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, the wheel has numbers labeled 1, 2, 6, 9, 3, 7. Wait, the Asymptote code labels some numbers on the wheel. Let me parse that. The Asymptote code has labels: "1" at (0.8,0.5), which is probably in the upper right quadrant, "2" at (0.8,-0.5) lower right, "6" at (-0.8,0.5) upper left, "9" at (-0.8,-0.5) lower left, "3" at (-0.7,0) which is left of center, and "7" at (0.7,0) right of center. So the numbers on the wheel are 1, 2, 3, 6, 7, 9. Wait, but there are six numbers on the wheel? Let me check the Asymptote code again. The circle is drawn with two lines crossing: horizontal and vertical, so dividing the circle into four quadrants, but also there's a vertical line from (0,2) to (0,-2), which is the vertical diameter, and horizontal line from (1.7,1) to (-1.7,-1) and (1.7,-1) to (-1.7,1). Wait, maybe the wheel is divided into six segments? Let me think.

The Asymptote code draws a circle, then draws three lines: the vertical diameter (from (0,2) to (0,-2)), and two lines that cross at the center: one is (1.7,1) to (-1.7,-1), which is a line from upper right to lower left, and (1.7,-1) to (-1.7,1), which is a line from lower right to upper left. So these lines divide the circle into six sectors? Each sector would be 60 degrees? Wait, if there are three lines crossing at the center, but one is vertical, and two are diagonals? Let me count. The vertical line divides the circle into top and bottom. Then the two diagonal lines divide each half into three parts? Maybe. So total of six sectors? Let me check labels. The labels are 1, 2, 6, 9, 3, 7. Each sector is labeled with those numbers. So there are six numbers on the wheel: 1, 2, 3, 6, 7, 9. Each equally likely. So when the wheel is spun, it can land on any of these six numbers with equal probability. Therefore, each spin has 6 possible outcomes, each with probability 1/6.

Now, the problem says the wheel is spun twice. The first number is divided by 4, giving a remainder which marks the columns of the checkerboard. The second number is divided by 5, giving a remainder which marks the rows of the checkerboard. Then the checker is placed on the square where column and row meet. The question is the probability that the checker is placed on a shaded square.

First, I need to figure out the structure of the checkerboard. The Asymptote code draws a checkerboard with columns and rows. From the Asymptote code: the checkerboard is drawn as a rectangle from (3,-2) to (6,2), divided into columns and rows. The columns are labeled 1, 2, 3 at the bottom (from (3,-2) to (6,-2)), and rows are labeled 1, 2, 3, 4 on the left side (from (3,-2) to (3,2)). The shaded squares are at positions (column, row) as follows: the first shaded square is at column 3, row 1? Wait, looking at the Asymptote code:

fill((3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7)); This is a square from x=3 to 4, y=0 to 1, which would be column 3 (since columns are labeled 1,2,3 at the bottom), and row 2? Wait, rows are labeled 1,2,3,4 on the left. Wait, the rows are vertical? Wait, no. Let me think.

The checkerboard is drawn with columns going from left to right, labeled 1,2,3 at the bottom. So column 1 is the leftmost column, column 2 next, column 3 rightmost. Rows are going from bottom to top, labeled 1,2,3,4 on the left side. So row 1 is the bottom row, row 2 above it, row 3 above, row 4 top. So each square is identified by column (1-3) and row (1-4). Then the shaded squares are as follows:

- The first shaded square is (3,0)--(3,1)--(4,1)--(4,0)--cycle: this is the square in column 3, row 2? Wait, the coordinates: x from 3 to 4 (which is column 3?), and y from 0 to 1. If rows are labeled 1, 2, 3, 4 on the left, then row 1 is from y=-2 to y=-1, row 2 from -1 to 0, row 3 from 0 to 1, row 4 from 1 to 2. Wait, the Asymptote code draws columns with labels on the bottom (x-axis) at the bottom, and rows on the left (y-axis) on the left. So for example, the label "1" at (3.5,-2) is for the bottom row (row 1), and "2" at (3,-1.5) is row 2? Wait, the 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 at the bottom (x-axis) as 1, 2, 3, each at positions 3.5,-2; 4.5,-2; 5.5,-2. Wait, but the columns are from x=3 to x=6? The checkerboard is from (3,-2) to (6,2). So columns are vertical? Wait, no. Wait, columns are marked horizontally? Wait, the columns are labeled at the bottom. So columns are vertical? Or horizontal?

Wait, the Asymptote code draws the checkerboard as a rectangle from (3,-2) to (6,2). So it spans from x=3 to x=6 (width 3 units) and y=-2 to y=2 (height 4 units). Then horizontal lines are drawn at y=-1, 0, 1, dividing the checkerboard into rows. Similarly vertical lines at x=4 and x=5 dividing into columns. 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 vertical lines are at x=4 and x=5, dividing the width (from x=3 to x=6) into three columns: column 1 from x=3 to x=4, column 2 from x=4 to x=5, column 3 from x=5 to x=6. Similarly, horizontal lines at y=-1, 0, 1 divide the height (from y=-2 to y=2) into four rows: row 1 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. The labels for columns are at the bottom (on the x-axis) at positions (3.5,-2), (4.5,-2), (5.5,-2), labeled 1, 2, 3. The labels for rows are on the left side (y-axis) at positions (3,-1.5), (3,-0.5), (3,0.5), (3,1.5), labeled 1, 2, 3, 4. So each square is identified by column (1-3) and row (1-4). So columns are vertical, rows are horizontal.

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 4 (column 1?), y=0 to 1 (row 3? Since row 3 is from y=0 to 1). Wait, but the column from x=3 to 4 is column 1. Wait, but the label for column 1 is at 3.5,-2, which is the middle of the column 1's bottom edge. So column 1 is x=3 to x=4. Similarly, column 2 is x=4 to x=5, column 3 is x=5 to x=6. Similarly, rows: row 1 is from y=-2 to y=-1, row 2 from -1 to 0, row 3 from 0 to 1, row 4 from 1 to 2.

