Explanation:
Okay, so I need to figure out the probability that when I spin this wheel twice, the checker ends up on a shaded square of the checkerboard. Let me start by understanding the problem step by step.

First, there's a wheel with numbers on it. When spun twice, each spin gives a number. The first number is divided by 4, and the remainder determines the column of the checkerboard. The second number is divided by 5, and the remainder determines the row. Then, the checker is placed at the intersection of that column and row. The question is asking for the probability that this square is shaded.

Let me visualize the Asymptote figure. The wheel has numbers labeled 1, 2, 6, 9, 3, 7. Wait, the Asymptote code labels the numbers on the wheel as follows: "1" at (0.8,0.5), "2" at (0.8,-0.5), "6" at (-0.8,0.5), "9" at (-0.8,-0.5), "3" at (-0.7,0), and "7" at (0.7,0). So the wheel is divided into six sections with numbers 1, 2, 3, 6, 7, 9. Each number is equally likely, so each has a probability of 1/6.

Wait, but when you spin the wheel, each number is equally likely? The problem says "each number on the wheel being equally likely." So there are six numbers, so each has probability 1/6. Got it.

Now, the first number is divided by 4, giving a remainder of 1, 2, 3. Wait, the problem says "determining one of the remainders 1,2,3 marking the columns of the checkerboard." Wait, dividing by 4, the possible remainders are 0, 1, 2, 3. But the problem says the remainders are 1, 2, 3. So does that mean that the remainder 0 is excluded? Or maybe the numbers on the wheel are such that when divided by 4, they give remainders 1, 2, 3? Let me check.

Looking at the numbers on the wheel: 1, 2, 3, 6, 7, 9. Let me compute each number modulo 4:

- 1 mod 4 = 1
- 2 mod 4 = 2
- 3 mod 4 = 3
- 6 mod 4 = 2 (since 6 divided by 4 is 1 with remainder 2)
- 7 mod 4 = 3 (7 divided by 4 is 1 with remainder 3)
- 9 mod 4 = 1 (9 divided by 4 is 2 with remainder 1)

So the possible remainders when dividing the first number by 4 are 1, 2, 3. Wait, but 6 and 7 give remainders 2 and 3, and 1, 2, 3, 9 give 1, 2, 3, 1. So the possible remainders are 1, 2, 3. However, note that some numbers result in the same remainder. For example, 1 and 9 both give remainder 1. Similarly, 2 and 6 both give remainder 2, and 3 and 7 both give remainder 3. Therefore, when we spin the wheel once, the probability of getting a remainder of 1 is 2/6 = 1/3, remainder 2 is 2/6 = 1/3, and remainder 3 is 2/6 = 1/3. Wait, that's interesting. Each remainder 1, 2, 3 has equal probability? Because there are two numbers for each remainder. So even though there are six numbers, each remainder is equally likely. So the first spin gives column numbers 1, 2, 3 each with probability 1/3.

Similarly, the second number is divided by 5, giving remainders 1, 2, 3, 4. Let me check the numbers on the wheel again. The numbers are 1, 2, 3, 6, 7, 9. Let me compute each modulo 5:

- 1 mod 5 = 1
- 2 mod 5 = 2
- 3 mod 5 = 3
- 6 mod 5 = 1
- 7 mod 5 = 2
- 9 mod 5 = 4

So the remainders when divided by 5 are 1, 2, 3, 1, 2, 4. Therefore, the possible remainders are 1, 2, 3, 4. Let's count how many numbers give each remainder:

- Remainder 1: numbers 1 and 6 → 2 numbers
- Remainder 2: numbers 2 and 7 → 2 numbers
- Remainder 3: number 3 → 1 number
- Remainder 4: number 9 → 1 number

Therefore, the probabilities for the second spin (row) are:

- Remainder 1: 2/6 = 1/3
- Remainder 2: 2/6 = 1/3
- Remainder 3: 1/6
- Remainder 4: 1/6

So the rows have different probabilities. So the columns are each 1/3, rows are 1/3, 1/3, 1/6, 1/6.

Now, the checkerboard has columns 1, 2, 3 (from the first spin) and rows 1, 2, 3, 4 (from the second spin). So the checkerboard is 3 columns by 4 rows, making 12 squares. The shaded squares are specified in the Asymptote code. Let me try to figure out which squares are shaded.

Looking at the Asymptote code:

The shaded squares are:

- (3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7); This is a square from column 3, row 0? Wait, but rows are labeled 1, 2, 3, 4. Wait, the Asymptote code draws the checkerboard with labels:

The labels for columns are on the left side: "1", "2", "3" on the left, and for rows at the bottom: "1", "2", "3", "4". Wait, the Asymptote code says:

label("1",(3.5,-2),S);
label("2",(4.5,-2),S);
label("3",(5.5,-2),S);
label("1",(3,-1.5),W);
label("2",(3,-0.5),W);
label("3",(3,0.5),W);
label("4",(3,1.5),W);

So the columns are labeled 1, 2, 3 on the bottom (x-axis), and rows are labeled 1, 2, 3, 4 on the left (y-axis). Wait, but the coordinates in the Asymptote code are a bit confusing. Let me try to parse it.

