Okay, so I need to figure out the probability that the checker is placed on a shaded square of the checkerboard. Let me try to break this down step by step.

First, the wheel is spun twice. Each spin gives a number, and these numbers are divided by 4 and 5 respectively to determine the column and row on the checkerboard. Then, the checker is placed at the intersection of that column and row. I need to find the probability that this square is shaded.

Let me start by understanding the components involved here.

1. **The Wheel and Spinning:**
   - The wheel has numbers, but the Asymptote code shows labels 1, 2, 3, 6, 7, 9. Wait, let me check the Asymptote code again. It labels "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), "7" at (0.7,0). So, the wheel has numbers 1, 2, 3, 6, 7, 9. Is that all? Let me count: 1,2,3,6,7,9. That's six numbers. So the wheel has 6 numbers, each equally likely. When spun, the pointer randomly points to one of these numbers.

2. **First Spin: Divided by 4 for Columns:**
   - The first number is divided by 4, and the remainder determines the column. The possible remainders when divided by 4 are 1, 2, or 3. Wait, but dividing 1,2,3,6,7,9 by 4 gives remainders:

     - 1 divided by 4 is 1 (remainder 1)
     - 2 divided by 4 is 0.5, but since we're talking about integer division, 2 divided by 4 is 0 with remainder 2? Wait, actually, when you divide a number by 4, the remainder is the number modulo 4. So:

     - 1 mod 4 = 1
     - 2 mod 4 = 2
     - 3 mod 4 = 3
     - 6 mod 4 = 2 (since 6 = 4*1 + 2)
     - 7 mod 4 = 3 (since 7 = 4*1 + 3)
     - 9 mod 4 = 1 (since 9 = 4*2 + 1)

     So possible remainders are 1, 2, 3. Each of these corresponds to columns. So the columns are labeled 1, 2, 3.

3. **Second Spin: Divided by 5 for Rows:**
   - The second number is divided by 5, and the remainder determines the row. The possible remainders when divided by 5 are 1, 2, 3, 4. Let's check:

     - 1 mod 5 = 1
     - 2 mod 5 = 2
     - 3 mod 5 = 3
     - 6 mod 5 = 1 (since 6 = 5*1 + 1)
     - 7 mod 5 = 2 (since 7 = 5*1 + 2)
     - 9 mod 5 = 4 (since 9 = 5*1 + 4)

     So possible remainders are 1, 2, 3, 4. These correspond to rows labeled 1, 2, 3, 4.

4. **Checkerboard Setup:**
   - The Asymptote code draws a checkerboard with columns labeled 1,2,3 and rows labeled 1,2,3,4. The checker is placed at the intersection of the column and row determined by the two spins. The checkerboard has shaded squares, and I need to find the probability that the checker lands on a shaded square.

5. **Analyzing the Checkerboard:**
   - From the Asymptote code, the checkerboard is drawn with specific shaded squares. Let's try to visualize it. The code draws a grid from (3,-2) to (6,2), divided into columns and rows. It shades certain squares. Let me parse the Asymptote code:

     - The main grid is from (3,-2) to (6,2), so each cell is 1 unit wide and 1 unit tall.
     - The shaded squares are:
       - (3,0) to (3,1) to (4,1) to (4,0): This is the square from (3,0) to (4,1), which is the square in column 1 (since x from 3 to 4) and row 0 (y from 0 to 1). Wait, but rows are labeled 1 to 4. Hmm, maybe the y-coordinate corresponds to row 1 at the top. Let me check.

       Looking at the Asymptote code:

       - The labels for rows are: "1" at (3,-1.5), "2" at (3,-0.5), "3" at (3,0.5), "4" at (3,1.5). So the rows are labeled from 1 (bottom) to 4 (top). So the y-coordinate increases upwards. Therefore, the square from (3,0) to (4,1) would be column 1 (x from 3 to 4) and row 2 (y from 0 to 1). Wait, no. Wait, the rows are labeled 1 at the bottom (y=-1.5) up to 4 at the top (y=1.5). So the rows are arranged vertically, with row 1 at the bottom, row 4 at the top. The y-coordinates for the centers of the rows would be:

         - Row 1: centered at y=-1.5
         - Row 2: centered at y=-0.5
         - Row 3: centered at y=0.5
         - Row 4: centered at y=1.5

       So the shaded squares in the Asymptote code are:

         - (3,0)--(3,1)--(4,1)--(4,0): This square is in the first column (x from 3 to 4) and spans rows from y=0 to y=1. Since row 2 is centered at y=-0.5 (bottom) and row 3 at y=0.5 (middle). So y=0 to y=1 would cover parts of row 2 and row 3. Wait, maybe the grid is such that each cell is a unit square. Let me think.