So the first shaded square is from x=3 to 4 (column 1), y=0 to 1 (row 3). Wait, but in the Asymptote code, that square is filled. Then another shaded square is (3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle: that is column 1 (x=3 to 4), row 1 (y=-2 to -1). Then (4,1)--(4,2)--(5,2)--(5,1)--cycle: column 2 (x=4-5), row 4 (y=1-2). Then (4,-1)--(4,0)--(5,0)--(5,-1)--cycle: column 2 (x=4-5), row 2 (y=-1-0). Then (5,0)--(5,1)--(6,1)--(6,0)--cycle: column 3 (x=5-6), row 3 (y=0-1). Then (5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle: column 3 (x=5-6), row 1 (y=-2--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 3
- Column 3, Row 1

Wait, that's six shaded squares? Let me count:

1. (column 1, row 1)
2. (column 1, row 3)
3. (column 2, row 2)
4. (column 2, row 4)
5. (column 3, row 3)
6. (column 3, row 1)

Yes, six shaded squares. So in total, the checkerboard has 3 columns and 4 rows, making 12 squares. Out of which 6 are shaded. Therefore, if all squares were equally likely, the probability would be 6/12 = 1/2. But the probability isn't necessarily uniform because the selection of column and row is determined by the numbers on the wheel, which have different remainders when divided by 4 and 5.

Wait, so the first spin determines the column: the first number is divided by 4, giving remainder 1, 2, or 3? Wait, dividing by 4, the possible remainders are 0, 1, 2, 3. But the problem says "determining one of the remainders 1,2,3 marking the columns of the checkerboard". Wait, so maybe the numbers on the wheel when divided by 4 can't give 0? Or maybe they are adjusted?

Wait, the problem says: "The first number is divided by 4, determining one of the remainders 1,2,3 marking the columns of the checkerboard." So, the remainder is 1, 2, or 3. But when you divide a number by 4, the possible remainders are 0,1,2,3. So how does this happen? Is it possible that the numbers on the wheel are such that when divided by 4, they give remainders 1, 2, 3? Let's check the numbers on the wheel: 1, 2, 3, 6, 7, 9.

Let me compute each number mod 4:

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

So the possible remainders are 1, 2, 3. Each remainder occurs as follows:

- Remainder 1: numbers 1 and 9
- Remainder 2: numbers 2 and 6
- Remainder 3: numbers 3 and 7

Therefore, each remainder 1, 2, 3 occurs with probability 2/6 = 1/3. Wait, but the problem says "each number on the wheel being equally likely". Therefore, each number has probability 1/6. But when we take the remainder mod 4, the distribution is not uniform? Wait, but the problem says that the first number divided by 4 gives one of the remainders 1, 2, 3. So even though 0 is a possible remainder, it's not shown? Wait, but the numbers on the wheel when divided by 4 give remainders 1, 2, 3. So maybe the numbers on the wheel are constructed such that none of them are divisible by 4? Let me check. The numbers are 1, 2, 3, 6, 7, 9. None of these are divisible by 4. 4 divides 4, 8, 12, etc. So none of the numbers on the wheel are congruent to 0 mod 4. Therefore, when divided by 4, the remainders are 1, 2, 3. So columns: column 1 corresponds to remainder 1, column 2 to remainder 2, column 3 to remainder 3. Therefore, column is determined by remainder mod 4, and since remainders are 1,2,3, we have columns 1,2,3. Each remainder has probability:

- Remainder 1: numbers 1 and 9 (two numbers) => probability 2/6 = 1/3
- Remainder 2: numbers 2 and 6 (two numbers) => 1/3
- Remainder 3: numbers 3 and 7 (two numbers) => 1/3

So each column is equally likely? Wait, no. Wait, each remainder 1,2,3 occurs with probability 2/6 = 1/3. So each column (1, 2, 3) has probability 1/3. Wait, but that would mean that each column is equally likely. However, since each number is equally likely, but the remainders are not equally distributed? Wait, no, since each remainder occurs twice. So yes, each column is equally likely with probability 1/3. Wait, but 1/3 for each column? Wait, there are three columns, each with probability 1/3. So that's uniform.

But let me think again. The probability that the first spin gives remainder 1 is 2/6 = 1/3, same for remainder 2 and 3. So columns are equally likely, 1/3 each. Similarly for the rows: the second number is divided by 5, giving remainder 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 mod 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 dividing by 5 are: 1, 2, 3, 1, 2, 4. Therefore, the possible remainders are 1,2,3,4. Let's see how many times each occurs:

- 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 rows (determined by second number divided by 5) are:

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

So rows are not equally likely. The rows 1, 2 have probability 1/3 each, while rows 3 and 4 have probability 1/6 each.

So the columns are uniform (each column 1,2,3 has probability 1/3), and the rows have probabilities: row 1 (1/3), row 2 (1/3), row 3 (1/6), row 4 (1/6).

Therefore, the total number of possible outcomes for (column, row) is 3 columns × 4 rows, but with different probabilities. However, since the columns and rows are determined by two independent spins, the combined probability distribution is the product of column probabilities and row probabilities.

Therefore, the probability of landing on a particular square (column, row) is (column probability) × (row probability).

Our goal is to find the probability that the square is shaded. The shaded squares are:

As established earlier:

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

So we need to compute the probability for each of these shaded squares and sum them up.

Let me list them:

1. Column 1, Row 1: P(column 1) × P(row 1) = (1/3) × (1/3) = 1/9
2. Column 1, Row 3: P(column 1) × P(row 3) = (1/3) × (1/6) = 1/18
3. Column 2, Row 2: P(column 2) × P(row 2) = (1/3) × (1/3) = 1/9
4. Column 2, Row 4: P(column 2) × P(row 4) = (1/3) × (1/6) = 1/18
5. Column 3, Row 3: P(column 3) × P(row 3) = (1/3) × (1/6) = 1/18
6. Column 3, Row 1: P(column 3) × P(row 1) = (1/3) × (1/3) = 1/9

Now, summing all these probabilities:

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

Let me convert all to eighteenths:

1/9 = 2/18

So:

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

Wait, so the total probability is 1/2? But let me check again.

Wait, adding up:

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

Let me compute step by step:

First term: 1/9

Second term: +1/18 = 1/9 + 1/18 = 2/18 + 1/18 = 3/18

Third term: +1/9 = 3/18 + 2/18 = 5/18

Fourth term: +1/18 = 5/18 + 1/18 = 6/18

Fifth term: +1/18 = 6/18 + 1/18 = 7/18

Sixth term: +1/9 = 7/18 + 2/18 = 9/18 = 1/2.

