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 as follows: on the wheel, there are 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), and "7" at (0.7,0). So the wheel is divided into six sections, each labeled with these numbers. So the numbers on the wheel are 1, 2, 3, 6, 7, 9. Each of these numbers is equally likely when the wheel is spun. Therefore, each number has a probability of 1/6.

Now, when the wheel is spun twice, the first spin determines the column. The first number is divided by 4, and the remainder is 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 when dividing by 4 are 0, 1, 2, 3. But the problem says the remainders are 1, 2, 3. So does that mean that 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, 6 mod 4 is 2, 7 mod 4 is 3, 9 mod 4 is 1. So the possible remainders are 1, 2, 3. Therefore, the column numbers are 1, 2, 3. So the columns are labeled 1, 2, 3. Similarly, the second number is divided by 5, giving remainders 1, 2, 3, 4. Let me check the numbers on the wheel again. Wait, the second number is from the same wheel? Wait, the problem says "the numbers indicated by the pointer are randomly determined (with each number on the wheel being equally likely)." So both spins are from the same wheel, which has numbers 1, 2, 3, 6, 7, 9. Therefore, the second number is also one of these numbers. So when we divide the second number by 5, what are the possible remainders?

Let me compute each number modulo 5:

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

So the possible remainders when dividing the second number by 5 are 1, 2, 3, 4. Therefore, the rows are labeled 1, 2, 3, 4. Therefore, the checkerboard has 3 columns (1, 2, 3) and 4 rows (1, 2, 3, 4). Therefore, the checkerboard is a 3x4 grid. Let me confirm that with the Asymptote code.

Looking at the Asymptote code, the checkerboard is drawn from (3,-2) to (6,2). The columns are labeled 1, 2, 3 on the bottom (S) and rows are labeled 1, 2, 3, 4 on the left (W). The shaded squares are:

- (3,0) to (4,0) etc. Wait, the Asymptote code fills several squares:

fill((3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7)); This is the square at column 3, row 1? Wait, coordinates might be a bit confusing. Let me think.

The checkerboard is drawn from (3,-2) to (6,2). The columns are along the x-axis, and rows along the y-axis. The labels for columns are on the bottom (S) at (3.5,-2), (4.5,-2), (5.5,-2) labeled 1, 2, 3. The labels for rows are on the left (W) at (3,-1.5), (3,-0.5), (3,0.5), (3,1.5) labeled 1, 2, 3, 4. Therefore, the columns are from x=3 to x=6, divided into three columns: column 1 is from x=3 to x=4, column 2 from x=4 to x=5, column 3 from x=5 to x=6. Similarly, rows are from y=-2 to y=2, divided into four rows: row 1 is from y=-2 to y=-1, row 2 from y=-1 to y=0, row 3 from y=0 to y=1, row 4 from y=1 to y=2. Therefore, each square is a unit square. The shaded squares are:

1. (3,0)--(3,1)--(4,1)--(4,0): This is column 1 (x=3 to 4) and row 3 (y=0 to 1). Wait, row labels are on the left. Let me check:

The row labels are at (3,-1.5) labeled 1, which would correspond to the row from y=-2 to y=-1. Similarly, (3,-0.5) is row 2 (y=-1 to 0), (3,0.5) is row 3 (y=0 to 1), and (3,1.5) is row 4 (y=1 to 2). Therefore, the square from (3,0) to (4,1) is column 1 (x=3-4) and row 3 (y=0-1). Similarly, the other shaded squares:

fill((3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle,gray(0.7)): column 1 (x=3-4), row 1 (y=-2 to -1).

fill((4,1)--(4,2)--(5,2)--(5,1)--cycle,gray(0.7)): column 2 (x=4-5), row 4 (y=1-2).

fill((4,-1)--(4,0)--(5,0)--(5,-1)--cycle,gray(0.7)): column 2 (x=4-5), row 2 (y=-1 to 0).

fill((5,0)--(5,1)--(6,1)--(6,0)--cycle,gray(0.7)): column 3 (x=5-6), row 3 (y=0-1).

fill((5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle,gray(0.7)): column 3 (x=5-6), row 1 (y=-2 to -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

Wait, let me count again:

First shaded square: column 1, row 1? Wait, the first shaded square is (3,0) to (4,1). Wait, coordinates are (x,y). The square from (3,0) to (4,1) is column 1 (x=3-4) and row 3 (y=0-1). Wait, but the label for row 3 is at (3,0.5). So that square is row 3. Similarly, the square (3,-2) to (4,-1) is column 1, row 1. The square (4,1) to (5,2) is column 2, row 4. The square (4,-1) to (5,0) is column 2, row 2. The square (5,0) to (6,1) is column 3, row 3. The square (5,-2) to (6,-1) is column 3, row 1. So total shaded squares are:

(1,1), (1,3), (2,2), (2,4), (3,1), (3,3). So that's 6 shaded squares. Let me check if there are more. The Asymptote code has five fill commands, but maybe I miscounted? Let me check again:

First fill: column 1, row 3

Second fill: column 1, row 1

Third fill: column 2, row 4

Fourth fill: column 2, row 2

Fifth fill: column 3, row 3

Sixth fill: column 3, row 1

Wait, the Asymptote code has five fill commands? Let me check:

Original Asymptote code:

fill((3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7)); // column 1, row 3

fill((3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle,gray(0.7)); // column 1, row 1

fill((4,1)--(4,2)--(5,2)--(5,1)--cycle,gray(0.7)); // column 2, row 4

fill((4,-1)--(4,0)--(5,0)--(5,-1)--cycle,gray(0.7)); // column 2, row 2

fill((5,0)--(5,1)--(6,1)--(6,0)--cycle,gray(0.7)); // column 3, row 3

fill((5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle,gray(0.7)); // column 3, row 1

Wait, actually, there are six fill commands. So six shaded squares. So the shaded squares are:

(1,1), (1,3), (2,2), (2,4), (3,1), (3,3). So that's 6 squares. The checkerboard is 3 columns and 4 rows, so total squares are 3*4=12. Therefore, 6 shaded squares. Therefore, if all squares are equally likely, the probability would be 6/12=1/2. But wait, that can't be right because the probability isn't necessarily uniform. Because the numbers on the wheel aren't equally likely to result in each column or row. Wait, no, actually, the probability of each column and row depends on the numbers on the wheel. So even though there are 12 squares, each square doesn't have the same probability. Therefore, I need to calculate the probability for each square and sum them up.

