines with slopes 2, -2, 1/2, -1/2 have three points.

What about other slopes? For example, slope 3? Points (0,0) and (1,3): (1,3) is outside.

Slope -3? Similarly, points (0,2) and (1,-1): outside.

Slope 1/3? Points (0,0) and (3,1): outside.

So, no.

How about slope 3/2? Points (0,0) and (2,3): (2,3) is outside.

Slope 2/3? Points (0,0) and (3,2): outside.

So, no.

Therefore, there are no other lines with three points.

Therefore, total number of colinear triplets is 8.

Wait, but hold on. Wait, I think I remember that in some problems, there are more lines. Wait, let me check again.

Wait, for example, in the 3x3 grid, there are also lines that go through the center and two other points. Wait, but those are already counted in the diagonals or the horizontal/vertical.

For example, the center point (1,1) is on the two main diagonals.

But if I take the line through (0,1), (1,1), (2,1), that's horizontal.

Similarly, (1,0), (1,1), (1,2) is vertical.

So, all lines through the center are already counted.

Wait, but maybe there are other lines? For example, take points (0,0), (1,1), (2,2) – that's the main diagonal.

How about points (0,0), (1,0.5), (2,1)? Not on grid.

No.

Alternatively, maybe there are lines that pass through three points not aligned with the grid or diagonals?

Wait, for example, take points (0,0), (1,1), (2,2) – already counted.

How about points (0,1), (1,2), (2,3)? (2,3) is outside.

No.

How about points (1,0), (1,1), (1,2)? Vertical, counted.

How about points (0,0), (0,1), (0,2)? Vertical, counted.

So, I think I have to accept that there are only 8 lines with three points. Therefore, 8 colinear triplets.

But wait, hold on. Wait, I think I made a mistake here. Because in the 3x3 grid, there are more lines with three points. Wait, for example, the line that goes through (0,0), (1,1), (2,2) is one. The line through (0,2), (1,1), (2,0) is another. Then, horizontal lines: 3, vertical lines: 3. So, 3 + 3 + 2 = 8.

But I think that's correct. Wait, but let me check with another approach.

Total number of lines in a 3x3 grid. Wait, but maybe I can calculate the number of lines that contain exactly three points.

But maybe I can think of all possible lines.

Alternatively, maybe I can think of all possible triplets and see how many are colinear.

But that would be tedious, but maybe manageable.

Total number of triplets is 84.

Number of colinear triplets: ?

If I can find that, then I can compute the probability.

But if I think that there are 8 lines, each with exactly one triplet, then 8 colinear triplets.

But I have a feeling that maybe there are more. Wait, for example, in the horizontal lines, each line has three points, so there is only one triplet per line. Similarly for vertical and diagonals.

But maybe there are other lines with three points?

Wait, let me think of the line that goes through (0,0), (1,2), and (2,1). Wait, earlier I thought they are not colinear. Let me check again.

Let me calculate the equation of the line passing through (0,0) and (1,2). The slope is 2, so the equation is y = 2x. Now, does (2,1) lie on this line? Plugging x=2, y=2*2=4. But (2,1) has y=1 ≠ 4. So, no.

How about the line through (0,1) and (1,2). Slope is 1. Equation is y - 1 = 1*(x - 0) => y = x + 1. Does (2,1) lie on this line? Plug x=2: y=2 + 1 = 3 ≠ 1. No.

How about the line through (0,0) and (2,1). Slope is 1/2. Equation is y = (1/2)x. Does (1, 0.5) lie on this line? Yes, but (1, 0.5) is not a grid point.

How about the line through (0,1) and (2,0). Slope is (0 - 1)/(2 - 0) = -1/2. Equation is y - 1 = -1/2 x => y = -1/2 x + 1. Does (1, 0.5) lie on this line? Yes, but again, not a grid point.

So, no.

Wait, maybe there are lines with three points that are not aligned with the grid or diagonals?

Wait, for example, take points (0,0), (1,1), (2,2). That's the main diagonal.

How about points (0,0), (1,0), (2,0). Horizontal.

How about points (0,0), (0,1), (0,2). Vertical.

How about points (0,1), (1,1), (2,1). Horizontal.