Yes, same result. Therefore, the probability is 1/2. Wait, but the total number of squares is 12, and shaded squares are 6. But the probabilities aren't uniform. However, in this case, the total probability ends up being 1/2. But is that correct?

Wait, let me think again. Because columns are uniform, but rows are not. However, the shaded squares have symmetric properties?

For example, column 1 has shaded squares in row 1 and row 3. Column 2 has shaded squares in row 2 and row 4. Column 3 has shaded squares in row 1 and row 3. But rows 1 and 3 have different probabilities than rows 2 and 4. However, the total ends up being 1/2.

Alternatively, maybe I made a mistake in identifying the shaded squares? Let me check again:

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

Fill((3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7)): column 1 (x 3-4), row 3 (y 0-1). Correct.

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

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

Fill((4,-1)--(4,0)--(5,0)--(5,-1)--cycle,gray(0.7)): column 2 (x 4-5), row 2 (y -1-0). Correct.

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

Fill((5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle,gray(0.7)): column 3 (x 5-6), row 1 (y -2 to -1). Correct.

Yes, that's six shaded squares. So the calculation seems correct. Then adding up the probabilities gives 1/2.

But wait, let me check if there's another way to think about this. Since columns are uniform, each column has probability 1/3, and rows have probabilities 1/3, 1/3, 1/6, 1/6. The shaded squares are:

For column 1: row 1 and row 3. So probability (1/3)*(1/3) + (1/3)*(1/6) = 1/9 + 1/18 = 1/6.

For column 2: row 2 and row 4. Similarly, (1/3)*(1/3) + (1/3)*(1/6) = same as column 1: 1/6.

For column 3: row 1 and row 3. Same as column 1 and 2: 1/6.

Total probability: 1/6 + 1/6 + 1/6 = 1/2. Yep, same result. So probability 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 all 6 numbers have equal probability. Therefore, total number of possible outcomes is 6*6=36. But since the column and row are determined by mod 4 and mod 5, the mapping isn't one-to-one? Wait, each spin has 6 outcomes, so two spins have 36 total outcomes. However, the columns and rows are determined by the remainders. However, different numbers can result in the same column or row. So maybe I can compute the number of outcomes leading to each shaded square, and then divide by 36.

Alternatively, since we already calculated using probabilities, but let me confirm with counting. Let me try that approach.

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

First, shaded squares:

1. Column 1, Row 1: column 1 corresponds to first spin remainder 1: which is numbers 1 and 9. Row 1 corresponds to second spin remainder 1: numbers 1 and 6. Therefore, number of outcomes: 2 (for first spin) × 2 (for second spin) = 4.

2. Column 1, Row 3: column 1 (numbers 1,9) × row 3 (number 3). Therefore, 2 × 1 = 2.

3. Column 2, Row 2: column 2 (numbers 2,6) × row 2 (numbers 2,7). Therefore, 2 × 2 = 4.

4. Column 2, Row 4: column 2 (numbers 2,6) × row 4 (number 9). Therefore, 2 × 1 = 2.

5. Column 3, Row 3: column 3 (numbers 3,7) × row 3 (number 3). Therefore, 2 × 1 = 2.

6. Column 3, Row 1: column 3 (numbers 3,7) × row 1 (numbers 1,6). Therefore, 2 × 2 = 4.

Total number of favorable outcomes:

4 + 2 + 4 + 2 + 2 + 4 = 4+2=6; 6+4=10; 10+2=12; 12+2=14; 14+4=18. So 18 favorable outcomes out of 36 total. Therefore, probability is 18/36 = 1/2. Same result. So the answer is 1/2.

But let me check once more to ensure I haven't made a mistake in counting the favorable outcomes.

For shaded square column 1, row 1: first spin must be 1 or 9 (remainders 1 mod 4), second spin must be 1 or 6 (remainders 1 mod 5). So numbers: (1,1), (1,6), (9,1), (9,6). That's 4 outcomes. Correct.

Column 1, row 3: first spin 1 or 9; second spin 3. So (1,3), (9,3). 2 outcomes. Correct.

Column 2, row 2: first spin 2 or 6; second spin 2 or 7. So (2,2), (2,7), (6,2), (6,7). 4 outcomes. Correct.

Column 2, row 4: first spin 2 or 6; second spin 9. (2,9), (6,9). 2 outcomes. Correct.

Column 3, row 3: first spin 3 or 7; second spin 3. (3,3), (7,3). 2 outcomes. Correct.

Column 3, row 1: first spin 3 or 7; second spin 1 or 6. (3,1), (3,6), (7,1), (7,6). 4 outcomes. Correct.

Total: 4 + 2 + 4 + 2 + 2 + 4 = 18. 18/36 = 1/2. So that seems correct.

Therefore, the probability is 1/2. But wait, wait, the problem says "the checkerboard shown". Is there any chance that I miscounted the shaded squares? Let me check again the Asymptote code:

The Asymptote code has six fill commands:

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

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

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

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

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

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

Yes, six shaded squares. So no mistake here.

Therefore, the answer is 1/2.

Wait, but the problem is from an Olympiad, perhaps? And maybe the answer is different? Let me check again.

Wait, another way: since columns are uniform, and rows have probabilities. The shaded squares are symmetric in some way. Let me think about the entire grid.

Columns: 1, 2, 3

Rows: 1, 2, 3, 4

Shaded squares:

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

So for column 1, shaded rows 1 and 3.

Column 2, shaded rows 2 and 4.

Column 3, shaded rows 1 and 3.

So for each column, half of the rows (if rows were uniform) would be shaded. But rows are not uniform. However, for columns 1 and 3, shaded rows are 1 and 3, which have probabilities 1/3 and 1/6. So for column 1:

P(shaded) = P(row1) + P(row3) = 1/3 + 1/6 = 1/2

Similarly, column 3: same as column 1.

Column 2: shaded rows 2 and 4. P(shaded) = P(row2) + P(row4) = 1/3 + 1/6 = 1/2

Therefore, since each column has probability 1/3, the total probability is:

P(column1)*P(shaded in column1) + P(column2)*P(shaded in column2) + P(column3)*P(shaded in column3) = (1/3)*(1/2) + (1/3)*(1/2) + (1/3)*(1/2) = 1/2.

Yes, same result. So regardless of the approach, it's 1/2. Therefore, the probability is 1/2.