So first, let me figure out the probability distribution for the column and row.

First, columns are determined by the first spin divided by 4. The first spin can result in numbers 1, 2, 3, 6, 7, 9, each with probability 1/6. As we saw earlier, their remainders mod 4 are:

Number: 1, 2, 3, 6, 7, 9

Remainder mod 4: 1, 2, 3, 2, 3, 1

Therefore, the column numbers:

- Column 1: numbers 1 and 9, which are two numbers. So probability of column 1 is 2/6 = 1/3.

- Column 2: numbers 2 and 6, which are two numbers. Probability 2/6 = 1/3.

- Column 3: numbers 3 and 7, which are two numbers. Probability 2/6 = 1/3.

Wait, so each column has equal probability? 1/3 each? Because there are two numbers for each column. So column 1: 1 and 9; column 2: 2 and 6; column 3: 3 and 7. Therefore, each column has probability 2/6 = 1/3. So columns are equally likely.

Now for the rows. The second spin is also numbers 1, 2, 3, 6, 7, 9. Divided by 5, the remainders are:

Number: 1, 2, 3, 6, 7, 9

Remainder mod 5: 1, 2, 3, 1, 2, 4

Therefore, the row numbers:

- Row 1: numbers 1 and 6. Two numbers. Probability 2/6 = 1/3.

- Row 2: numbers 2 and 7. Two numbers. Probability 2/6 = 1/3.

- Row 3: number 3. One number. Probability 1/6.

- Row 4: number 9. One number. Probability 1/6.

Therefore, rows are not equally likely. Rows 1 and 2 have probability 1/3 each, rows 3 and 4 have probability 1/6 each.

Therefore, the probability of each square is the product of the column probability and the row probability. Since the two spins are independent.

Therefore, to find the total probability of landing on a shaded square, we need to find all shaded squares, compute their individual probabilities, and sum them up.

First, let me list the shaded squares again:

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

So let me compute the probability for each of these squares.

First, square (1,1): column 1, row 1.

Probability column 1: 1/3.

Probability row 1: 1/3.

Therefore, probability for (1,1): 1/3 * 1/3 = 1/9.

Square (1,3): column 1, row 3.

Probability column 1: 1/3.

Probability row 3: 1/6.

Therefore, probability: 1/3 * 1/6 = 1/18.

Square (2,2): column 2, row 2.

Probability column 2: 1/3.

Probability row 2: 1/3.

Probability: 1/3 * 1/3 = 1/9.

Square (2,4): column 2, row 4.

Probability column 2: 1/3.

Probability row 4: 1/6.

Probability: 1/3 * 1/6 = 1/18.

Square (3,1): column 3, row 1.

Probability column 3: 1/3.

Probability row 1: 1/3.

Probability: 1/3 * 1/3 = 1/9.

Square (3,3): column 3, row 3.

Probability column 3: 1/3.

Probability row 3: 1/6.

Probability: 1/3 * 1/6 = 1/18.

Now, let's sum all these probabilities:

For (1,1): 1/9

(1,3): 1/18

(2,2): 1/9

(2,4): 1/18

(3,1): 1/9

(3,3): 1/18

So adding them up:

First, convert all to eighteenths:

1/9 = 2/18

So:

(1,1): 2/18

(1,3): 1/18

(2,2): 2/18

(2,4): 1/18

(3,1): 2/18

(3,3): 1/18

Total: 2 + 1 + 2 + 1 + 2 + 1 = 9/18 = 1/2.

Wait, so the total probability is 9/18 = 1/2? But that seems high. Let me check again.

Wait, 2/18 + 1/18 + 2/18 + 1/18 + 2/18 + 1/18 = (2+1+2+1+2+1)/18 = 9/18 = 1/2. So 1/2. But let me think again. The total number of squares is 12, but the probabilities aren't uniform. However, the shaded squares are 6 out of 12, but their probabilities are not all equal. For example, (1,1) has probability 1/9, while (1,3) has 1/18. So even though there are 6 shaded squares, their total probability is 1/2. Is that correct?

Alternatively, maybe I made a mistake in calculating the probabilities. Let me check again.

First, columns:

Each column (1,2,3) has probability 1/3. Rows:

Row 1: 1/3, row 2: 1/3, row 3: 1/6, row 4: 1/6.

So for each square, column * row.

Shaded squares:

(1,1): 1/3 * 1/3 = 1/9

(1,3): 1/3 * 1/6 = 1/18

(2,2): 1/3 * 1/3 = 1/9

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

(3,1): 1/3 * 1/3 = 1/9

(3,3): 1/3 * 1/6 = 1/18

Adding them up:

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

Convert to eighteenths:

2/18 + 1/18 + 2/18 + 1/18 + 2/18 + 1/18 = 9/18 = 1/2. So that's 50% chance. But let me think again. Maybe there's a different approach.

Alternatively, since the column and row are independent, maybe we can compute the probability by considering the possible combinations.

But let me check if the total probability over all squares is 1. Let me check that.

Total probability should be 1. Let me compute the total probability for all squares.

Total columns: 3, each with 1/3. Total rows: 4, with probabilities 1/3, 1/3, 1/6, 1/6. So total probability is sum over columns and rows:

Sum_{columns} [column_prob * sum_{rows} row_prob] = sum_{columns} column_prob * 1 = 1. Since sum of row probabilities is 1. Therefore, total probability is 1. So that checks out.