The checkerboard is drawn from (3,-2) to (6,2). The vertical lines are at x=3, x=4, x=5, and horizontal lines at y=-2, y=-1, y=0, y=1, y=2. Wait, the code draws:

draw((3,-2)--(3,2)--(6,2)--(6,-2)--cycle,linewidth(0.7));
draw((3,-1)--(6,-1),linewidth(0.7));
draw((3,0)--(6,0),linewidth(0.7));
draw((3,1)--(6,1),linewidth(0.7));
draw((4,-2)--(4,2),linewidth(0.7));
draw((5,-2)--(5,2),linewidth(0.7));

So the grid is divided into columns between x=3 to x=4, x=4 to x=5, x=5 to x=6 (so columns 1, 2, 3) and rows between y=-2 to y=-1, y=-1 to y=0, y=0 to y=1, y=1 to y=2 (rows 1, 2, 3, 4). Wait, but the labels for rows are on the left side. The label "1" is at (3,-1.5), which is the middle of the row from y=-2 to y=-1, so that's row 1. Similarly, "2" is at (3,-0.5), middle of y=-1 to y=0, row 2. "3" at (3,0.5), middle of y=0 to y=1, row 3. "4" at (3,1.5), middle of y=1 to y=2, row 4. So rows are numbered from bottom to top as 1, 2, 3, 4. Columns are from left to right as 1, 2, 3. So the square at column 1, row 1 is the square from (3,-2) to (4,-1). Similarly, column 1, row 2 is (3,-1) to (4,0), etc.

Now, the shaded squares are:

fill((3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7)); This is the square from x=3 to x=4, y=0 to y=1. That's column 1, row 3? Wait, y=0 to y=1 is row 3? Wait, row 1 is y=-2 to y=-1, row 2 is y=-1 to y=0, row 3 is y=0 to y=1, row 4 is y=1 to y=2. So the square from y=0 to y=1 is row 3. And x=3 to x=4 is column 1. So that's column 1, row 3.

Similarly, fill((3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle,gray(0.7)); That's x=3 to x=4, y=-2 to y=-1, which is column 1, row 1.

fill((4,1)--(4,2)--(5,2)--(5,1)--cycle,gray(0.7)); That's x=4 to x=5, y=1 to y=2, which is column 2, row 4.

fill((4,-1)--(4,0)--(5,0)--(5,-1)--cycle,gray(0.7)); That's x=4 to x=5, y=-1 to y=0, which is column 2, row 2.

fill((5,0)--(5,1)--(6,1)--(6,0)--cycle,gray(0.7)); That's x=5 to x=6, y=0 to y=1, which is column 3, row 3.

fill((5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle,gray(0.7)); That's x=5 to x=6, y=-2 to y=-1, which is column 3, row 1.

So the shaded squares are:

- Column 1, Row 1
- Column 1, Row 3
- Column 2, Row 2
- Column 2, Row 4
- Column 3, Row 1
- Column 3, Row 3

So total of 6 shaded squares. Let me confirm:

First shaded square: column 1, row 1.

Second: column 1, row 3.

Third: column 2, row 2.

Fourth: column 2, row 4.

Fifth: column 3, row 1.

Sixth: column 3, row 3.

Yes, that's six squares. So the checkerboard has 3 columns and 4 rows, so 12 squares total. 6 shaded, 6 unshaded. Wait, but maybe not? Let me check again.

Wait, the Asymptote code has six fill commands, each for a square. So yes, six shaded squares. So the probability would be the number of shaded squares divided by total number of squares? Wait, no, because the probability isn't uniform over the squares. Because the columns and rows have different probabilities. So even though there are 6 shaded squares, each square doesn't have the same probability of being selected. Therefore, I can't just do 6/12. I need to calculate the probability for each shaded square and sum them up.

So, the approach is:

1. Determine the probability of each column (1, 2, 3) and each row (1, 2, 3, 4).

2. For each shaded square, compute the probability of landing on that square, which is (probability of column) * (probability of row).

3. Sum all those probabilities to get the total probability of landing on a shaded square.

So let me first note the probabilities for columns and rows.

Columns:

- Column 1: remainder 1 when divided by 4. As calculated earlier, probability is 1/3.

- Column 2: remainder 2 when divided by 4. Probability is 1/3.

- Column 3: remainder 3 when divided by 4. Probability is 1/3.

Rows:

- Row 1: remainder 1 when divided by 5. Probability is 2/6 = 1/3.

Wait, earlier I had for the second spin (rows):

Remainder 1: 2/6 = 1/3

Remainder 2: 2/6 = 1/3

Remainder 3: 1/6

Remainder 4: 1/6

So rows:

Row 1: 1/3

Row 2: 1/3

Row 3: 1/6

Row 4: 1/6

Therefore, for each shaded square, we need to multiply the column probability by the row probability.

Let me list the shaded squares again:

1. Column 1, Row 1: (1/3) * (1/3) = 1/9

2. Column 1, Row 3: (1/3) * (1/6) = 1/18

3. Column 2, Row 2: (1/3) * (1/3) = 1/9

4. Column 2, Row 4: (1/3) * (1/6) = 1/18

5. Column 3, Row 1: (1/3) * (1/3) = 1/9

6. Column 3, Row 3: (1/3) * (1/6) = 1/18

Now, let me sum these up:

First, the three squares with column 1, 2, 3 and row 1: each is 1/9, 1/9, 1/9. So that's 3*(1/9) = 1/3.

Then, the three squares with column 1, 2, 3 and row 3: each is 1/18, 1/18, 1/18. So that's 3*(1/18) = 1/6.

So total probability is 1/3 + 1/6 = 1/2.

Wait, that seems straightforward. But let me check again.

Alternatively, adding all six probabilities:

1/9 + 1/18 + 1/9 + 1/18 + 1/9 + 1/18

Let me convert them all to eighteenths:

1/9 = 2/18

So:

2/18 + 1/18 + 2/18 + 1/18 + 2/18 + 1/18 = (2 + 1 + 2 + 1 + 2 + 1)/18 = 9/18 = 1/2.

Yes, so the total probability is 1/2. Therefore, the probability is 1/2.

But let me verify once more to make sure I didn't make a mistake in identifying the shaded squares or their probabilities.

First, columns are each 1/3. Rows: row 1 and 2 are 1/3 each, rows 3 and 4 are 1/6 each.