But let me check one more time with actual numbers. Total outcomes 36. Shaded outcomes 18. So 18/36=1/2.

Yes. Therefore, I think the answer is 1/2, which is 0.5, so the probability is 1/2.

Wait, but the problem is presented with an Asymptote figure, which might have some detail I didn't consider? Let me check once again.

Wait, the numbers on the wheel are 1, 2, 6, 9, 3, 7. Each spin, so two spins. The first spin determines column via mod 4. So numbers 1, 2, 3, 6, 7, 9. Each number corresponds to column 1, 2, 3 as follows:

- 1,9: column 1

- 2,6: column 2

- 3,7: column 3

Yes. Second spin determines row via mod 5. Numbers 1,2,3,6,7,9:

- 1,6: row 1

- 2,7: row 2

- 3: row 3

- 9: row 4

Therefore, the mapping is correct.

Therefore, all steps are correct. Therefore, answer is 1/2.

But wait, the problem is from an Olympiad, and perhaps the answer is different? Wait, maybe I made a mistake in the mapping of the shaded squares? Let me check once again:

Looking at the Asymptote code's shaded squares:

First shaded square is column 1, row 3 (as per coordinates). Then column 1, row 1. Column 2, row 4. Column 2, row 2. Column 3, row 3. Column 3, row 1. So that's six squares. Therefore, yes.

Alternatively, maybe the rows are not 1-4 but something else? Let me check:

The rows are labeled 1,2,3,4 on the left, which correspond to the y-axis. The row labels are on the left side at positions:

label("1",(3.5,-2),S); - this is for row 1, at the bottom.

label("2",(4.5,-2),S); - row 2?

Wait, no. Wait, the labels for columns are on the bottom (x-axis), so label("1",(3.5,-2),S) is for column 1. Similarly, label("2",(4.5,-2),S) is column 2, label("3",(5.5,-2),S) is column 3. Then labels for rows are on the left (y-axis):

label("1",(3,-1.5),W); - this is for row 1.

label("2",(3,-0.5),W); - row 2.

label("3",(3,0.5),W); - row 3.

label("4",(3,1.5),W); - row 4.

Therefore, row 1 is the bottom row (from y=-2 to y=-1). Row 2 is from y=-1 to y=0. Row 3 from y=0 to y=1. Row 4 from y=1 to y=2. Therefore, coordinates:

- For example, a square in column 1, row 1 is from x=3-4, y=-2--1.

So the shaded square (3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle is indeed column 1, row 1. Similarly for others. So correct.

Therefore, I think my answer is correct: 1/2.

But to make sure, let me check with an example. Suppose we compute the total number of shaded squares as 6, and total squares 12. But the probability isn't uniform, so 6/12 = 1/2 isn't necessarily the probability. However, in our case, due to the distribution of the rows and columns, the probability ends up being 1/2. Therefore, the answer is 1/2.

Alternatively, maybe the problem is designed that way. Therefore, I think the answer is 1/2.

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

To determine the probability that the checker ends up on a shaded square of the checkerboard, we need to analyze how the numbers on the wheel determine the placement of the checker on the grid.

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### Step 1: Understand the Wheel and Its Outcomes

The spin of the wheel results in one of the following numbers: **1, 2, 6, 9, 3, 7**. Each number is equally likely, so each has a **probability of $ \frac{1}{6} $**.

- **First Spin**: Used to determine the **column** of the checkerboard. The number is divided by **4**, resulting in a remainder. Since none of the numbers are divisible by 4, the remainder can only be **1, 2, or 3**.
- **Second Spin**: Used to determine the **row** of the checkerboard. The number is divided by **5**, which can produce remainders **1, 2, 3, or 4**.

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### Step 2: Map Numbers to Columns and Rows

- **Column Mapping**:
  - Numbers that result in remainder **1**: 1, 9 → 2 possibilities → probability = $ \frac{2}{6} = \frac{1}{3} $
  - Numbers that result in remainder **2**: 2, 6 → 2 possibilities → probability = $ \frac{1}{3} $
  - Numbers that result in remainder **3**: 3, 7 → 2 possibilities → probability = $ \frac{1}{3} $

- **Row Mapping**:
  - Remainder **1**: 1, 6 → 2 possibilities → probability = $ \frac{1}{3} $
  - Remainder **2**: 2, 7 → 2 possibilities → probability = $ \frac{1}{3} $
  - Remainder **3**: 3 → 1 possibility → probability = $ \frac{1}{6} $
  - Remainder **4**: 9 → 1 possibility → probability = $ \frac{1}{6} $

Thus, the probability for each column is $ \frac{1}{3} $, and for rows, it is:
- Row 1: $ \frac{1}{3} $
- Row 2: $ \frac{1}{3} $
- Row 3: $ \frac{1}{6} $
- Row 4: $ \frac{1}{6} $

---

### Step 3: Identify Shaded Squares

From the figure, the shaded squares are:
- (1,1), (1,3), (2,2), (2,4), (3,3), (3,1)

There are **6 shaded squares**.

---

### Step 4: Calculate Probability of Each Shaded Square

Using the probabilities for rows and columns, we compute the probability of landing on each of these squares:

| Square | Column | Row | Probability | Calculation |
|--------|--------|-----|-------------|-------------|
| (1,1)  | 1      | 1   | $ \frac{1}{3} \times \frac{1}{3} $ | $ \frac{1}{9} $ |
| (1,3)  | 1      | 3   | $ \frac{1}{3} \times \frac{1}{6} $ | $ \frac{1}{18} $ |
| (2,2)  | 2      | 2   | $ \frac{1}{3} \times \frac{1}{3} $ | $ \frac{1}{9} $ |
| (2,4)  | 2      | 4   | $ \frac{1}{3} \times \frac{1}{6} $ | $ \frac{1}{18} $ |
| (3,3)  | 3      | 3   | $ \frac{1}{3} \times \frac{1}{6} $ | $ \frac{1}{18} $ |
| (3,1)  | 3      | 1   | $ \frac{1}{3} \times \frac{1}{3} $ | $ \frac{1}{9} $ |

---

### Step 5: Add Up Probabilities

Now, sum up the probabilities of all shaded squares:

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

---

### 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, the wheel has numbers labeled 1, 2, 6, 9, 3, 7. Wait, the Asymptote code labels some numbers on the wheel. Let me parse that. The Asymptote code has labels: "1" at (0.8,0.5), which is probably in the upper right quadrant, "2" at (0.8,-0.5) lower right, "6" at (-0.8,0.5) upper left, "9" at (-0.8,-0.5) lower left, "3" at (-0.7,0) which is left of center, and "7" at (0.7,0) right of center. So the numbers on the wheel are 1, 2, 3, 6, 7, 9. Wait, but there are six numbers on the wheel? Let me check the Asymptote code again. The circle is drawn with two lines crossing: horizontal and vertical, so dividing the circle into four quadrants, but also there's a vertical line from (0,2) to (0,-2), which is the vertical diameter, and horizontal line from (1.7,1) to (-1.7,-1) and (1.7,-1) to (-1.7,1). Wait, maybe the wheel is divided into six segments? Let me think.