Alternatively, if I compute the total probability for all squares:

Each column has 1/3, and for each column, the rows have probabilities 1/3, 1/3, 1/6, 1/6. Therefore, for each column, total probability is 1/3*(1/3 + 1/3 + 1/6 + 1/6) = 1/3*(2/3 + 1/3) = 1/3*(1) = 1/3. Therefore, total over all columns is 3*(1/3) = 1. Correct.

So the calculation for shaded squares seems correct. Therefore, the probability is 1/2. But let me check again with another approach.

Alternative approach: Let's think about all possible outcomes of the two spins. Each spin has 6 possibilities, so total outcomes are 6*6=36. Each outcome has probability 1/36.

But we need to map each outcome to a square, then count how many of those outcomes result in a shaded square.

But since the column is determined by first spin mod 4, and row by second spin mod 5. Let me see.

First, let me list all possible first spins and their column:

First spin numbers: 1, 2, 3, 6, 7, 9.

Columns:

1: numbers 1, 9

2: numbers 2, 6

3: numbers 3, 7

Similarly, second spin numbers: same numbers, rows:

Row 1: numbers 1, 6

Row 2: numbers 2, 7

Row 3: number 3

Row 4: number 9

Therefore, for each first spin (which determines column) and second spin (which determines row), we can find the square.

But since each outcome is (first spin, second spin), with 6*6=36 possibilities. Let me count how many of these 36 outcomes result in a shaded square.

First, let me note that the shaded squares are:

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

So for each shaded square, count the number of (first spin, second spin) pairs that result in that square.

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

Column 1: first spin is 1 or 9.

Row 1: second spin is 1 or 6.

Therefore, number of outcomes: 2 (for column 1) * 2 (for row 1) = 4.

Similarly, square (1,3): column 1, row 3.

Column 1: 1 or 9.

Row 3: second spin is 3.

Therefore, number of outcomes: 2 * 1 = 2.

Square (2,2): column 2, row 2.

Column 2: 2 or 6.

Row 2: 2 or 7.

Number of outcomes: 2 * 2 = 4.

Square (2,4): column 2, row 4.

Column 2: 2 or 6.

Row 4: 9.

Number of outcomes: 2 * 1 = 2.

Square (3,1): column 3, row 1.

Column 3: 3 or 7.

Row 1: 1 or 6.

Number of outcomes: 2 * 2 = 4.

Square (3,3): column 3, row 3.

Column 3: 3 or 7.

Row 3: 3.

Number of outcomes: 2 * 1 = 2.

Total number of favorable outcomes: 4 + 2 + 4 + 2 + 4 + 2 = 18.

Total possible outcomes: 36.

Therefore, probability is 18/36 = 1/2. So same result as before. Therefore, the probability is 1/2. Therefore, the answer is 1/2. Wait, but the problem says "the numbers indicated by the pointer are randomly determined (with each number on the wheel being equally likely)". So each spin is independent, each number has probability 1/6, so total 36 outcomes, each with probability 1/36. Therefore, 18 favorable outcomes, so 18/36=1/2. Therefore, the probability is 1/2. Therefore, the answer is 1/2. But let me check once more.

Alternatively, maybe I made a mistake in mapping the numbers to the rows and columns. Let me check for square (1,1). First spin is 1 or 9 (column 1). Second spin is 1 or 6 (row 1). So the pairs are (1,1), (1,6), (9,1), (9,6). Each of these is a valid outcome. Similarly for others. So 4 outcomes for (1,1). Similarly for others. So total 18. So 18/36=1/2. Therefore, the answer is 1/2. But let me check if the shaded squares are indeed those six. Let me check the Asymptote code again.

The Asymptote code fills:

First square: column 1, row 3 (since (3,0) to (4,1) is column 1, row 3). Second square: column 1, row 1. Third: column 2, row 4. Fourth: column 2, row 2. Fifth: column 3, row 3. Sixth: column 3, row 1. So yes, those six squares. Therefore, the count is correct. Therefore, the probability is 1/2. Therefore, the answer is 1/2. But wait, the problem says "the checkerboard shown". Maybe I misread the shaded squares? Let me check again.

Looking at the Asymptote code:

The first fill is (3,0)--(3,1)--(4,1)--(4,0)--cycle. So that's the square from x=3 to 4 (column 1) and y=0 to 1 (row 3). Then (3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle: column 1, row 1. Then (4,1)--(4,2)--(5,2)--(5,1)--cycle: column 2, row 4. Then (4,-1)--(4,0)--(5,0)--(5,-1)--cycle: column 2, row 2. Then (5,0)--(5,1)--(6,1)--(6,0)--cycle: column 3, row 3. Then (5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle: column 3, row 1. So yes, those six squares. Therefore, the count is correct. Therefore, the answer is 1/2. Therefore, the probability is 1/2. Therefore, the answer is \boxed{\dfrac{1}{2}}.

But wait, let me check once more with another approach. Let me think about the probability of column and row.

Since columns are 1/3 each, rows are 1/3, 1/3, 1/6, 1/6. The shaded squares are:

For column 1: rows 1 and 3.

For column 2: rows 2 and 4.

For column 3: rows 1 and 3.

Therefore, the probability for column 1: 1/3, and for rows 1 and 3: 1/3 + 1/6 = 1/2. Therefore, probability for column 1 and shaded rows: 1/3 * 1/2 = 1/6.

Similarly, column 2: 1/3, and shaded rows 2 and 4: 1/3 + 1/6 = 1/2. Therefore, 1/3 * 1/2 = 1/6.

Column 3: same as column 1: 1/3 * 1/2 = 1/6.

Total probability: 1/6 + 1/6 + 1/6 = 1/2. Same result. Therefore, this confirms it again. Therefore, the answer is 1/2.

But wait, the problem says "the numbers indicated by the pointer are randomly determined (with each number on the wheel being equally likely)". So each spin is independent, and each number has probability 1/6. Therefore, the answer is 1/2. Therefore, I think that is correct.

But let me check once more with actual enumeration. Let me list all 36 outcomes and count the shaded ones. Although that's tedious, but maybe for some squares.