How about points (1,0), (1,1), (1,2). Vertical.

How about points (0,2), (1,1), (2,0). Diagonal.

How about points (0,0), (1,1), (2,2). Diagonal.

How about points (0,2), (1,2), (2,2). Horizontal.

How about points (0,0), (0,1), (0,2). Vertical.

Wait, so all the lines with three points are already accounted for.

Therefore, total number of colinear triplets is 8.

But wait, I think I remember that in some references, the number is higher. Wait, let me check with another approach.

Let me think of all possible lines.

First, horizontal lines: 3 lines, each with 3 points.

Vertical lines: 3 lines, each with 3 points.

Diagonals with slope 1: 1 line with 3 points.

Diagonals with slope -1: 1 line with 3 points.

But wait, are there other diagonals? For example, in a 3x3 grid, there are also diagonals of length 2. But they don't have three points.

Wait, but maybe there are other lines with three points?

Wait, for example, take the line that goes through (0,0), (1,1), (2,2). That's slope 1.

Another line with slope 1: if you start at (0,1), then the next point would be (1,2), but then the third point would be (2,3), which is outside.

Similarly, starting at (1,0), next point (2,1), then (3,2) outside.

So, only one line with slope 1 that has three points.

Same with slope -1.

So, total lines with three points: 3 + 3 + 1 + 1 = 8.

Therefore, 8 colinear triplets.

But then, if that's the case, the probability would be 8 / 84 = 2 / 21 ≈ 0.095.

But I have a feeling that the answer is different. Wait, let me check with an example.

Wait, let me think of all possible lines.

Wait, for example, in the 3x3 grid, there are also lines that pass through three points if you consider different slopes. Wait, but we saw that there are none.

Alternatively, maybe I made a mistake in counting the number of lines.

Wait, let me think of all possible lines that can be formed with three points.

Each line is uniquely determined by two points. So, maybe I can count the number of lines that have exactly three points.

But how?

Alternatively, think of all possible pairs of points and see which ones are colinear with a third point.

But that might take time, but let me try.

Total number of pairs: C(9, 2) = 36.

For each pair, check if there is a third point on the line.

But this is tedious, but maybe manageable.

Alternatively, think of how many lines have exactly three points.

We know that there are 8 lines with three points.

But maybe there are more lines with exactly two points?

But the question is about lines that have at least three points.

But since we already considered that there are only 8 lines with three points, then the number of colinear triplets is 8.

But let me check with an example.

Take the line y = x. It has three points: (0,0), (1,1), (2,2).

Similarly, y = -x + 2: (0,2), (1,1), (2,0).

Horizontal lines: y = 0, y = 1, y = 2.

Vertical lines: x = 0, x = 1, x = 2.

So, that's 3 + 3 + 2 = 8 lines.

Therefore, 8 lines, each contributing one triplet.

Therefore, 8 colinear triplets.

Therefore, probability is 8 / 84.

Simplify that: divide numerator and denominator by 4: 2 / 21.

But wait, I think I remember that the answer is different. Wait, let me check with another approach.

Wait, maybe I missed some lines. For example, in the 3x3 grid, there are also lines that go through the center and two other points. Wait, but those are already counted in the diagonals or horizontal/vertical.

Wait, for example, the line through (0,1), (1,1), (2,1) is horizontal.

The line through (1,0), (1,1), (1,2) is vertical.

The line through (0,0), (1,1), (2,2) is diagonal.

The line through (0,2), (1,1), (2,0) is diagonal.

So, all lines through the center are already counted.

Wait, but what about lines that pass through the center and two other points not aligned with the grid? For example, (0,1), (1,1), (2,1) is horizontal.

(1,0), (1,1), (1,2) is vertical.

(0,0), (1,1), (2,2) is diagonal.

(0,2), (1,1), (2,0) is diagonal.

So, no other lines.

Therefore, I think my original count of 8 is correct.

But let me check online... Wait, no, I can't. But maybe I can think of another way.

Wait, suppose I think of all possible lines.

In a 3x3 grid, the number of lines that contain exactly three points is 8.

But maybe there are more lines with three points?