Shaded squares:

- Column 1, Row 1: 1/3 * 1/3 = 1/9

- Column 1, Row 3: 1/3 * 1/6 = 1/18

- Column 2, Row 2: 1/3 * 1/3 = 1/9

- Column 2, Row 4: 1/3 * 1/6 = 1/18

- Column 3, Row 1: 1/3 * 1/3 = 1/9

- Column 3, Row 3: 1/3 * 1/6 = 1/18

Yes, that's correct. Then adding them up gives 1/9 + 1/18 + 1/9 + 1/18 + 1/9 + 1/18 = 3*(1/9) + 3*(1/18) = 1/3 + 1/6 = 1/2. So 1/2 is the probability.

But let me think again: is there another way to approach this problem?

Alternatively, since the column and row are determined independently, maybe we can model the entire probability space.

Each spin is independent, so the total number of possible outcomes is 6 (for the first spin) * 6 (for the second spin) = 36. However, since the probabilities aren't uniform across the squares, because the remainders have different probabilities. Wait, but maybe we can think in terms of the possible remainders.

But perhaps it's easier to think in terms of the probabilities for columns and rows. Since columns are each 1/3, rows are 1/3, 1/3, 1/6, 1/6. So the total probability is the sum over all shaded squares of P(column) * P(row). Which is exactly what I did.

Alternatively, maybe we can think of the probability as follows:

The probability that the column is 1, 2, or 3, and the row is 1, 2, 3, or 4, with the shaded squares being specific combinations.

But since the columns and rows are independent, the total probability is the sum over shaded squares of P(column) * P(row). Which again gives 1/2.

Alternatively, maybe I can check by enumerating all possibilities. But since there are 6 numbers on the wheel, each spin has 6 possibilities, so total 36 possible outcomes. However, since the numbers on the wheel have different remainders, each outcome (first spin, second spin) has different probabilities. Therefore, enumerating all 36 outcomes would be tedious, but maybe manageable.

Let me try.

First, list all possible first spins (numbers 1, 2, 3, 6, 7, 9) and their remainders mod 4:

Numbers:

1: remainder 1

2: remainder 2

3: remainder 3

6: remainder 2

7: remainder 3

9: remainder 1

Similarly, second spins (same numbers) with remainders mod 5:

Numbers:

1: remainder 1

2: remainder 2

3: remainder 3

6: remainder 1

7: remainder 2

9: remainder 4

So for each first spin, we can note its column (remainder mod 4), and for each second spin, note its row (remainder mod 5). Then, for each combination, check if the square is shaded.

But since each number has a probability of 1/6, each combination (first number, second number) has probability 1/6 * 1/6 = 1/36. However, some combinations will result in the same column and row. For example, first spin 1 and second spin 1 will result in column 1, row 1. Similarly, first spin 9 and second spin 6 will result in column 1 (since 9 mod 4 = 1) and row 1 (since 6 mod 5 = 1). So multiple combinations can lead to the same square.

But since we need the total probability, we can calculate for each square, how many combinations lead to it, multiply by 1/36, and sum over shaded squares.

Alternatively, since we already know the probabilities for columns and rows, maybe this approach is redundant, but let me try for verification.

First, let's find for each square (column, row), how many (first number, second number) pairs result in that square.

For example, square (column 1, row 1):

Column 1 is achieved by first numbers 1, 9 (since they have remainder 1 mod 4). Row 1 is achieved by second numbers 1, 6 (remainder 1 mod 5). Therefore, number of combinations: 2 (first numbers) * 2 (second numbers) = 4. So probability for this square is 4/36 = 1/9.

Similarly, square (column 1, row 3):

Column 1: first numbers 1, 9. Row 3: second numbers 3. So combinations: 2 * 1 = 2. Probability: 2/36 = 1/18.

Square (column 2, row 2):

Column 2: first numbers 2, 6. Row 2: second numbers 2, 7. Combinations: 2 * 2 = 4. Probability: 4/36 = 1/9.

Square (column 2, row 4):

Column 2: first numbers 2, 6. Row 4: second numbers 9. Combinations: 2 * 1 = 2. Probability: 2/36 = 1/18.

Square (column 3, row 1):

Column 3: first numbers 3, 7. Row 1: second numbers 1, 6. Combinations: 2 * 2 = 4. Probability: 4/36 = 1/9.

Square (column 3, row 3):

Column 3: first numbers 3, 7. Row 3: second numbers 3. Combinations: 2 * 1 = 2. Probability: 2/36 = 1/18.

So adding up all shaded squares:

1/9 + 1/18 + 1/9 + 1/18 + 1/9 + 1/18 = same as before, which is 1/2. Therefore, this confirms the previous result.

Therefore, the probability is 1/2.

But let me check if there are any other shaded squares. Wait, the Asymptote code had six shaded squares, which we accounted for. So that seems correct.

Alternatively, maybe I can think of the entire grid and compute the probability.

The grid has columns 1, 2, 3 and rows 1, 2, 3, 4. Let me make a table of probabilities for each square.

Columns:

- Column 1: 1/3

- Column 2: 1/3

- Column 3: 1/3

Rows:

- Row 1: 1/3

- Row 2: 1/3

- Row 3: 1/6

- Row 4: 1/6

So the probability for each square is column probability * row probability.