The Asymptote code draws a circle, then draws three lines: the vertical diameter (from (0,2) to (0,-2)), and two lines that cross at the center: one is (1.7,1) to (-1.7,-1), which is a line from upper right to lower left, and (1.7,-1) to (-1.7,1), which is a line from lower right to upper left. So these lines divide the circle into six sectors? Each sector would be 60 degrees? Wait, if there are three lines crossing at the center, but one is vertical, and two are diagonals? Let me count. The vertical line divides the circle into top and bottom. Then the two diagonal lines divide each half into three parts? Maybe. So total of six sectors? Let me check labels. The labels are 1, 2, 6, 9, 3, 7. Each sector is labeled with those numbers. So there are six numbers on the wheel: 1, 2, 3, 6, 7, 9. Each equally likely. So when the wheel is spun, it can land on any of these six numbers with equal probability. Therefore, each spin has 6 possible outcomes, each with probability 1/6.

Now, the problem says the wheel is spun twice. The first number is divided by 4, giving a remainder which marks the columns of the checkerboard. The second number is divided by 5, giving a remainder which marks the rows of the checkerboard. Then the checker is placed on the square where column and row meet. The question is the probability that the checker is placed on a shaded square.

First, I need to figure out the structure of the checkerboard. The Asymptote code draws a checkerboard with columns and rows. From the Asymptote code: the checkerboard is drawn as a rectangle from (3,-2) to (6,2), divided into columns and rows. The columns are labeled 1, 2, 3 at the bottom (from (3,-2) to (6,-2)), and rows are labeled 1, 2, 3, 4 on the left side (from (3,-2) to (3,2)). The shaded squares are at positions (column, row) as follows: the first shaded square is at column 3, row 1? Wait, looking at the Asymptote code:

fill((3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7)); This is a square from x=3 to 4, y=0 to 1, which would be column 3 (since columns are labeled 1,2,3 at the bottom), and row 2? Wait, rows are labeled 1,2,3,4 on the left. Wait, the rows are vertical? Wait, no. Let me think.

The checkerboard is drawn with columns going from left to right, labeled 1,2,3 at the bottom. So column 1 is the leftmost column, column 2 next, column 3 rightmost. Rows are going from bottom to top, labeled 1,2,3,4 on the left side. So row 1 is the bottom row, row 2 above it, row 3 above, row 4 top. So each square is identified by column (1-3) and row (1-4). Then the shaded squares are as follows:

- The first shaded square is (3,0)--(3,1)--(4,1)--(4,0)--cycle: this is the square in column 3, row 2? Wait, the coordinates: x from 3 to 4 (which is column 3?), and y from 0 to 1. If rows are labeled 1, 2, 3, 4 on the left, then row 1 is from y=-2 to y=-1, row 2 from -1 to 0, row 3 from 0 to 1, row 4 from 1 to 2. Wait, the Asymptote code draws columns with labels on the bottom (x-axis) at the bottom, and rows on the left (y-axis) on the left. So for example, the label "1" at (3.5,-2) is for the bottom row (row 1), and "2" at (3,-1.5) is row 2? Wait, the 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 at the bottom (x-axis) as 1, 2, 3, each at positions 3.5,-2; 4.5,-2; 5.5,-2. Wait, but the columns are from x=3 to x=6? The checkerboard is from (3,-2) to (6,2). So columns are vertical? Wait, no. Wait, columns are marked horizontally? Wait, the columns are labeled at the bottom. So columns are vertical? Or horizontal?

Wait, the Asymptote code draws the checkerboard as a rectangle from (3,-2) to (6,2). So it spans from x=3 to x=6 (width 3 units) and y=-2 to y=2 (height 4 units). Then horizontal lines are drawn at y=-1, 0, 1, dividing the checkerboard into rows. Similarly vertical lines at x=4 and x=5 dividing into columns. 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 vertical lines are at x=4 and x=5, dividing the width (from x=3 to x=6) into three columns: column 1 from x=3 to x=4, column 2 from x=4 to x=5, column 3 from x=5 to x=6. Similarly, horizontal lines at y=-1, 0, 1 divide the height (from y=-2 to y=2) into four rows: row 1 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. The labels for columns are at the bottom (on the x-axis) at positions (3.5,-2), (4.5,-2), (5.5,-2), labeled 1, 2, 3. The labels for rows are on the left side (y-axis) at positions (3,-1.5), (3,-0.5), (3,0.5), (3,1.5), labeled 1, 2, 3, 4. So each square is identified by column (1-3) and row (1-4). So columns are vertical, rows are horizontal.

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 4 (column 1?), y=0 to 1 (row 3? Since row 3 is from y=0 to 1). Wait, but the column from x=3 to 4 is column 1. Wait, but the label for column 1 is at 3.5,-2, which is the middle of the column 1's bottom edge. So column 1 is x=3 to x=4. Similarly, column 2 is x=4 to x=5, column 3 is x=5 to x=6. Similarly, rows: row 1 is from y=-2 to y=-1, row 2 from -1 to 0, row 3 from 0 to 1, row 4 from 1 to 2.