First, list all first spins (column) and second spins (row):

First spins (column):

1: numbers 1, 9

2: numbers 2, 6

3: numbers 3, 7

Second spins (row):

1: numbers 1, 6

2: numbers 2, 7

3: number 3

4: number 9

So for each first spin and second spin, we can note the square.

Let me make a table:

First spin: 1 (column 1)

Second spin: 1 (row 1) → square (1,1) shaded

Second spin: 2 (row 2) → square (1,2) not shaded

Second spin: 3 (row 3) → square (1,3) shaded

Second spin: 6 (row 1) → square (1,1) shaded

Second spin: 7 (row 2) → square (1,2) not shaded

Second spin: 9 (row 4) → square (1,4) not shaded

So for first spin 1, second spins:

Shaded: 1, 3, 6 (three outcomes? Wait, second spin numbers are 1,2,3,6,7,9. Wait, for first spin 1, second spin can be 1,2,3,6,7,9. So:

Second spin 1: row 1 → shaded

Second spin 2: row 2 → not

Second spin 3: row 3 → shaded

Second spin 6: row 1 → shaded

Second spin 7: row 2 → not

Second spin 9: row 4 → not

Therefore, for first spin 1, shaded outcomes are when second spin is 1, 3, 6. That's three outcomes. Wait, but earlier I thought it was 4. Wait, what's happening here?

Wait, earlier I said for square (1,1): first spin 1 or 9, second spin 1 or 6. So for first spin 1, second spin 1 or 6. So two outcomes. Similarly, for first spin 9, second spin 1 or 6. So total 4 outcomes for square (1,1). But square (1,3) is when first spin 1 or 9, second spin 3. So two outcomes. Therefore, total for column 1 and shaded rows: 4 + 2 = 6 outcomes? Wait, no. Wait, square (1,1) is two outcomes (first spin 1, second spin 1; first spin 1, second spin 6; first spin 9, second spin 1; first spin 9, second spin 6). That's four outcomes. Square (1,3) is first spin 1 or 9, second spin 3. That's two outcomes. So total for column 1 and shaded rows: 4 + 2 = 6 outcomes. Similarly for column 2 and shaded rows, and column 3 and shaded rows. Wait, but earlier I said total 18 outcomes. Let me check.

Wait, for column 1:

Shaded rows are 1 and 3. So:

For row 1: second spin 1 or 6. So for column 1, first spin 1 or 9, second spin 1 or 6: 2*2=4 outcomes.

For row 3: second spin 3. So column 1, first spin 1 or 9, second spin 3: 2*1=2 outcomes.

Total for column 1: 4 + 2 = 6.

Similarly, column 2:

Shaded rows are 2 and 4.

Row 2: second spin 2 or 7. So column 2, first spin 2 or 6, second spin 2 or 7: 2*2=4.

Row 4: second spin 9. So column 2, first spin 2 or 6, second spin 9: 2*1=2.

Total for column 2: 4 + 2 = 6.

Column 3:

Shaded rows are 1 and 3.

Row 1: second spin 1 or 6. Column 3, first spin 3 or 7, second spin 1 or 6: 2*2=4.

Row 3: second spin 3. Column 3, first spin 3 or 7, second spin 3: 2*1=2.

Total for column 3: 4 + 2 = 6.

Total shaded outcomes: 6 + 6 + 6 = 18. Therefore, 18/36=1/2. Therefore, correct. So earlier when I was checking for first spin 1, I thought there were three shaded outcomes, but actually, for first spin 1, second spin 1, 3, 6 are shaded. But second spin 1, 3, 6 are three different numbers. However, each second spin has a probability of 1/6. But since first spin is fixed as 1, there are 6 possible second spins. So for first spin 1, there are three shaded outcomes. Similarly, for first spin 9, three shaded outcomes. So total 6. Similarly for column 2 and 3. Therefore, total 18. Therefore, 18/36=1/2. Therefore, the answer is 1/2. Therefore, I think this is correct.

But let me check with specific examples. For example, the square (1,1) is achieved by (1,1), (1,6), (9,1), (9,6). So four outcomes. Similarly, square (1,3) is (1,3), (9,3). Two outcomes. So total for column 1: 4 + 2 = 6. Similarly for others. Therefore, 18 total. Therefore, 1/2. Therefore, the answer is 1/2. Therefore, I think that is correct.

But just to be thorough, let me check if there are any other shaded squares. The Asymptote code has six shaded squares, which we accounted for. Therefore, no other squares. Therefore, the answer is 1/2. Therefore, I think the probability is 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.

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

The wheel has six numbers: **1, 2, 3, 6, 7, 9**, each equally likely. When the wheel is spun twice, the first spin determines the **column** and the second spin determines the **row**.

- **Column Determination**: The first 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).
- **Row Determination**: The second number is divided by 5, and the remainder determines the row. The possible remainders are **1, 2, 3, 4**.

Thus, the checkerboard is a **3x4 grid** with columns labeled 1, 2, 3 and rows labeled 1, 2, 3, 4.

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### **Step 2: Mapping Numbers to Columns and Rows**

We compute the remainders of the numbers on the wheel modulo 4 and 5:

- **Modulo 4 (Columns)**:
  - 1 → 1
  - 2 → 2
  - 3 → 3
  - 6 → 2
  - 7 → 3
  - 9 → 1
  - So, **columns 1, 2, 3** each have **2 numbers** → **probability 1/3** for each column.

- **Modulo 5 (Rows)**:
  - 1 → 1
  - 2 → 2
  - 3 → 3
  - 6 → 1
  - 7 → 2
  - 9 → 4
  - So, **rows 1 and 2** each have **2 numbers** → **probability 1/3** for each.
  - **Rows 3 and 4** each have **1 number** → **probability 1/6** for each.

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

From the Asymptote diagram, the shaded squares are:

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

These are the six shaded squares in the 3x4 grid.

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

Each square's probability is 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**

Convert all to a common denominator (18):

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

So the total probability is:

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

---

### **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 as follows: on the wheel, there are 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), and "7" at (0.7,0). So the wheel is divided into six sections, each labeled with these numbers. So the numbers on the wheel are 1, 2, 3, 6, 7, 9. Each of these numbers is equally likely when the wheel is spun. Therefore, each number has a probability of 1/6.