So:

For column 1:

- Row 1: 1/3 * 1/3 = 1/9

- Row 2: 1/3 * 1/3 = 1/9

- Row 3: 1/3 * 1/6 = 1/18

- Row 4: 1/3 * 1/6 = 1/18

For column 2:

- Row 1: 1/3 * 1/3 = 1/9

- Row 2: 1/3 * 1/3 = 1/9

- Row 3: 1/3 * 1/6 = 1/18

- Row 4: 1/3 * 1/6 = 1/18

For column 3:

- Row 1: 1/3 * 1/3 = 1/9

- Row 2: 1/3 * 1/3 = 1/9

- Row 3: 1/3 * 1/6 = 1/18

- Row 4: 1/3 * 1/6 = 1/18

Now, shaded squares are:

- Column 1, Row 1: 1/9

- Column 1, Row 3: 1/18

- Column 2, Row 2: 1/9

- Column 2, Row 4: 1/18

- Column 3, Row 1: 1/9

- Column 3, Row 3: 1/18

So adding those up:

(1/9 + 1/18 + 1/9 + 1/18 + 1/9 + 1/18) = same as before, 1/2.

Alternatively, if I sum all probabilities for the grid, they should add up to 1. Let me check:

Total probability = sum over all squares of P(column) * P(row).

For column 1:

Rows 1-4: 1/9 + 1/9 + 1/18 + 1/18 = (2/18 + 2/18 + 1/18 + 1/18) = 6/18 = 1/3

Similarly for column 2: same as column 1: 1/3

Column 3: same as column 1: 1/3

Total: 1/3 + 1/3 + 1/3 = 1. Correct.

So that checks out. Therefore, the shaded squares have total probability 1/2.

Therefore, the answer is 1/2. Wait, but let me check once more if I didn't miscount the shaded squares.

Looking back at the Asymptote code, the shaded squares are:

- (3,0)--(3,1)--(4,1)--(4,0)--cycle: column 1, row 3? Wait, earlier I thought that was column 1, row 3. Wait, but in the grid, rows are numbered from bottom to top as 1, 2, 3, 4. So y=-2 to y=-1 is row 1, y=-1 to y=0 is row 2, y=0 to y=1 is row 3, y=1 to y=2 is row 4. So the square from y=0 to y=1 is row 3. So column 1, row 3 is correct. Similarly, the square from y=-2 to y=-1 is row 1. So column 1, row 1. Then column 2, row 2 is y=-1 to y=0? Wait, no. Wait, column 2 is x=4 to x=5. The square (4,1)--(4,2)--(5,2)--(5,1) is column 2, row 4 (y=1 to y=2). Similarly, (4,-1)--(4,0)--(5,0)--(5,-1) is column 2, row 2 (y=-1 to y=0). Similarly, column 3, row 1 is (5,-2)--(5,-1)--(6,-1)--(6,-2), which is column 3, row 1. Column 3, row 3 is (5,0)--(5,1)--(6,1)--(6,0), which is column 3, row 3. So yes, those are the six shaded squares. So that's correct.

Therefore, the probability is indeed 1/2. Therefore, the answer is 1/2, which is 0.5. So boxed answer is \boxed{\dfrac{1}{2}}.

**Final Answer**
\boxed{\dfrac{1}{2}}

To determine the probability that the checker is placed on a shaded square of the checkerboard, we analyze the process step by step.

---

### **Step 1: Understanding the Wheel and Its Outcomes**

The wheel has six numbers: **1, 2, 3, 6, 7, 9**. Each number is equally likely, so the probability of each number being selected is **1/6**.

- **First spin (column):** The number is divided by 4, and the **remainder** determines the column. The possible remainders are **1, 2, 3** (since 0 is not among the possible remainders from the numbers on the wheel).

  - **Remainder 1:** Numbers 1 and 9 → 2 outcomes → probability = **2/6 = 1/3**
  - **Remainder 2:** Numbers 2 and 6 → 2 outcomes → probability = **1/3**
  - **Remainder 3:** Numbers 3 and 7 → 2 outcomes → probability = **1/3**

- **Second spin (row):** The number is divided by 5, and the **remainder** determines the row. The possible remainders are **1, 2, 3, 4**.

  - **Remainder 1:** Numbers 1 and 6 → 2 outcomes → probability = **1/3**
  - **Remainder 2:** Numbers 2 and 7 → 2 outcomes → probability = **1/3**
  - **Remainder 3:** Number 3 → 1 outcome → probability = **1/6**
  - **Remainder 4:** Number 9 → 1 outcome → probability = **1/6**

---

### **Step 2: Mapping to the Checkerboard**

The checkerboard has:

- **Columns:** 1, 2, 3 (each with probability **1/3**)
- **Rows:** 1, 2, 3, 4 (with probabilities **1/3, 1/3, 1/6, 1/6** respectively)

Each square is uniquely identified by a column and a row. The checker is placed on the square where the column and row intersect.

---

### **Step 3: Identifying the Shaded Squares**

From the Asymptote code, the shaded squares are:

- (1,1), (1,3), (2,2), (2,4), (3,1), (3,3)

So, there are **6 shaded squares**.

---

### **Step 4: Calculating the Probability of Each Shaded Square**

Each shaded square has a probability equal to the product of the probability of its column and the probability of its row.

- (1,1): $ \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} $
- (1,3): $ \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} $
- (2,2): $ \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} $
- (2,4): $ \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} $
- (3,1): $ \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} $
- (3,3): $ \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} $

---

### **Step 5: Summing the Probabilities**

Add the probabilities of all shaded squares:

$$
\frac{1}{9} + \frac{1}{18} + \frac{1}{9} + \frac{1}{18} + \frac{1}{9} + \frac{1}{18}
$$

Convert all to a common denominator (18):

$$
\frac{2}{18} + \frac{1}{18} + \frac{2}{18} + \frac{1}{18} + \frac{2}{18} + \frac{1}{18} = \frac{9}{18} = \frac{1}{2}
$$

---

### **Final Answer**

$$
\boxed{\dfrac{1}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the probability that when I spin this wheel twice, the checker ends up on a shaded square of the checkerboard. Let me start by understanding the problem step by step.