So the first shaded square is from x=3 to 4 (column 1), y=0 to 1 (row 3). Wait, but in the Asymptote code, that square is filled. Then another shaded square is (3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle: that is column 1 (x=3 to 4), row 1 (y=-2 to -1). Then (4,1)--(4,2)--(5,2)--(5,1)--cycle: column 2 (x=4-5), row 4 (y=1-2). Then (4,-1)--(4,0)--(5,0)--(5,-1)--cycle: column 2 (x=4-5), row 2 (y=-1-0). Then (5,0)--(5,1)--(6,1)--(6,0)--cycle: column 3 (x=5-6), row 3 (y=0-1). Then (5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle: column 3 (x=5-6), row 1 (y=-2--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 3
- Column 3, Row 1

Wait, that's six shaded squares? Let me count:

1. (column 1, row 1)
2. (column 1, row 3)
3. (column 2, row 2)
4. (column 2, row 4)
5. (column 3, row 3)
6. (column 3, row 1)

Yes, six shaded squares. So in total, the checkerboard has 3 columns and 4 rows, making 12 squares. Out of which 6 are shaded. Therefore, if all squares were equally likely, the probability would be 6/12 = 1/2. But the probability isn't necessarily uniform because the selection of column and row is determined by the numbers on the wheel, which have different remainders when divided by 4 and 5.

Wait, so the first spin determines the column: the first number is divided by 4, giving remainder 1, 2, or 3? Wait, dividing by 4, the possible remainders are 0, 1, 2, 3. But the problem says "determining one of the remainders 1,2,3 marking the columns of the checkerboard". Wait, so maybe the numbers on the wheel when divided by 4 can't give 0? Or maybe they are adjusted?

Wait, the problem says: "The first number is divided by 4, determining one of the remainders 1,2,3 marking the columns of the checkerboard." So, the remainder is 1, 2, or 3. But when you divide a number by 4, the possible remainders are 0,1,2,3. So how does this happen? Is it possible that the numbers on the wheel are such that when divided by 4, they give remainders 1, 2, 3? Let's check the numbers on the wheel: 1, 2, 3, 6, 7, 9.

Let me compute each number mod 4:

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

So the possible remainders are 1, 2, 3. Each remainder occurs as follows:

- Remainder 1: numbers 1 and 9
- Remainder 2: numbers 2 and 6
- Remainder 3: numbers 3 and 7

Therefore, each remainder 1, 2, 3 occurs with probability 2/6 = 1/3. Wait, but the problem says "each number on the wheel being equally likely". Therefore, each number has probability 1/6. But when we take the remainder mod 4, the distribution is not uniform? Wait, but the problem says that the first number divided by 4 gives one of the remainders 1, 2, 3. So even though 0 is a possible remainder, it's not shown? Wait, but the numbers on the wheel when divided by 4 give remainders 1, 2, 3. So maybe the numbers on the wheel are constructed such that none of them are divisible by 4? Let me check. The numbers are 1, 2, 3, 6, 7, 9. None of these are divisible by 4. 4 divides 4, 8, 12, etc. So none of the numbers on the wheel are congruent to 0 mod 4. Therefore, when divided by 4, the remainders are 1, 2, 3. So columns: column 1 corresponds to remainder 1, column 2 to remainder 2, column 3 to remainder 3. Therefore, column is determined by remainder mod 4, and since remainders are 1,2,3, we have columns 1,2,3. Each remainder has probability:

- Remainder 1: numbers 1 and 9 (two numbers) => probability 2/6 = 1/3
- Remainder 2: numbers 2 and 6 (two numbers) => 1/3
- Remainder 3: numbers 3 and 7 (two numbers) => 1/3

So each column is equally likely? Wait, no. Wait, each remainder 1,2,3 occurs with probability 2/6 = 1/3. So each column (1, 2, 3) has probability 1/3. Wait, but that would mean that each column is equally likely. However, since each number is equally likely, but the remainders are not equally distributed? Wait, no, since each remainder occurs twice. So yes, each column is equally likely with probability 1/3. Wait, but 1/3 for each column? Wait, there are three columns, each with probability 1/3. So that's uniform.

But let me think again. The probability that the first spin gives remainder 1 is 2/6 = 1/3, same for remainder 2 and 3. So columns are equally likely, 1/3 each. Similarly for the rows: the second number is divided by 5, giving remainder 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 mod 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 dividing by 5 are: 1, 2, 3, 1, 2, 4. Therefore, the possible remainders are 1,2,3,4. Let's see how many times each occurs:

- 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 rows (determined by second number divided by 5) are:

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

So rows are not equally likely. The rows 1, 2 have probability 1/3 each, while rows 3 and 4 have probability 1/6 each.

So the columns are uniform (each column 1,2,3 has probability 1/3), and the rows have probabilities: row 1 (1/3), row 2 (1/3), row 3 (1/6), row 4 (1/6).

Therefore, the total number of possible outcomes for (column, row) is 3 columns × 4 rows, but with different probabilities. However, since the columns and rows are determined by two independent spins, the combined probability distribution is the product of column probabilities and row probabilities.

Therefore, the probability of landing on a particular square (column, row) is (column probability) × (row probability).

Our goal is to find the probability that the square is shaded. The shaded squares are:

As established earlier:

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

So we need to compute the probability for each of these shaded squares and sum them up.

Let me list them:

1. Column 1, Row 1: P(column 1) × P(row 1) = (1/3) × (1/3) = 1/9
2. Column 1, Row 3: P(column 1) × P(row 3) = (1/3) × (1/6) = 1/18
3. Column 2, Row 2: P(column 2) × P(row 2) = (1/3) × (1/3) = 1/9
4. Column 2, Row 4: P(column 2) × P(row 4) = (1/3) × (1/6) = 1/18
5. Column 3, Row 3: P(column 3) × P(row 3) = (1/3) × (1/6) = 1/18
6. Column 3, Row 1: P(column 3) × P(row 1) = (1/3) × (1/3) = 1/9

Now, summing all these probabilities:

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

Let me convert all to eighteenths:

1/9 = 2/18

So:

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

Wait, so the total probability is 1/2? But let me check again.

Wait, adding up:

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

Let me compute step by step:

First term: 1/9

Second term: +1/18 = 1/9 + 1/18 = 2/18 + 1/18 = 3/18

Third term: +1/9 = 3/18 + 2/18 = 5/18

Fourth term: +1/18 = 5/18 + 1/18 = 6/18

Fifth term: +1/18 = 6/18 + 1/18 = 7/18

Sixth term: +1/9 = 7/18 + 2/18 = 9/18 = 1/2.