Now, when the wheel is spun twice, the first spin determines the column. The first number is divided by 4, and the remainder is 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 when dividing by 4 are 0, 1, 2, 3. But the problem says the remainders are 1, 2, 3. So does that mean that 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, 6 mod 4 is 2, 7 mod 4 is 3, 9 mod 4 is 1. So the possible remainders are 1, 2, 3. Therefore, the column numbers are 1, 2, 3. So the columns are labeled 1, 2, 3. Similarly, the second number is divided by 5, giving remainders 1, 2, 3, 4. Let me check the numbers on the wheel again. Wait, the second number is from the same wheel? Wait, the problem says "the numbers indicated by the pointer are randomly determined (with each number on the wheel being equally likely)." So both spins are from the same wheel, which has numbers 1, 2, 3, 6, 7, 9. Therefore, the second number is also one of these numbers. So when we divide the second number by 5, what are the possible remainders?

Let me compute each number modulo 5:

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

So the possible remainders when dividing the second number by 5 are 1, 2, 3, 4. Therefore, the rows are labeled 1, 2, 3, 4. Therefore, the checkerboard has 3 columns (1, 2, 3) and 4 rows (1, 2, 3, 4). Therefore, the checkerboard is a 3x4 grid. Let me confirm that with the Asymptote code.

Looking at the Asymptote code, the checkerboard is drawn from (3,-2) to (6,2). The columns are labeled 1, 2, 3 on the bottom (S) and rows are labeled 1, 2, 3, 4 on the left (W). The shaded squares are:

- (3,0) to (4,0) etc. Wait, the Asymptote code fills several squares:

fill((3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7)); This is the square at column 3, row 1? Wait, coordinates might be a bit confusing. Let me think.

The checkerboard is drawn from (3,-2) to (6,2). The columns are along the x-axis, and rows along the y-axis. The labels for columns are on the bottom (S) at (3.5,-2), (4.5,-2), (5.5,-2) labeled 1, 2, 3. The labels for rows are on the left (W) at (3,-1.5), (3,-0.5), (3,0.5), (3,1.5) labeled 1, 2, 3, 4. Therefore, the columns are from x=3 to x=6, divided into three columns: column 1 is from x=3 to x=4, column 2 from x=4 to x=5, column 3 from x=5 to x=6. Similarly, rows are from y=-2 to y=2, divided into four rows: row 1 is from y=-2 to y=-1, row 2 from y=-1 to y=0, row 3 from y=0 to y=1, row 4 from y=1 to y=2. Therefore, each square is a unit square. The shaded squares are:

1. (3,0)--(3,1)--(4,1)--(4,0): This is column 1 (x=3 to 4) and row 3 (y=0 to 1). Wait, row labels are on the left. Let me check:

The row labels are at (3,-1.5) labeled 1, which would correspond to the row from y=-2 to y=-1. Similarly, (3,-0.5) is row 2 (y=-1 to 0), (3,0.5) is row 3 (y=0 to 1), and (3,1.5) is row 4 (y=1 to 2). Therefore, the square from (3,0) to (4,1) is column 1 (x=3-4) and row 3 (y=0-1). Similarly, the other shaded squares:

fill((3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle,gray(0.7)): column 1 (x=3-4), row 1 (y=-2 to -1).

fill((4,1)--(4,2)--(5,2)--(5,1)--cycle,gray(0.7)): column 2 (x=4-5), row 4 (y=1-2).

fill((4,-1)--(4,0)--(5,0)--(5,-1)--cycle,gray(0.7)): column 2 (x=4-5), row 2 (y=-1 to 0).

fill((5,0)--(5,1)--(6,1)--(6,0)--cycle,gray(0.7)): column 3 (x=5-6), row 3 (y=0-1).

fill((5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle,gray(0.7)): column 3 (x=5-6), row 1 (y=-2 to -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

Wait, let me count again:

First shaded square: column 1, row 1? Wait, the first shaded square is (3,0) to (4,1). Wait, coordinates are (x,y). The square from (3,0) to (4,1) is column 1 (x=3-4) and row 3 (y=0-1). Wait, but the label for row 3 is at (3,0.5). So that square is row 3. Similarly, the square (3,-2) to (4,-1) is column 1, row 1. The square (4,1) to (5,2) is column 2, row 4. The square (4,-1) to (5,0) is column 2, row 2. The square (5,0) to (6,1) is column 3, row 3. The square (5,-2) to (6,-1) is column 3, row 1. So total shaded squares are:

(1,1), (1,3), (2,2), (2,4), (3,1), (3,3). So that's 6 shaded squares. Let me check if there are more. The Asymptote code has five fill commands, but maybe I miscounted? Let me check again:

First fill: column 1, row 3

Second fill: column 1, row 1

Third fill: column 2, row 4

Fourth fill: column 2, row 2

Fifth fill: column 3, row 3

Sixth fill: column 3, row 1

Wait, the Asymptote code has five fill commands? Let me check:

Original Asymptote code:

fill((3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7)); // column 1, row 3

fill((3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle,gray(0.7)); // column 1, row 1

fill((4,1)--(4,2)--(5,2)--(5,1)--cycle,gray(0.7)); // column 2, row 4

fill((4,-1)--(4,0)--(5,0)--(5,-1)--cycle,gray(0.7)); // column 2, row 2

fill((5,0)--(5,1)--(6,1)--(6,0)--cycle,gray(0.7)); // column 3, row 3

fill((5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle,gray(0.7)); // column 3, row 1

Wait, actually, there are six fill commands. So six shaded squares. So the shaded squares are:

(1,1), (1,3), (2,2), (2,4), (3,1), (3,3). So that's 6 squares. The checkerboard is 3 columns and 4 rows, so total squares are 3*4=12. Therefore, 6 shaded squares. Therefore, if all squares are equally likely, the probability would be 6/12=1/2. But wait, that can't be right because the probability isn't necessarily uniform. Because the numbers on the wheel aren't equally likely to result in each column or row. Wait, no, actually, the probability of each column and row depends on the numbers on the wheel. So even though there are 12 squares, each square doesn't have the same probability. Therefore, I need to calculate the probability for each square and sum them up.