       The grid is drawn from (3,-2) to (6,2), so the entire grid is 3 units wide (x from 3 to 6) and 4 units tall (y from -2 to 2). Divided into columns and rows:

         - Columns: 3 columns, each 1 unit wide (x from 3-4, 4-5, 5-6)
         - Rows: 5 rows, each 1 unit tall (y from -2 to -1, -1 to 0, 0 to 1, 1 to 2). But the labels are only for rows 1 to 4, which are at y=-1.5 (row 1), -0.5 (row 2), 0.5 (row 3), 1.5 (row 4). So the rows are centered at these y-coordinates.

       The shaded squares are:

         - (3,0) to (3,1) to (4,1) to (4,0): This is the square in the first column (x 3-4) and between y=0 and y=1. Which rows does this cover? Since row 3 is centered at y=0.5 and row 4 at y=1.5. So y=0 to y=1 would overlap with row 3 (from y=0 to y=1 is from the center of row 3 at 0.5, so from 0.5 - 0.5 = 0 to 0.5 + 0.5 = 1). So this square is in row 3, column 1.

         - Similarly, (3,-2) to (3,-1) to (4,-1) to (4,-2): This is column 1 (x 3-4), y from -2 to -1. The row 1 is centered at y=-1.5, so y=-2 to y=-1 would cover row 1.

         - (4,1) to (4,2) to (5,2) to (5,1): Column 2 (x 4-5), y 1 to 2. Row 4 is centered at y=1.5, so y=1 to 2 covers row 4.

         - (4,-1) to (4,0) to (5,0) to (5,-1): Column 2 (x 4-5), y -1 to 0. Row 2 is centered at y=-0.5, so this covers row 2.

         - (5,0) to (5,1) to (6,1) to (6,0): Column 3 (x 5-6), y 0 to 1. This is row 3.

         - (5,-2) to (5,-1) to (6,-1) to (6,-2): Column 3 (x 5-6), y -2 to -1. This is row 1.

       So, the shaded squares are:

         - Column 1, Row 3

         - Column 1, Row 1

         - Column 2, Row 4

         - Column 2, Row 2

         - Column 3, Row 3

         - Column 3, Row 1

       So, in total, there are 6 shaded squares on a 3x4 grid (columns 1-3, rows 1-4). Each square is equally likely? Wait, no. Because the grid is divided into 3 columns and 4 rows, each cell is a square. But the total number of cells is 3*4=12. However, the checkerboard is a 3x4 grid, but the Asymptote code shades 6 squares. So 6 out of 12 squares are shaded? That would be 50%. But let me check again.

       Wait, columns are 3 (1,2,3) and rows are 4 (1,2,3,4). So total cells 12. The checkerboard pattern is such that in a 3x4 grid, how many shaded squares are there? If it's a checkerboard pattern, alternating colors, then in each row, the colors alternate. Since there are 3 columns, the pattern would repeat every two rows. Let's see:

       If row 1 starts with shaded, then row 2 starts with unshaded, row 3 starts with shaded, etc. For 3 columns:

       - Row 1: shaded, unshaded, shaded

       - Row 2: unshaded, shaded, unshaded

       - Row 3: shaded, unshaded, shaded

       - Row 4: unshaded, shaded, unshaded

       So in each pair of rows, the shaded squares are 2 per row. For 4 rows, that's 2*2=4 shaded squares. But according to the Asymptote code, there are 6 shaded squares. Wait, maybe the checkerboard pattern is different.

       Wait, looking back at the Asymptote code, the shaded squares are specifically:

         - (3,0) to (4,1): Column 1, Rows 3 and 1? Wait, no. Wait, the first shaded square is (3,0)--(3,1)--(4,1)--(4,0). So that's a square from (3,0) to (4,1). So it's a square that spans columns 3-4 and rows 0-1. But since columns are labeled 1-3 and rows 1-4, maybe the coordinates are offset. Maybe the Asymptote code is using a coordinate system where (0,0) is the center of the wheel, but the grid starts at (3,-2). This might be complicating things.

       Alternatively, maybe the checkerboard is such that each cell is a square, and the shaded squares are those where the sum of the column and row indices is odd. Let's check:

       If columns are 1,2,3 and rows 1,2,3,4.