First, there's a wheel with numbers on it. When spun twice, each spin gives a number. The first number is divided by 4, and the remainder determines the column of the checkerboard. The second number is divided by 5, and the remainder determines the row. Then, the checker is placed at the intersection of that column and row. The question is asking for the probability that this square is shaded.

Let me visualize the Asymptote figure. The wheel has numbers labeled 1, 2, 6, 9, 3, 7. Wait, the Asymptote code labels the numbers on the wheel as follows: "1" at (0.8,0.5), "2" at (0.8,-0.5), "6" at (-0.8,0.5), "9" at (-0.8,-0.5), "3" at (-0.7,0), and "7" at (0.7,0). So the wheel is divided into six sections with numbers 1, 2, 3, 6, 7, 9. Each number is equally likely, so each has a probability of 1/6.

Wait, but when you spin the wheel, each number is equally likely? The problem says "each number on the wheel being equally likely." So there are six numbers, so each has probability 1/6. Got it.

Now, the first number is divided by 4, giving a remainder of 1, 2, 3. Wait, the problem says "determining one of the remainders 1,2,3 marking the columns of the checkerboard." Wait, dividing by 4, the possible remainders are 0, 1, 2, 3. But the problem says the remainders are 1, 2, 3. So does that mean that the remainder 0 is excluded? Or maybe the numbers on the wheel are such that when divided by 4, they give remainders 1, 2, 3? Let me check.

Looking at the numbers on the wheel: 1, 2, 3, 6, 7, 9. Let me compute each number modulo 4:

- 1 mod 4 = 1
- 2 mod 4 = 2
- 3 mod 4 = 3
- 6 mod 4 = 2 (since 6 divided by 4 is 1 with remainder 2)
- 7 mod 4 = 3 (7 divided by 4 is 1 with remainder 3)
- 9 mod 4 = 1 (9 divided by 4 is 2 with remainder 1)

So the possible remainders when dividing the first number by 4 are 1, 2, 3. Wait, but 6 and 7 give remainders 2 and 3, and 1, 2, 3, 9 give 1, 2, 3, 1. So the possible remainders are 1, 2, 3. However, note that some numbers result in the same remainder. For example, 1 and 9 both give remainder 1. Similarly, 2 and 6 both give remainder 2, and 3 and 7 both give remainder 3. Therefore, when we spin the wheel once, the probability of getting a remainder of 1 is 2/6 = 1/3, remainder 2 is 2/6 = 1/3, and remainder 3 is 2/6 = 1/3. Wait, that's interesting. Each remainder 1, 2, 3 has equal probability? Because there are two numbers for each remainder. So even though there are six numbers, each remainder is equally likely. So the first spin gives column numbers 1, 2, 3 each with probability 1/3.

Similarly, the second number is divided by 5, giving remainders 1, 2, 3, 4. Let me check the numbers on the wheel again. The numbers are 1, 2, 3, 6, 7, 9. Let me compute each modulo 5:

- 1 mod 5 = 1
- 2 mod 5 = 2
- 3 mod 5 = 3
- 6 mod 5 = 1
- 7 mod 5 = 2
- 9 mod 5 = 4

So the remainders when divided by 5 are 1, 2, 3, 1, 2, 4. Therefore, the possible remainders are 1, 2, 3, 4. Let's count how many numbers give each remainder:

- Remainder 1: numbers 1 and 6 → 2 numbers
- Remainder 2: numbers 2 and 7 → 2 numbers
- Remainder 3: number 3 → 1 number
- Remainder 4: number 9 → 1 number

Therefore, the probabilities for the second spin (row) are:

- Remainder 1: 2/6 = 1/3
- Remainder 2: 2/6 = 1/3
- Remainder 3: 1/6
- Remainder 4: 1/6

So the rows have different probabilities. So the columns are each 1/3, rows are 1/3, 1/3, 1/6, 1/6.

Now, the checkerboard has columns 1, 2, 3 (from the first spin) and rows 1, 2, 3, 4 (from the second spin). So the checkerboard is 3 columns by 4 rows, making 12 squares. The shaded squares are specified in the Asymptote code. Let me try to figure out which squares are shaded.

Looking at the Asymptote code:

The shaded squares are:

- (3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7); This is a square from column 3, row 0? Wait, but rows are labeled 1, 2, 3, 4. Wait, the Asymptote code draws the checkerboard with labels:

The labels for columns are on the left side: "1", "2", "3" on the left, and for rows at the bottom: "1", "2", "3", "4". Wait, the Asymptote code says:

label("1",(3.5,-2),S);
label("2",(4.5,-2),S);
label("3",(5.5,-2),S);
label("1",(3,-1.5),W);
label("2",(3,-0.5),W);
label("3",(3,0.5),W);
label("4",(3,1.5),W);

So the columns are labeled 1, 2, 3 on the bottom (x-axis), and rows are labeled 1, 2, 3, 4 on the left (y-axis). Wait, but the coordinates in the Asymptote code are a bit confusing. Let me try to parse it.

The checkerboard is drawn from (3,-2) to (6,2). The vertical lines are at x=3, x=4, x=5, and horizontal lines at y=-2, y=-1, y=0, y=1, y=2. Wait, the code draws:

draw((3,-2)--(3,2)--(6,2)--(6,-2)--cycle,linewidth(0.7));
draw((3,-1)--(6,-1),linewidth(0.7));
draw((3,0)--(6,0),linewidth(0.7));
draw((3,1)--(6,1),linewidth(0.7));
draw((4,-2)--(4,2),linewidth(0.7));
draw((5,-2)--(5,2),linewidth(0.7));

So the grid is divided into columns between x=3 to x=4, x=4 to x=5, x=5 to x=6 (so columns 1, 2, 3) and rows between y=-2 to y=-1, y=-1 to y=0, y=0 to y=1, y=1 to y=2 (rows 1, 2, 3, 4). Wait, but the labels for rows are on the left side. The label "1" is at (3,-1.5), which is the middle of the row from y=-2 to y=-1, so that's row 1. Similarly, "2" is at (3,-0.5), middle of y=-1 to y=0, row 2. "3" at (3,0.5), middle of y=0 to y=1, row 3. "4" at (3,1.5), middle of y=1 to y=2, row 4. So rows are numbered from bottom to top as 1, 2, 3, 4. Columns are from left to right as 1, 2, 3. So the square at column 1, row 1 is the square from (3,-2) to (4,-1). Similarly, column 1, row 2 is (3,-1) to (4,0), etc.