So first, let me figure out the probability distribution for the column and row.

First, columns are determined by the first spin divided by 4. The first spin can result in numbers 1, 2, 3, 6, 7, 9, each with probability 1/6. As we saw earlier, their remainders mod 4 are:

Number: 1, 2, 3, 6, 7, 9

Remainder mod 4: 1, 2, 3, 2, 3, 1

Therefore, the column numbers:

- Column 1: numbers 1 and 9, which are two numbers. So probability of column 1 is 2/6 = 1/3.

- Column 2: numbers 2 and 6, which are two numbers. Probability 2/6 = 1/3.

- Column 3: numbers 3 and 7, which are two numbers. Probability 2/6 = 1/3.

Wait, so each column has equal probability? 1/3 each? Because there are two numbers for each column. So column 1: 1 and 9; column 2: 2 and 6; column 3: 3 and 7. Therefore, each column has probability 2/6 = 1/3. So columns are equally likely.

Now for the rows. The second spin is also numbers 1, 2, 3, 6, 7, 9. Divided by 5, the remainders are:

Number: 1, 2, 3, 6, 7, 9

Remainder mod 5: 1, 2, 3, 1, 2, 4

Therefore, the row numbers:

- Row 1: numbers 1 and 6. Two numbers. Probability 2/6 = 1/3.

- Row 2: numbers 2 and 7. Two numbers. Probability 2/6 = 1/3.

- Row 3: number 3. One number. Probability 1/6.

- Row 4: number 9. One number. Probability 1/6.

Therefore, rows are not equally likely. Rows 1 and 2 have probability 1/3 each, rows 3 and 4 have probability 1/6 each.

Therefore, the probability of each square is the product of the column probability and the row probability. Since the two spins are independent.

Therefore, to find the total probability of landing on a shaded square, we need to find all shaded squares, compute their individual probabilities, and sum them up.

First, let me list the shaded squares again:

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

So let me compute the probability for each of these squares.

First, square (1,1): column 1, row 1.

Probability column 1: 1/3.

Probability row 1: 1/3.

Therefore, probability for (1,1): 1/3 * 1/3 = 1/9.

Square (1,3): column 1, row 3.

Probability column 1: 1/3.

Probability row 3: 1/6.

Therefore, probability: 1/3 * 1/6 = 1/18.

Square (2,2): column 2, row 2.

Probability column 2: 1/3.

Probability row 2: 1/3.

Probability: 1/3 * 1/3 = 1/9.

Square (2,4): column 2, row 4.

Probability column 2: 1/3.

Probability row 4: 1/6.

Probability: 1/3 * 1/6 = 1/18.

Square (3,1): column 3, row 1.

Probability column 3: 1/3.

Probability row 1: 1/3.

Probability: 1/3 * 1/3 = 1/9.

Square (3,3): column 3, row 3.

Probability column 3: 1/3.

Probability row 3: 1/6.

Probability: 1/3 * 1/6 = 1/18.

Now, let's sum all these probabilities:

For (1,1): 1/9

(1,3): 1/18

(2,2): 1/9

(2,4): 1/18

(3,1): 1/9

(3,3): 1/18

So adding them up:

First, convert all to eighteenths:

1/9 = 2/18

So:

(1,1): 2/18

(1,3): 1/18

(2,2): 2/18

(2,4): 1/18

(3,1): 2/18

(3,3): 1/18

Total: 2 + 1 + 2 + 1 + 2 + 1 = 9/18 = 1/2.

Wait, so the total probability is 9/18 = 1/2? But that seems high. Let me check again.

Wait, 2/18 + 1/18 + 2/18 + 1/18 + 2/18 + 1/18 = (2+1+2+1+2+1)/18 = 9/18 = 1/2. So 1/2. But let me think again. The total number of squares is 12, but the probabilities aren't uniform. However, the shaded squares are 6 out of 12, but their probabilities are not all equal. For example, (1,1) has probability 1/9, while (1,3) has 1/18. So even though there are 6 shaded squares, their total probability is 1/2. Is that correct?

Alternatively, maybe I made a mistake in calculating the probabilities. Let me check again.

First, columns:

Each column (1,2,3) has probability 1/3. Rows:

Row 1: 1/3, row 2: 1/3, row 3: 1/6, row 4: 1/6.

So for each square, column * row.

Shaded squares:

(1,1): 1/3 * 1/3 = 1/9

(1,3): 1/3 * 1/6 = 1/18

(2,2): 1/3 * 1/3 = 1/9

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

(3,1): 1/3 * 1/3 = 1/9

(3,3): 1/3 * 1/6 = 1/18

Adding them up:

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

Convert to eighteenths:

2/18 + 1/18 + 2/18 + 1/18 + 2/18 + 1/18 = 9/18 = 1/2. So that's 50% chance. But let me think again. Maybe there's a different approach.

Alternatively, since the column and row are independent, maybe we can compute the probability by considering the possible combinations.

But let me check if the total probability over all squares is 1. Let me check that.

Total probability should be 1. Let me compute the total probability for all squares.

Total columns: 3, each with 1/3. Total rows: 4, with probabilities 1/3, 1/3, 1/6, 1/6. So total probability is sum over columns and rows:

Sum_{columns} [column_prob * sum_{rows} row_prob] = sum_{columns} column_prob * 1 = 1. Since sum of row probabilities is 1. Therefore, total probability is 1. So that checks out.

Alternatively, if I compute the total probability for all squares:

Each column has 1/3, and for each column, the rows have probabilities 1/3, 1/3, 1/6, 1/6. Therefore, for each column, total probability is 1/3*(1/3 + 1/3 + 1/6 + 1/6) = 1/3*(2/3 + 1/3) = 1/3*(1) = 1/3. Therefore, total over all columns is 3*(1/3) = 1. Correct.