       For each cell (column, row):

       - (1,1): sum 2 (even) → unshaded

       - (1,2): sum 3 (odd) → shaded

       - (1,3): sum 4 (even) → unshaded

       - (1,4): sum 5 (odd) → shaded

       - (2,1): sum 3 (odd) → shaded

       - (2,2): sum 4 (even) → unshaded

       - (2,3): sum 5 (odd) → shaded

       - (2,4): sum 6 (even) → unshaded

       - (3,1): sum 4 (even) → unshaded

       - (3,2): sum 5 (odd) → shaded

       - (3,3): sum 6 (even) → unshaded

       - (3,4): sum 7 (odd) → shaded

       So in this case, shaded squares would be:

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

       Which are 6 squares, matching the Asymptote code's shading. So the checkerboard is shaded when column + row is odd. Therefore, the probability is the number of squares where column + row is odd divided by total squares (12). So 6/12=1/2. But wait, the problem states that the checker is placed on a shaded square. But is the probability simply 1/2? Wait, no. Because the columns and rows are determined by the spins, which have different probabilities depending on the numbers on the wheel.

       Wait, hold on. The columns are determined by first spin divided by 4, which gives 1,2,3 each with certain probabilities. The rows are determined by second spin divided by 5, giving 1,2,3,4 each with certain probabilities. But the wheel numbers have different probabilities. So the mapping from wheel numbers to column and row is not uniform.

       Therefore, the probability is not simply 1/2. Instead, I need to calculate the probability that (column + row) is odd, considering the probabilities of each column and row.

       So, to approach this, I need to:

       1. Determine the probability distribution for columns (1,2,3) based on the first spin.

       2. Determine the probability distribution for rows (1,2,3,4) based on the second spin.

       3. For each possible column and row, check if (column + row) is odd.

       4. Multiply the probabilities of each (column, row) pair where column + row is odd and sum them all.

       So, let's break this down step by step.

6. **Calculating Column Probabilities:**
   - The first spin gives numbers from the wheel: 1,2,3,6,7,9. Each number is equally likely, so 6 outcomes, each with probability 1/6.
   - The column is determined by the remainder when divided by 4:
     - 1 mod 4 = 1 → column 1
     - 2 mod 4 = 2 → column 2
     - 3 mod 4 = 3 → column 3
     - 6 mod 4 = 2 → column 2
     - 7 mod 4 = 3 → column 3
     - 9 mod 4 = 1 → column 1
   - So, columns and their counts:
     - Column 1: numbers 1 and 9 → 2 numbers
     - Column 2: numbers 2 and 6 → 2 numbers
     - Column 3: numbers 3 and 7 → 2 numbers
   - Therefore, probability of each column:
     - P(column 1) = 2/6 = 1/3
     - P(column 2) = 2/6 = 1/3
     - P(column 3) = 2/6 = 1/3
   - Wait, that's interesting. Each column has equal probability? Because each column has 2 numbers out of 6, so 1/3 each.

7. **Calculating Row Probabilities:**
   - The second spin gives numbers from the wheel: 1,2,3,6,7,9. Each number is equally likely, so 6 outcomes, each with probability 1/6.
   - The row is determined by the remainder when divided by 5:
     - 1 mod 5 = 1 → row 1
     - 2 mod 5 = 2 → row 2
     - 3 mod 5 = 3 → row 3
     - 6 mod 5 = 1 → row 1
     - 7 mod 5 = 2 → row 2
     - 9 mod 5 = 4 → row 4
   - So, rows and their counts:
     - Row 1: numbers 1 and 6 → 2 numbers
     - Row 2: numbers 2 and 7 → 2 numbers
     - Row 3: number 3 → 1 number
     - Row 4: number 9 → 1 number
   - Therefore, probabilities of each row:
     - P(row 1) = 2/6 = 1/3
     - P(row 2) = 2/6 = 1/3
     - P(row 3) = 1/6
     - P(row 4) = 1/6

8. **Determining Shaded Squares:**
   - As established earlier, a square is shaded if column + row is odd.
   - Therefore, for each column (1,2,3) and row (1,2,3,4), we need to check if their sum is odd.

9. **Listing All Possible Column-Row Pairs and Checking Shading:**
   Let's make a table:

   | Column | Row | Sum | Shaded? |
   |--------|-----|-----|---------|
   | 1      | 1   | 2   | No      |
   | 1      | 2   | 3   | Yes     |
   | 1      | 3   | 4   | No      |
   | 1      | 4   | 5   | Yes     |
   | 2      | 1   | 3   | Yes     |
   | 2      | 2   | 4   | No      |
   | 2      | 3   | 5   | Yes     |
   | 2      | 4   | 6   | No      |
   | 3      | 1   | 4   | No      |
   | 3      | 2   | 5   | Yes     |
   | 3      | 3   | 6   | No      |
   | 3      | 4   | 7   | Yes     |

   So shaded pairs are:

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

   Which are 6 pairs, as before.

10. **Calculating Probability for Each Shaded Pair:**
    For each of these 6 pairs, calculate the joint probability P(column, row) = P(column) * P(row), since the spins are independent.

    Then sum all these probabilities.