Now, the shaded squares are:

fill((3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7)); This is the square from x=3 to x=4, y=0 to y=1. That's column 1, row 3? Wait, y=0 to y=1 is row 3? Wait, row 1 is y=-2 to y=-1, row 2 is y=-1 to y=0, row 3 is y=0 to y=1, row 4 is y=1 to y=2. So the square from y=0 to y=1 is row 3. And x=3 to x=4 is column 1. So that's column 1, row 3.

Similarly, fill((3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle,gray(0.7)); That's x=3 to x=4, y=-2 to y=-1, which is column 1, row 1.

fill((4,1)--(4,2)--(5,2)--(5,1)--cycle,gray(0.7)); That's x=4 to x=5, y=1 to y=2, which is column 2, row 4.

fill((4,-1)--(4,0)--(5,0)--(5,-1)--cycle,gray(0.7)); That's x=4 to x=5, y=-1 to y=0, which is column 2, row 2.

fill((5,0)--(5,1)--(6,1)--(6,0)--cycle,gray(0.7)); That's x=5 to x=6, y=0 to y=1, which is column 3, row 3.

fill((5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle,gray(0.7)); That's x=5 to x=6, y=-2 to y=-1, which is column 3, row 1.

So the shaded squares are:

- Column 1, Row 1
- Column 1, Row 3
- Column 2, Row 2
- Column 2, Row 4
- Column 3, Row 1
- Column 3, Row 3

So total of 6 shaded squares. Let me confirm:

First shaded square: column 1, row 1.

Second: column 1, row 3.

Third: column 2, row 2.

Fourth: column 2, row 4.

Fifth: column 3, row 1.

Sixth: column 3, row 3.

Yes, that's six squares. So the checkerboard has 3 columns and 4 rows, so 12 squares total. 6 shaded, 6 unshaded. Wait, but maybe not? Let me check again.

Wait, the Asymptote code has six fill commands, each for a square. So yes, six shaded squares. So the probability would be the number of shaded squares divided by total number of squares? Wait, no, because the probability isn't uniform over the squares. Because the columns and rows have different probabilities. So even though there are 6 shaded squares, each square doesn't have the same probability of being selected. Therefore, I can't just do 6/12. I need to calculate the probability for each shaded square and sum them up.

So, the approach is:

1. Determine the probability of each column (1, 2, 3) and each row (1, 2, 3, 4).

2. For each shaded square, compute the probability of landing on that square, which is (probability of column) * (probability of row).

3. Sum all those probabilities to get the total probability of landing on a shaded square.

So let me first note the probabilities for columns and rows.

Columns:

- Column 1: remainder 1 when divided by 4. As calculated earlier, probability is 1/3.

- Column 2: remainder 2 when divided by 4. Probability is 1/3.

- Column 3: remainder 3 when divided by 4. Probability is 1/3.

Rows:

- Row 1: remainder 1 when divided by 5. Probability is 2/6 = 1/3.

Wait, earlier I had for the second spin (rows):

Remainder 1: 2/6 = 1/3

Remainder 2: 2/6 = 1/3

Remainder 3: 1/6

Remainder 4: 1/6

So rows:

Row 1: 1/3

Row 2: 1/3

Row 3: 1/6

Row 4: 1/6

Therefore, for each shaded square, we need to multiply the column probability by the row probability.

Let me list the shaded squares again:

1. Column 1, Row 1: (1/3) * (1/3) = 1/9

2. Column 1, Row 3: (1/3) * (1/6) = 1/18

3. Column 2, Row 2: (1/3) * (1/3) = 1/9

4. Column 2, Row 4: (1/3) * (1/6) = 1/18

5. Column 3, Row 1: (1/3) * (1/3) = 1/9

6. Column 3, Row 3: (1/3) * (1/6) = 1/18

Now, let me sum these up:

First, the three squares with column 1, 2, 3 and row 1: each is 1/9, 1/9, 1/9. So that's 3*(1/9) = 1/3.

Then, the three squares with column 1, 2, 3 and row 3: each is 1/18, 1/18, 1/18. So that's 3*(1/18) = 1/6.

So total probability is 1/3 + 1/6 = 1/2.

Wait, that seems straightforward. But let me check again.

Alternatively, adding all six probabilities:

1/9 + 1/18 + 1/9 + 1/18 + 1/9 + 1/18

Let me convert them all to eighteenths:

1/9 = 2/18

So:

2/18 + 1/18 + 2/18 + 1/18 + 2/18 + 1/18 = (2 + 1 + 2 + 1 + 2 + 1)/18 = 9/18 = 1/2.

Yes, so the total probability is 1/2. Therefore, the probability is 1/2.

But let me verify once more to make sure I didn't make a mistake in identifying the shaded squares or their probabilities.

First, columns are each 1/3. Rows: row 1 and 2 are 1/3 each, rows 3 and 4 are 1/6 each.