Yes, same result. Therefore, the probability is 1/2. Wait, but the total number of squares is 12, and shaded squares are 6. But the probabilities aren't uniform. However, in this case, the total probability ends up being 1/2. But is that correct?

Wait, let me think again. Because columns are uniform, but rows are not. However, the shaded squares have symmetric properties?

For example, column 1 has shaded squares in row 1 and row 3. Column 2 has shaded squares in row 2 and row 4. Column 3 has shaded squares in row 1 and row 3. But rows 1 and 3 have different probabilities than rows 2 and 4. However, the total ends up being 1/2.

Alternatively, maybe I made a mistake in identifying the shaded squares? Let me check again:

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

Fill((3,0)--(3,1)--(4,1)--(4,0)--cycle,gray(0.7)): column 1 (x 3-4), row 3 (y 0-1). Correct.

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

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

Fill((4,-1)--(4,0)--(5,0)--(5,-1)--cycle,gray(0.7)): column 2 (x 4-5), row 2 (y -1-0). Correct.

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

Fill((5,-2)--(5,-1)--(6,-1)--(6,-2)--cycle,gray(0.7)): column 3 (x 5-6), row 1 (y -2 to -1). Correct.

Yes, that's six shaded squares. So the calculation seems correct. Then adding up the probabilities gives 1/2.

But wait, let me check if there's another way to think about this. Since columns are uniform, each column has probability 1/3, and rows have probabilities 1/3, 1/3, 1/6, 1/6. The shaded squares are:

For column 1: row 1 and row 3. So probability (1/3)*(1/3) + (1/3)*(1/6) = 1/9 + 1/18 = 1/6.

For column 2: row 2 and row 4. Similarly, (1/3)*(1/3) + (1/3)*(1/6) = same as column 1: 1/6.

For column 3: row 1 and row 3. Same as column 1 and 2: 1/6.

Total probability: 1/6 + 1/6 + 1/6 = 1/2. Yep, same result. So probability 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 all 6 numbers have equal probability. Therefore, total number of possible outcomes is 6*6=36. But since the column and row are determined by mod 4 and mod 5, the mapping isn't one-to-one? Wait, each spin has 6 outcomes, so two spins have 36 total outcomes. However, the columns and rows are determined by the remainders. However, different numbers can result in the same column or row. So maybe I can compute the number of outcomes leading to each shaded square, and then divide by 36.

Alternatively, since we already calculated using probabilities, but let me confirm with counting. Let me try that approach.

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

First, shaded squares:

1. Column 1, Row 1: column 1 corresponds to first spin remainder 1: which is numbers 1 and 9. Row 1 corresponds to second spin remainder 1: numbers 1 and 6. Therefore, number of outcomes: 2 (for first spin) × 2 (for second spin) = 4.

2. Column 1, Row 3: column 1 (numbers 1,9) × row 3 (number 3). Therefore, 2 × 1 = 2.

3. Column 2, Row 2: column 2 (numbers 2,6) × row 2 (numbers 2,7). Therefore, 2 × 2 = 4.

4. Column 2, Row 4: column 2 (numbers 2,6) × row 4 (number 9). Therefore, 2 × 1 = 2.

5. Column 3, Row 3: column 3 (numbers 3,7) × row 3 (number 3). Therefore, 2 × 1 = 2.

6. Column 3, Row 1: column 3 (numbers 3,7) × row 1 (numbers 1,6). Therefore, 2 × 2 = 4.

Total number of favorable outcomes:

4 + 2 + 4 + 2 + 2 + 4 = 4+2=6; 6+4=10; 10+2=12; 12+2=14; 14+4=18. So 18 favorable outcomes out of 36 total. Therefore, probability is 18/36 = 1/2. Same result. So the answer is 1/2.

But let me check once more to ensure I haven't made a mistake in counting the favorable outcomes.

For shaded square column 1, row 1: first spin must be 1 or 9 (remainders 1 mod 4), second spin must be 1 or 6 (remainders 1 mod 5). So numbers: (1,1), (1,6), (9,1), (9,6). That's 4 outcomes. Correct.

Column 1, row 3: first spin 1 or 9; second spin 3. So (1,3), (9,3). 2 outcomes. Correct.

Column 2, row 2: first spin 2 or 6; second spin 2 or 7. So (2,2), (2,7), (6,2), (6,7). 4 outcomes. Correct.

Column 2, row 4: first spin 2 or 6; second spin 9. (2,9), (6,9). 2 outcomes. Correct.

Column 3, row 3: first spin 3 or 7; second spin 3. (3,3), (7,3). 2 outcomes. Correct.

Column 3, row 1: first spin 3 or 7; second spin 1 or 6. (3,1), (3,6), (7,1), (7,6). 4 outcomes. Correct.

Total: 4 + 2 + 4 + 2 + 2 + 4 = 18. 18/36 = 1/2. So that seems correct.

Therefore, the probability is 1/2. But wait, wait, the problem says "the checkerboard shown". Is there any chance that I miscounted the shaded squares? Let me check again the Asymptote code:

The Asymptote code has six fill commands:

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

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

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

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

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

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

Yes, six shaded squares. So no mistake here.

Therefore, the answer is 1/2.

Wait, but the problem is from an Olympiad, perhaps? And maybe the answer is different? Let me check again.

Wait, another way: since columns are uniform, and rows have probabilities. The shaded squares are symmetric in some way. Let me think about the entire grid.

Columns: 1, 2, 3

Rows: 1, 2, 3, 4

Shaded squares:

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

So for column 1, shaded rows 1 and 3.

Column 2, shaded rows 2 and 4.

Column 3, shaded rows 1 and 3.

So for each column, half of the rows (if rows were uniform) would be shaded. But rows are not uniform. However, for columns 1 and 3, shaded rows are 1 and 3, which have probabilities 1/3 and 1/6. So for column 1:

P(shaded) = P(row1) + P(row3) = 1/3 + 1/6 = 1/2

Similarly, column 3: same as column 1.

Column 2: shaded rows 2 and 4. P(shaded) = P(row2) + P(row4) = 1/3 + 1/6 = 1/2

Therefore, since each column has probability 1/3, the total probability is:

P(column1)*P(shaded in column1) + P(column2)*P(shaded in column2) + P(column3)*P(shaded in column3) = (1/3)*(1/2) + (1/3)*(1/2) + (1/3)*(1/2) = 1/2.