So the calculation for shaded squares seems correct. Therefore, the probability is 1/2. But let me check again with another approach.

Alternative approach: Let's think about all possible outcomes of the two spins. Each spin has 6 possibilities, so total outcomes are 6*6=36. Each outcome has probability 1/36.

But we need to map each outcome to a square, then count how many of those outcomes result in a shaded square.

But since the column is determined by first spin mod 4, and row by second spin mod 5. Let me see.

First, let me list all possible first spins and their column:

First spin numbers: 1, 2, 3, 6, 7, 9.

Columns:

1: numbers 1, 9

2: numbers 2, 6

3: numbers 3, 7

Similarly, second spin numbers: same numbers, rows:

Row 1: numbers 1, 6

Row 2: numbers 2, 7

Row 3: number 3

Row 4: number 9

Therefore, for each first spin (which determines column) and second spin (which determines row), we can find the square.

But since each outcome is (first spin, second spin), with 6*6=36 possibilities. Let me count how many of these 36 outcomes result in a shaded square.

First, let me note that the shaded squares are:

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

So for each shaded square, count the number of (first spin, second spin) pairs that result in that square.

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

Column 1: first spin is 1 or 9.

Row 1: second spin is 1 or 6.

Therefore, number of outcomes: 2 (for column 1) * 2 (for row 1) = 4.

Similarly, square (1,3): column 1, row 3.

Column 1: 1 or 9.

Row 3: second spin is 3.

Therefore, number of outcomes: 2 * 1 = 2.

Square (2,2): column 2, row 2.

Column 2: 2 or 6.

Row 2: 2 or 7.

Number of outcomes: 2 * 2 = 4.

Square (2,4): column 2, row 4.

Column 2: 2 or 6.

Row 4: 9.

Number of outcomes: 2 * 1 = 2.

Square (3,1): column 3, row 1.

Column 3: 3 or 7.

Row 1: 1 or 6.

Number of outcomes: 2 * 2 = 4.

Square (3,3): column 3, row 3.

Column 3: 3 or 7.

Row 3: 3.

Number of outcomes: 2 * 1 = 2.

Total number of favorable outcomes: 4 + 2 + 4 + 2 + 4 + 2 = 18.

Total possible outcomes: 36.

Therefore, probability is 18/36 = 1/2. So same result as before. Therefore, the probability is 1/2. Therefore, the answer is 1/2. Wait, but the problem says "the numbers indicated by the pointer are randomly determined (with each number on the wheel being equally likely)". So each spin is independent, each number has probability 1/6, so total 36 outcomes, each with probability 1/36. Therefore, 18 favorable outcomes, so 18/36=1/2. Therefore, the probability is 1/2. Therefore, the answer is 1/2. But let me check once more.

Alternatively, maybe I made a mistake in mapping the numbers to the rows and columns. Let me check for square (1,1). First spin is 1 or 9 (column 1). Second spin is 1 or 6 (row 1). So the pairs are (1,1), (1,6), (9,1), (9,6). Each of these is a valid outcome. Similarly for others. So 4 outcomes for (1,1). Similarly for others. So total 18. So 18/36=1/2. Therefore, the answer is 1/2. But let me check if the shaded squares are indeed those six. Let me check the Asymptote code again.

The Asymptote code fills:

First square: column 1, row 3 (since (3,0) to (4,1) is column 1, row 3). Second square: column 1, row 1. Third: column 2, row 4. Fourth: column 2, row 2. Fifth: column 3, row 3. Sixth: column 3, row 1. So yes, those six squares. Therefore, the count is correct. Therefore, the probability is 1/2. Therefore, the answer is 1/2. But wait, the problem says "the checkerboard shown". Maybe I misread the shaded squares? Let me check again.

Looking at the Asymptote code:

The first fill is (3,0)--(3,1)--(4,1)--(4,0)--cycle. So that's the square from x=3 to 4 (column 1) and y=0 to 1 (row 3). Then (3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle: column 1, row 1. Then (4,1)--(4,2)--(5,2)--(5,1)--cycle: column 2, row 4. Then (4,-1)--(4,0)--(5,0)--(5,-1)--cycle: column 2, row 2. Then (5,0)--(5,1)--(6,1)--(6,0)--cycle: column 3, row 3. Then (5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle: column 3, row 1. So yes, those six squares. Therefore, the count is correct. Therefore, the answer is 1/2. Therefore, the probability is 1/2. Therefore, the answer is \boxed{\dfrac{1}{2}}.

But wait, let me check once more with another approach. Let me think about the probability of column and row.

Since columns are 1/3 each, rows are 1/3, 1/3, 1/6, 1/6. The shaded squares are:

For column 1: rows 1 and 3.

For column 2: rows 2 and 4.

For column 3: rows 1 and 3.

Therefore, the probability for column 1: 1/3, and for rows 1 and 3: 1/3 + 1/6 = 1/2. Therefore, probability for column 1 and shaded rows: 1/3 * 1/2 = 1/6.

Similarly, column 2: 1/3, and shaded rows 2 and 4: 1/3 + 1/6 = 1/2. Therefore, 1/3 * 1/2 = 1/6.

Column 3: same as column 1: 1/3 * 1/2 = 1/6.

Total probability: 1/6 + 1/6 + 1/6 = 1/2. Same result. Therefore, this confirms it again. Therefore, the answer is 1/2.

But wait, the problem says "the numbers indicated by the pointer are randomly determined (with each number on the wheel being equally likely)". So each spin is independent, and each number has probability 1/6. Therefore, the answer is 1/2. Therefore, I think that is correct.

But let me check once more with actual enumeration. Let me list all 36 outcomes and count the shaded ones. Although that's tedious, but maybe for some squares.

First, list all first spins (column) and second spins (row):

First spins (column):

1: numbers 1, 9

2: numbers 2, 6

3: numbers 3, 7

Second spins (row):

1: numbers 1, 6

2: numbers 2, 7

3: number 3

4: number 9

So for each first spin and second spin, we can note the square.