Shaded squares:

- Column 1, Row 1: 1/3 * 1/3 = 1/9

- Column 1, Row 3: 1/3 * 1/6 = 1/18

- Column 2, Row 2: 1/3 * 1/3 = 1/9

- Column 2, Row 4: 1/3 * 1/6 = 1/18

- Column 3, Row 1: 1/3 * 1/3 = 1/9

- Column 3, Row 3: 1/3 * 1/6 = 1/18

Yes, that's correct. Then adding them up gives 1/9 + 1/18 + 1/9 + 1/18 + 1/9 + 1/18 = 3*(1/9) + 3*(1/18) = 1/3 + 1/6 = 1/2. So 1/2 is the probability.

But let me think again: is there another way to approach this problem?

Alternatively, since the column and row are determined independently, maybe we can model the entire probability space.

Each spin is independent, so the total number of possible outcomes is 6 (for the first spin) * 6 (for the second spin) = 36. However, since the probabilities aren't uniform across the squares, because the remainders have different probabilities. Wait, but maybe we can think in terms of the possible remainders.

But perhaps it's easier to think in terms of the probabilities for columns and rows. Since columns are each 1/3, rows are 1/3, 1/3, 1/6, 1/6. So the total probability is the sum over all shaded squares of P(column) * P(row). Which is exactly what I did.

Alternatively, maybe we can think of the probability as follows:

The probability that the column is 1, 2, or 3, and the row is 1, 2, 3, or 4, with the shaded squares being specific combinations.

But since the columns and rows are independent, the total probability is the sum over shaded squares of P(column) * P(row). Which again gives 1/2.

Alternatively, maybe I can check by enumerating all possibilities. But since there are 6 numbers on the wheel, each spin has 6 possibilities, so total 36 possible outcomes. However, since the numbers on the wheel have different remainders, each outcome (first spin, second spin) has different probabilities. Therefore, enumerating all 36 outcomes would be tedious, but maybe manageable.

Let me try.

First, list all possible first spins (numbers 1, 2, 3, 6, 7, 9) and their remainders mod 4:

Numbers:

1: remainder 1

2: remainder 2

3: remainder 3

6: remainder 2

7: remainder 3

9: remainder 1

Similarly, second spins (same numbers) with remainders mod 5:

Numbers:

1: remainder 1

2: remainder 2

3: remainder 3

6: remainder 1

7: remainder 2

9: remainder 4

So for each first spin, we can note its column (remainder mod 4), and for each second spin, note its row (remainder mod 5). Then, for each combination, check if the square is shaded.

But since each number has a probability of 1/6, each combination (first number, second number) has probability 1/6 * 1/6 = 1/36. However, some combinations will result in the same column and row. For example, first spin 1 and second spin 1 will result in column 1, row 1. Similarly, first spin 9 and second spin 6 will result in column 1 (since 9 mod 4 = 1) and row 1 (since 6 mod 5 = 1). So multiple combinations can lead to the same square.

But since we need the total probability, we can calculate for each square, how many combinations lead to it, multiply by 1/36, and sum over shaded squares.

Alternatively, since we already know the probabilities for columns and rows, maybe this approach is redundant, but let me try for verification.

First, let's find for each square (column, row), how many (first number, second number) pairs result in that square.

For example, square (column 1, row 1):

Column 1 is achieved by first numbers 1, 9 (since they have remainder 1 mod 4). Row 1 is achieved by second numbers 1, 6 (remainder 1 mod 5). Therefore, number of combinations: 2 (first numbers) * 2 (second numbers) = 4. So probability for this square is 4/36 = 1/9.

Similarly, square (column 1, row 3):

Column 1: first numbers 1, 9. Row 3: second numbers 3. So combinations: 2 * 1 = 2. Probability: 2/36 = 1/18.

Square (column 2, row 2):

Column 2: first numbers 2, 6. Row 2: second numbers 2, 7. Combinations: 2 * 2 = 4. Probability: 4/36 = 1/9.

Square (column 2, row 4):

Column 2: first numbers 2, 6. Row 4: second numbers 9. Combinations: 2 * 1 = 2. Probability: 2/36 = 1/18.

Square (column 3, row 1):

Column 3: first numbers 3, 7. Row 1: second numbers 1, 6. Combinations: 2 * 2 = 4. Probability: 4/36 = 1/9.

Square (column 3, row 3):

Column 3: first numbers 3, 7. Row 3: second numbers 3. Combinations: 2 * 1 = 2. Probability: 2/36 = 1/18.

So adding up all shaded squares:

1/9 + 1/18 + 1/9 + 1/18 + 1/9 + 1/18 = same as before, which is 1/2. Therefore, this confirms the previous result.

Therefore, the probability is 1/2.

But let me check if there are any other shaded squares. Wait, the Asymptote code had six shaded squares, which we accounted for. So that seems correct.

Alternatively, maybe I can think of the entire grid and compute the probability.

The grid has columns 1, 2, 3 and rows 1, 2, 3, 4. Let me make a table of probabilities for each square.

Columns:

- Column 1: 1/3

- Column 2: 1/3

- Column 3: 1/3

Rows:

- Row 1: 1/3

- Row 2: 1/3

- Row 3: 1/6

- Row 4: 1/6

So the probability for each square is column probability * row probability.

So:

For column 1:

- Row 1: 1/3 * 1/3 = 1/9

- Row 2: 1/3 * 1/3 = 1/9

- Row 3: 1/3 * 1/6 = 1/18

- Row 4: 1/3 * 1/6 = 1/18

For column 2:

- Row 1: 1/3 * 1/3 = 1/9

- Row 2: 1/3 * 1/3 = 1/9

- Row 3: 1/3 * 1/6 = 1/18

- Row 4: 1/3 * 1/6 = 1/18

For column 3:

- Row 1: 1/3 * 1/3 = 1/9

- Row 2: 1/3 * 1/3 = 1/9

- Row 3: 1/3 * 1/6 = 1/18

- Row 4: 1/3 * 1/6 = 1/18

Now, shaded squares are:

- Column 1, Row 1: 1/9

- Column 1, Row 3: 1/18

- Column 2, Row 2: 1/9

- Column 2, Row 4: 1/18

- Column 3, Row 1: 1/9

- Column 3, Row 3: 1/18

So adding those up:

(1/9 + 1/18 + 1/9 + 1/18 + 1/9 + 1/18) = same as before, 1/2.