Wait, for example, take the line that goes through (0,0), (1,2), and (2,1). Wait, we saw that they are not colinear.

How about (0,0), (2,1), and (something)?

No.

Alternatively, think of the line that goes through (0,1), (1,2), and (2,3). But (2,3) is outside.

No.

Alternatively, think of the line that goes through (1,0), (1,1), (1,2). Vertical, already counted.

So, no.

Therefore, I think my count is correct.

Therefore, the probability is 8 / 84.

Simplify that: divide numerator and denominator by 4: 2 / 21.

But wait, let me check with another approach.

Suppose I think of all possible triplets.

Total triplets: 84.

Number of colinear triplets: ?

If I can think of all possible lines with three points, which are 8, each contributing one triplet, then 8.

But maybe there are more?

Wait, for example, in the horizontal lines, each line has three points, so there is only one triplet per line. Similarly for vertical and diagonals.

But maybe there are lines with more than three points? No, because the grid is 3x3, so maximum three points on a line.

Therefore, each line with three points contributes exactly one triplet.

Therefore, 8 triplets.

Therefore, probability is 8 / 84 = 2 / 21.

But I have a doubt because I remember that in some similar problems, the answer is different. Wait, for example, in a 3x3 grid, the number of lines is more. Wait, let me think again.

Wait, here's a different approach.

In a 3x3 grid, how many lines are there?

Well, the number of lines can be calculated as follows:

- Horizontal lines: 3

- Vertical lines: 3

- Diagonals with slope 1: 2 (the main diagonal and the ones that are shorter? Wait, no. Wait, in a 3x3 grid, there are two diagonals with three points: the main ones.

Wait, but actually, there are more diagonals with two points. But they don't contribute to triplets.

But if I think of all possible lines, regardless of how many points they have, but the question is only about lines that have at least three points.

So, only the 8 lines mentioned before.

Therefore, 8 lines with three points.

Therefore, 8 triplets.

Therefore, probability is 8 / 84 = 2 / 21.

But let me check with an example.

Suppose I list all the colinear triplets.

Horizontal lines:

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

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

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

Vertical lines:

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)

Yes, that's 8 triplets.

Therefore, the number of favorable outcomes is 8.

Total number of possible triplets is 84.

Therefore, probability is 8 / 84.

Simplify: divide numerator and denominator by 4: 2 / 21.

So, the probability is 2/21.

But wait, I think I remember that the answer is different. Wait, let me check again.

Wait, maybe I missed some lines. For example, are there lines with three points that are not horizontal, vertical, or diagonal?

Wait, for example, take the points (0,0), (1,2), and (2,1). Wait, we saw they are not colinear.

How about (0,1), (1,0), (2, -1)? No.

How about (0,1), (1,2), (2,3)? No.

How about (1,0), (2,1), (3,2)? No.

How about (1,1), (1,2), (1,3)? No.

So, no.

Alternatively, maybe there are lines with three points that are not aligned with the grid or diagonals.

Wait, for example, take points (0,0), (1,1), (2,2). That's diagonal.

How about points (0,0), (1,0), (2,0). Horizontal.

How about points (0,0), (0,1), (0,2). Vertical.

So, no.

Therefore, I think my count is correct.

Therefore, the probability is 8 / 84 = 2 / 21.

But let me check with another method.

Suppose I calculate the number of lines with three points.

As we said, 8 lines.

Each line has exactly one triplet.

Therefore, 8 triplets.

Alternatively, maybe there are more triplets?

Wait, for example, suppose there is a line with three points, but we have counted it once.

But no, each line is unique.

Therefore, 8.

Alternatively, maybe I can think of the number of lines with three points as follows:

In a 3x3 grid, the number of lines is calculated as follows:

For each possible slope, count the number of lines with that slope that contain at least three points.

Slope 0 (horizontal): 3 lines.

Slope infinity (vertical): 3 lines.

Slope 1: 1 line.

Slope -1: 1 line.

Other slopes: 0 lines.

Therefore, total lines: 3 + 3 + 1 + 1 = 8.

Therefore, 8 lines.

Therefore, 8 triplets.

Therefore, probability is 8 / 84 = 2 / 21.

Therefore, I think that is the answer.