Let me make a table:

First spin: 1 (column 1)

Second spin: 1 (row 1) → square (1,1) shaded

Second spin: 2 (row 2) → square (1,2) not shaded

Second spin: 3 (row 3) → square (1,3) shaded

Second spin: 6 (row 1) → square (1,1) shaded

Second spin: 7 (row 2) → square (1,2) not shaded

Second spin: 9 (row 4) → square (1,4) not shaded

So for first spin 1, second spins:

Shaded: 1, 3, 6 (three outcomes? Wait, second spin numbers are 1,2,3,6,7,9. Wait, for first spin 1, second spin can be 1,2,3,6,7,9. So:

Second spin 1: row 1 → shaded

Second spin 2: row 2 → not

Second spin 3: row 3 → shaded

Second spin 6: row 1 → shaded

Second spin 7: row 2 → not

Second spin 9: row 4 → not

Therefore, for first spin 1, shaded outcomes are when second spin is 1, 3, 6. That's three outcomes. Wait, but earlier I thought it was 4. Wait, what's happening here?

Wait, earlier I said for square (1,1): first spin 1 or 9, second spin 1 or 6. So for first spin 1, second spin 1 or 6. So two outcomes. Similarly, for first spin 9, second spin 1 or 6. So total 4 outcomes for square (1,1). But square (1,3) is when first spin 1 or 9, second spin 3. So two outcomes. Therefore, total for column 1 and shaded rows: 4 + 2 = 6 outcomes? Wait, no. Wait, square (1,1) is two outcomes (first spin 1, second spin 1; first spin 1, second spin 6; first spin 9, second spin 1; first spin 9, second spin 6). That's four outcomes. Square (1,3) is first spin 1 or 9, second spin 3. That's two outcomes. So total for column 1 and shaded rows: 4 + 2 = 6 outcomes. Similarly for column 2 and shaded rows, and column 3 and shaded rows. Wait, but earlier I said total 18 outcomes. Let me check.

Wait, for column 1:

Shaded rows are 1 and 3. So:

For row 1: second spin 1 or 6. So for column 1, first spin 1 or 9, second spin 1 or 6: 2*2=4 outcomes.

For row 3: second spin 3. So column 1, first spin 1 or 9, second spin 3: 2*1=2 outcomes.

Total for column 1: 4 + 2 = 6.

Similarly, column 2:

Shaded rows are 2 and 4.

Row 2: second spin 2 or 7. So column 2, first spin 2 or 6, second spin 2 or 7: 2*2=4.

Row 4: second spin 9. So column 2, first spin 2 or 6, second spin 9: 2*1=2.

Total for column 2: 4 + 2 = 6.

Column 3:

Shaded rows are 1 and 3.

Row 1: second spin 1 or 6. Column 3, first spin 3 or 7, second spin 1 or 6: 2*2=4.

Row 3: second spin 3. Column 3, first spin 3 or 7, second spin 3: 2*1=2.

Total for column 3: 4 + 2 = 6.

Total shaded outcomes: 6 + 6 + 6 = 18. Therefore, 18/36=1/2. Therefore, correct. So earlier when I was checking for first spin 1, I thought there were three shaded outcomes, but actually, for first spin 1, second spin 1, 3, 6 are shaded. But second spin 1, 3, 6 are three different numbers. However, each second spin has a probability of 1/6. But since first spin is fixed as 1, there are 6 possible second spins. So for first spin 1, there are three shaded outcomes. Similarly, for first spin 9, three shaded outcomes. So total 6. Similarly for column 2 and 3. Therefore, total 18. Therefore, 18/36=1/2. Therefore, the answer is 1/2. Therefore, I think this is correct.

But let me check with specific examples. For example, the square (1,1) is achieved by (1,1), (1,6), (9,1), (9,6). So four outcomes. Similarly, square (1,3) is (1,3), (9,3). Two outcomes. So total for column 1: 4 + 2 = 6. Similarly for others. Therefore, 18 total. Therefore, 1/2. Therefore, the answer is 1/2. Therefore, I think that is correct.

But just to be thorough, let me check if there are any other shaded squares. The Asymptote code has six shaded squares, which we accounted for. Therefore, no other squares. Therefore, the answer is 1/2. Therefore, I think the probability is 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 Outcomes**

The wheel has six numbers: **1, 2, 3, 6, 7, 9**, each equally likely. When the wheel is spun twice, the first spin determines the **column** and the second spin determines the **row**.

- **Column Determination**: The first 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).
- **Row Determination**: The second number is divided by 5, and the remainder determines the row. The possible remainders are **1, 2, 3, 4**.

Thus, the checkerboard is a **3x4 grid** with columns labeled 1, 2, 3 and rows labeled 1, 2, 3, 4.

---

### **Step 2: Mapping Numbers to Columns and Rows**

We compute the remainders of the numbers on the wheel modulo 4 and 5:

- **Modulo 4 (Columns)**:
  - 1 → 1
  - 2 → 2
  - 3 → 3
  - 6 → 2
  - 7 → 3
  - 9 → 1
  - So, **columns 1, 2, 3** each have **2 numbers** → **probability 1/3** for each column.

- **Modulo 5 (Rows)**:
  - 1 → 1
  - 2 → 2
  - 3 → 3
  - 6 → 1
  - 7 → 2
  - 9 → 4
  - So, **rows 1 and 2** each have **2 numbers** → **probability 1/3** for each.
  - **Rows 3 and 4** each have **1 number** → **probability 1/6** for each.

---

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

From the Asymptote diagram, the shaded squares are:

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

These are the six shaded squares in the 3x4 grid.

---

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

Each square's probability is 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**

Convert all to a common denominator (18):

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

So the total probability is:

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

---

### **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 \[
\frac{1}{3}\cdot \frac{1}{2} + \frac{2}{3}\cdot\frac{1}{2} =\boxed{\frac{1}{2}}.
\]