Alternatively, if I sum all probabilities for the grid, they should add up to 1. Let me check:

Total probability = sum over all squares of P(column) * P(row).

For column 1:

Rows 1-4: 1/9 + 1/9 + 1/18 + 1/18 = (2/18 + 2/18 + 1/18 + 1/18) = 6/18 = 1/3

Similarly for column 2: same as column 1: 1/3

Column 3: same as column 1: 1/3

Total: 1/3 + 1/3 + 1/3 = 1. Correct.

So that checks out. Therefore, the shaded squares have total probability 1/2.

Therefore, the answer is 1/2. Wait, but let me check once more if I didn't miscount the shaded squares.

Looking back at the Asymptote code, the shaded squares are:

- (3,0)--(3,1)--(4,1)--(4,0)--cycle: column 1, row 3? Wait, earlier I thought that was column 1, row 3. Wait, but in the grid, rows are numbered from bottom to top as 1, 2, 3, 4. So y=-2 to y=-1 is row 1, y=-1 to y=0 is row 2, y=0 to y=1 is row 3, y=1 to y=2 is row 4. So the square from y=0 to y=1 is row 3. So column 1, row 3 is correct. Similarly, the square from y=-2 to y=-1 is row 1. So column 1, row 1. Then column 2, row 2 is y=-1 to y=0? Wait, no. Wait, column 2 is x=4 to x=5. The square (4,1)--(4,2)--(5,2)--(5,1) is column 2, row 4 (y=1 to y=2). Similarly, (4,-1)--(4,0)--(5,0)--(5,-1) is column 2, row 2 (y=-1 to y=0). Similarly, column 3, row 1 is (5,-2)--(5,-1)--(6,-1)--(6,-2), which is column 3, row 1. Column 3, row 3 is (5,0)--(5,1)--(6,1)--(6,0), which is column 3, row 3. So yes, those are the six shaded squares. So that's correct.

Therefore, the probability is indeed 1/2. Therefore, the answer is 1/2, which is 0.5. So boxed answer is \boxed{\dfrac{1}{2}}.

**Final Answer**
\boxed{\dfrac{1}{2}}

To determine the probability that the checker is placed on a shaded square of the checkerboard, we analyze the process step by step.

---

### **Step 1: Understanding the Wheel and Its Outcomes**

The wheel has six numbers: **1, 2, 3, 6, 7, 9**. Each number is equally likely, so the probability of each number being selected is **1/6**.

- **First spin (column):** The number is divided by 4, and the **remainder** determines the column. The possible remainders are **1, 2, 3** (since 0 is not among the possible remainders from the numbers on the wheel).

  - **Remainder 1:** Numbers 1 and 9 → 2 outcomes → probability = **2/6 = 1/3**
  - **Remainder 2:** Numbers 2 and 6 → 2 outcomes → probability = **1/3**
  - **Remainder 3:** Numbers 3 and 7 → 2 outcomes → probability = **1/3**

- **Second spin (row):** The number is divided by 5, and the **remainder** determines the row. The possible remainders are **1, 2, 3, 4**.

  - **Remainder 1:** Numbers 1 and 6 → 2 outcomes → probability = **1/3**
  - **Remainder 2:** Numbers 2 and 7 → 2 outcomes → probability = **1/3**
  - **Remainder 3:** Number 3 → 1 outcome → probability = **1/6**
  - **Remainder 4:** Number 9 → 1 outcome → probability = **1/6**

---

### **Step 2: Mapping to the Checkerboard**

The checkerboard has:

- **Columns:** 1, 2, 3 (each with probability **1/3**)
- **Rows:** 1, 2, 3, 4 (with probabilities **1/3, 1/3, 1/6, 1/6** respectively)

Each square is uniquely identified by a column and a row. The checker is placed on the square where the column and row intersect.

---

### **Step 3: Identifying the Shaded Squares**

From the Asymptote code, the shaded squares are:

- (1,1), (1,3), (2,2), (2,4), (3,1), (3,3)

So, there are **6 shaded squares**.

---

### **Step 4: Calculating the Probability of Each Shaded Square**

Each shaded square has a probability equal to the product of the probability of its column and the probability of its row.

- (1,1): $ \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} $
- (1,3): $ \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} $
- (2,2): $ \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} $
- (2,4): $ \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} $
- (3,1): $ \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} $
- (3,3): $ \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} $

---

### **Step 5: Summing the Probabilities**

Add the probabilities of all shaded squares:

$$
\frac{1}{9} + \frac{1}{18} + \frac{1}{9} + \frac{1}{18} + \frac{1}{9} + \frac{1}{18}
$$

Convert all to a common denominator (18):

$$
\frac{2}{18} + \frac{1}{18} + \frac{2}{18} + \frac{1}{18} + \frac{2}{18} + \frac{1}{18} = \frac{9}{18} = \frac{1}{2}
$$

---

### **Final Answer**

$$
\boxed{\dfrac{1}{2}}
$$

Solution:
The first remainder is even with probability $2/6=1/3$ and odd with probability 2/3.  The second remainder is even with probability $3/6=1/2$ and odd with probability 1/2.  The parity of the first remainder and the parity of the second remainder are independent, since they're determined by separate spins of the wheel.

The shaded squares are those that indicate that both remainders are odd or both are even. Hence the square is shaded with probability \[
\frac{1}{3}\cdot \frac{1}{2} + \frac{2}{3}\cdot\frac{1}{2} =\boxed{\frac{1}{2}}.
\]