But to be thorough, let me check if there are any other lines.

For example, take the line that goes through (0,0), (1,1), (2,2).

Another line through (0,1), (1,2), but only two points.

Another line through (1,0), (2,1), only two points.

Similarly for others.

Therefore, no.

Therefore, I think my answer is correct.

Therefore, the probability is 2/21.

But let me check with another approach.

Suppose I calculate the total number of colinear triplets.

For horizontal lines: 3 lines, each with C(3,3) = 1 triplet. So, 3 * 1 = 3.

Vertical lines: same, 3 * 1 = 3.

Diagonals: 2 lines, each with 1 triplet. So, 2 * 1 = 2.

Total: 3 + 3 + 2 = 8.

Yes, same as before.

Therefore, 8.

Therefore, probability is 8 / 84 = 2 / 21.

Therefore, the answer is 2/21.

But wait, I just found a resource in my mind that says that in a 3x3 grid, there are 8 lines with three points, so the probability is 8 / 84 = 2 / 21.

But I also recall that sometimes people consider more lines, but in this case, I think it's correct.

Therefore, I think the answer is 2/21.

But to be 100% sure, let me think of all possible triplets.

Total triplets: 84.

Number of colinear triplets: 8.

Therefore, probability is 8/84 = 2/21.

Yes.

Therefore, the probability is 2/21.

**Final Answer**
The probability is \boxed{\dfrac{2}{21}}.

To determine the probability that three randomly chosen points from a 3×3 grid lie on the same straight line, we proceed in two main steps: **counting the total number of possible three-point sets**, and **counting the number of those sets that are colinear**.

---

### **Step 1: Total Number of Three-Point Sets**

The grid consists of 9 points. The number of ways to choose any 3 points from these 9 is given by the combination:

$$
\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
$$

So, there are **84** possible sets of three points.

---

### **Step 2: Counting Colinear Triplets**

We now count how many of these 84 sets lie on the same straight line. A line can contain at most 3 points in a 3×3 grid, and we need to find all such lines that contain exactly 3 points.

#### **Horizontal Lines**
There are 3 rows, each containing 3 points. Each row forms a horizontal line with exactly 3 points. So, we have:

- 3 horizontal lines → 3 colinear triplets

#### **Vertical Lines**
Similarly, there are 3 columns, each with 3 points. Each column forms a vertical line with exactly 3 points. So, we have:

- 3 vertical lines → 3 colinear triplets

#### **Diagonal Lines**
There are two main diagonals that pass through all 3 points:

- Diagonal from (0,0) to (1,1) to (2,2)
- Diagonal from (0,2) to (1,1) to (2,0)

So, we have:

- 2 diagonal lines → 2 colinear triplets

#### **Other Lines**
We now check if there are any other lines that pass through exactly 3 points. Consider lines with slopes like 2, 1/2, -1/2, or -2. For example:

- A line with slope 2 through (0,0) would pass through (1,2), but the next point (2,4) is outside the grid.
- A line with slope 1/2 through (0,0) would pass through (2,1), but the next point (4,2) is outside the grid.

Similarly, no other lines with different slopes pass through three grid points. Therefore, **no additional lines** with three points exist.

---

### **Total Number of Colinear Triplets**

Adding up all the lines that contain exactly 3 points:

$$
3 \text{ (horizontal)} + 3 \text{ (vertical)} + 2 \text{ (diagonal)} = 8 \text{ colinear triplets}
$$

---

### **Final Probability Calculation**

The probability that a randomly chosen set of three points lies on the same straight line is:

$$
\frac{\text{Number of colinear triplets}}{\text{Total number of triplets}} = \frac{8}{84} = \frac{2}{21}
$$

---

### **Final Answer**

$$
\boxed{\dfrac{2}{21}}
$$

Solution:
The number of three-point sets that can be chosen from the nine grid points is \[
\binom{9}{3} = \frac{9!}{3!\cdot 6!} = 84.
\]Eight of these sets consist of three collinear points: 3 sets of points lie on vertical lines, 3 on horizontal lines, and 2 on diagonals. Hence the probability is $8/84 = \boxed{\frac{2}{21}}$.