11. **Calculating Individual Probabilities:**

    Let's list the shaded column-row pairs:

    1. (1,2)
       - P(column 1) = 1/3
       - P(row 2) = 1/3
       - Joint probability: 1/3 * 1/3 = 1/9

    2. (1,4)
       - P(column 1) = 1/3
       - P(row 4) = 1/6
       - Joint probability: 1/3 * 1/6 = 1/18

    3. (2,1)
       - P(column 2) = 1/3
       - P(row 1) = 1/3
       - Joint probability: 1/3 * 1/3 = 1/9

    4. (2,3)
       - P(column 2) = 1/3
       - P(row 3) = 1/6
       - Joint probability: 1/3 * 1/6 = 1/18

    5. (3,2)
       - P(column 3) = 1/3
       - P(row 2) = 1/3
       - Joint probability: 1/3 * 1/3 = 1/9

    6. (3,4)
       - P(column 3) = 1/3
       - P(row 4) = 1/6
       - Joint probability: 1/3 * 1/6 = 1/18

12. **Summing All Probabilities:**
    Now, sum these joint probabilities:

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

    Let's convert all to 18 denominators:

    1/9 = 2/18

    So:

    2/18 + 1/18 + 2/18 + 1/18 + 2/18 + 1/18

    Adding them up:

    2 + 1 + 2 + 1 + 2 + 1 = 9

    So total probability = 9/18 = 1/2

    Wait, so the total probability is 1/2? But that can't be right because the columns and rows have different probabilities. Let me check the calculations again.

13. **Verification of Calculations:**

    Let's recompute the sum:

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

    Converting each to 18ths:

    1/9 = 2/18

    So:

    2/18 (from 1,2) +
    1/18 (from 1,4) +
    2/18 (from 2,1) +
    1/18 (from 2,3) +
    2/18 (from 3,2) +
    1/18 (from 3,4)

    Adding numerators: 2 + 1 + 2 + 1 + 2 + 1 = 9

    9/18 = 1/2

    So yes, the total is 1/2. But that seems counterintuitive because the row probabilities are not uniform. Let me check the shaded pairs again.

    Wait, in the table, we have 6 shaded pairs, each with their respective probabilities. The calculation seems correct. But why is the probability 1/2?

    Let me think differently. Since the checkerboard is shaded when column + row is odd, and since columns and rows are selected independently, the probability that column + row is odd is equal to the probability that one is even and the other is odd. If the columns and rows are independent, then:

    P(column odd) * P(row even) + P(column even) * P(row odd)

    Let's compute this:

    - P(column odd): columns are 1,2,3. Odd columns are 1,3. So P(column odd) = P(1) + P(3) = 1/3 + 1/3 = 2/3

    - P(column even): P(2) = 1/3

    - P(row even): rows are 1,2,3,4. Even rows are 2,4. P(row even) = P(2) + P(4) = 1/3 + 1/6 = 1/2

    - P(row odd): P(1) + P(3) = 1/3 + 1/6 = 1/2

    Therefore:

    P(odd column and even row) = (2/3) * (1/2) = 1/3

    P(even column and odd row) = (1/3) * (1/2) = 1/6

    Total probability = 1/3 + 1/6 = 1/2

    So that's another way to see it. Therefore, the probability is indeed 1/2. So even though the rows have different probabilities, the symmetry in the problem causes the total probability to be 1/2.

14. **Conclusion:**
    Therefore, despite the different probabilities for each row and column, the probability that the checker lands on a shaded square is 1/2.

    Wait, but let me confirm once more. Because the columns are each 1/3, and the rows are 1/3, 1/3, 1/6, 1/6. The shaded squares are symmetric in the sense that for each column, half of its rows are shaded. But since the column probabilities are uniform, the overall probability remains 1/2.

    Alternatively, since the shading is a checkerboard pattern, the probability is indeed 1/2, regardless of the distribution of columns and rows. Because in a checkerboard pattern, half the squares are shaded. However, this might not always be the case if the grid is not symmetric. But in this case, the grid is 3 columns by 4 rows, and the checkerboard pattern indeed results in exactly half of the squares being shaded.

    Wait, 3x4 grid has 12 squares, 6 shaded. So 6/12=1/2. Therefore, regardless of the probabilities, if the checkerboard is evenly split, the probability is 1/2. But in reality, the probabilities of landing on each square depend on the column and row probabilities, which are not uniform.

    However, in this case, the calculation shows that the probability is 1/2. So even though the column and row probabilities are different, the shading pattern causes the total probability to average out to 1/2.

    Therefore, the final answer is 1/2.

**Final Answer**
The probability that the checker is placed on a shaded square of the checkerboard is \boxed{\dfrac{1}{2}}.
Therefore, the probability that the checker is placed on a shaded square of the checkerboard is \boxed{\frac{1}{2}}.
</think>