Yes, same result. So regardless of the approach, it's 1/2. Therefore, the probability is 1/2.

But let me check one more time with actual numbers. Total outcomes 36. Shaded outcomes 18. So 18/36=1/2.

Yes. Therefore, I think the answer is 1/2, which is 0.5, so the probability is 1/2.

Wait, but the problem is presented with an Asymptote figure, which might have some detail I didn't consider? Let me check once again.

Wait, the numbers on the wheel are 1, 2, 6, 9, 3, 7. Each spin, so two spins. The first spin determines column via mod 4. So numbers 1, 2, 3, 6, 7, 9. Each number corresponds to column 1, 2, 3 as follows:

- 1,9: column 1

- 2,6: column 2

- 3,7: column 3

Yes. Second spin determines row via mod 5. Numbers 1,2,3,6,7,9:

- 1,6: row 1

- 2,7: row 2

- 3: row 3

- 9: row 4

Therefore, the mapping is correct.

Therefore, all steps are correct. Therefore, answer is 1/2.

But wait, the problem is from an Olympiad, and perhaps the answer is different? Wait, maybe I made a mistake in the mapping of the shaded squares? Let me check once again:

Looking at the Asymptote code's shaded squares:

First shaded square is column 1, row 3 (as per coordinates). Then column 1, row 1. Column 2, row 4. Column 2, row 2. Column 3, row 3. Column 3, row 1. So that's six squares. Therefore, yes.

Alternatively, maybe the rows are not 1-4 but something else? Let me check:

The rows are labeled 1,2,3,4 on the left, which correspond to the y-axis. The row labels are on the left side at positions:

label("1",(3.5,-2),S); - this is for row 1, at the bottom.

label("2",(4.5,-2),S); - row 2?

Wait, no. Wait, the labels for columns are on the bottom (x-axis), so label("1",(3.5,-2),S) is for column 1. Similarly, label("2",(4.5,-2),S) is column 2, label("3",(5.5,-2),S) is column 3. Then labels for rows are on the left (y-axis):

label("1",(3,-1.5),W); - this is for row 1.

label("2",(3,-0.5),W); - row 2.

label("3",(3,0.5),W); - row 3.

label("4",(3,1.5),W); - row 4.

Therefore, row 1 is the bottom row (from y=-2 to y=-1). Row 2 is from y=-1 to y=0. Row 3 from y=0 to y=1. Row 4 from y=1 to y=2. Therefore, coordinates:

- For example, a square in column 1, row 1 is from x=3-4, y=-2--1.

So the shaded square (3,-2)--(3,-1)--(4,-1)--(4,-2)--cycle is indeed column 1, row 1. Similarly for others. So correct.

Therefore, I think my answer is correct: 1/2.

But to make sure, let me check with an example. Suppose we compute the total number of shaded squares as 6, and total squares 12. But the probability isn't uniform, so 6/12 = 1/2 isn't necessarily the probability. However, in our case, due to the distribution of the rows and columns, the probability ends up being 1/2. Therefore, the answer is 1/2.

Alternatively, maybe the problem is designed that way. Therefore, I think the answer is 1/2.

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

To determine the probability that the checker ends up on a shaded square of the checkerboard, we need to analyze how the numbers on the wheel determine the placement of the checker on the grid.

---

### Step 1: Understand the Wheel and Its Outcomes

The spin of the wheel results in one of the following numbers: **1, 2, 6, 9, 3, 7**. Each number is equally likely, so each has a **probability of $ \frac{1}{6} $**.

- **First Spin**: Used to determine the **column** of the checkerboard. The number is divided by **4**, resulting in a remainder. Since none of the numbers are divisible by 4, the remainder can only be **1, 2, or 3**.
- **Second Spin**: Used to determine the **row** of the checkerboard. The number is divided by **5**, which can produce remainders **1, 2, 3, or 4**.

---

### Step 2: Map Numbers to Columns and Rows

- **Column Mapping**:
  - Numbers that result in remainder **1**: 1, 9 → 2 possibilities → probability = $ \frac{2}{6} = \frac{1}{3} $
  - Numbers that result in remainder **2**: 2, 6 → 2 possibilities → probability = $ \frac{1}{3} $
  - Numbers that result in remainder **3**: 3, 7 → 2 possibilities → probability = $ \frac{1}{3} $

- **Row Mapping**:
  - Remainder **1**: 1, 6 → 2 possibilities → probability = $ \frac{1}{3} $
  - Remainder **2**: 2, 7 → 2 possibilities → probability = $ \frac{1}{3} $
  - Remainder **3**: 3 → 1 possibility → probability = $ \frac{1}{6} $
  - Remainder **4**: 9 → 1 possibility → probability = $ \frac{1}{6} $

Thus, the probability for each column is $ \frac{1}{3} $, and for rows, it is:
- Row 1: $ \frac{1}{3} $
- Row 2: $ \frac{1}{3} $
- Row 3: $ \frac{1}{6} $
- Row 4: $ \frac{1}{6} $

---

### Step 3: Identify Shaded Squares

From the figure, the shaded squares are:
- (1,1), (1,3), (2,2), (2,4), (3,3), (3,1)

There are **6 shaded squares**.

---

### Step 4: Calculate Probability of Each Shaded Square

Using the probabilities for rows and columns, we compute the probability of landing on each of these squares:

| Square | Column | Row | Probability | Calculation |
|--------|--------|-----|-------------|-------------|
| (1,1)  | 1      | 1   | $ \frac{1}{3} \times \frac{1}{3} $ | $ \frac{1}{9} $ |
| (1,3)  | 1      | 3   | $ \frac{1}{3} \times \frac{1}{6} $ | $ \frac{1}{18} $ |
| (2,2)  | 2      | 2   | $ \frac{1}{3} \times \frac{1}{3} $ | $ \frac{1}{9} $ |
| (2,4)  | 2      | 4   | $ \frac{1}{3} \times \frac{1}{6} $ | $ \frac{1}{18} $ |
| (3,3)  | 3      | 3   | $ \frac{1}{3} \times \frac{1}{6} $ | $ \frac{1}{18} $ |
| (3,1)  | 3      | 1   | $ \frac{1}{3} \times \frac{1}{3} $ | $ \frac{1}{9} $ |

---

### Step 5: Add Up Probabilities

Now, sum up the probabilities of all shaded squares:

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

---

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