AMC 10 · 2012 · #20
Grade 8 probabilityA 3×3 square is partitioned into 9 unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated 90∘ clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability the grid is now entirely black?
Pick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A 3x3 grid has each of its 9 unit squares colored black or white independently, each color with probability 1/2. The whole grid is turned 90 degrees clockwise about its center. Then any square that is white but sits on a spot that used to hold a black square is repainted black; every other square keeps its color. Find the probability that afterward the entire grid is black.
Givens: The grid is 3 by 3, so it has 9 unit squares: one center, 4 corners, and 4 edge-middles.; Each square is independently black or white with probability 1/2, so there are 2^9 = 512 equally likely starting grids.; The grid is rotated 90 degrees clockwise about its center.; After the turn, a white square that lands where a black square used to be is repainted black; all other squares keep their color.
Unknowns: The probability that the whole 3x3 grid is black after the rotate-and-repaint step.
Understand
Restated: A 3x3 grid has each of its 9 unit squares colored black or white independently, each color with probability 1/2. The whole grid is turned 90 degrees clockwise about its center. Then any square that is white but sits on a spot that used to hold a black square is repainted black; every other square keeps its color. Find the probability that afterward the entire grid is black.
Givens: The grid is 3 by 3, so it has 9 unit squares: one center, 4 corners, and 4 edge-middles.; Each square is independently black or white with probability 1/2, so there are 2^9 = 512 equally likely starting grids.; The grid is rotated 90 degrees clockwise about its center.; After the turn, a white square that lands where a black square used to be is repainted black; all other squares keep their color.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #17 Visualize Spatial Relationships, #16 Change Focus / Count the Complement, #2 Make a Systematic List
The turn never mixes corners with edges with the center, so the 9 squares split into three independent groups: the lone center, a loop of 4 corners, and a loop of 4 edges. Once the painting rule is restated as a simple ban on certain white pairs, each group can be counted on its own and the counts multiplied. Breaking the grid into these separate pieces is what turns a tangled picture into three tiny counting jobs.
Execute — Answer: A
8.G.A.1 Step 1 See what the turn moves
- Picture turning the grid 90 degrees clockwise.
- The center stays put.
- The top-left corner goes to the top-right, then to the bottom-right, then to the bottom-left, then back: the 4 corners ride around one loop.
- In the same way the 4 edge-middles ride around their own separate loop.
- So the moving squares form two independent 4-square loops, and the center is alone.
💡 A quarter turn just slides each ring of squares one step around its own circle.
7.SP.C.7 Step 2 Restate when a square ends white
- Follow one square's fate.
- After the turn, some square from its loop lands on it.
- If the square that started on this spot was black, the spot is already black and stays black.
- If it started white, then the newcomer decides: the newcomer only gets repainted when it covers a spot that was black, so a white newcomer over a white spot stays white.
- So a spot ends white exactly when it started white AND the square rotating onto it also started white.
- The grid is all black precisely when no such white-on-white pair happens.
💡 One white square alone always gets covered or paints over; only two whites that chase each other around the loop survive.
7.SP.C.8 Step 3 Split into center, corners, edges
- Because a square and the one that rotates onto it always live in the same loop, the white-on-white danger never crosses between groups.
- So the three groups are independent, and the number of all-black starting grids is the product of the counts from each group.
- The center must start black (1 way).
- The corner loop and the edge loop each need to be counted for how many colorings avoid a bad white pair.
💡 Independent pieces multiply, so three small counts replace one giant count.
7.SP.C.8 Step 4 Count one 4-square loop
- Take the 4 corners in their loop; two squares are neighbors in the loop if one turns onto the other.
- We need colorings with no two neighbors both white.
- Count by how many whites there are: all black is 1 way; exactly one white works from any of the 4 spots, 4 ways; two whites works only when they sit opposite each other (not neighbors), which is 2 ways; three or four whites always force two neighbors white, 0 ways.
- That gives 1 + 4 + 2 = 7 good colorings.
- The edge loop works exactly the same way, so it also has 7.
💡 Around a loop of four, whites are only safe when they never sit side by side.
7.SP.C.5 Step 5 Multiply and divide
- Put the pieces together: 1 way for the center, 7 for the corners, 7 for the edges, so 1 x 7 x 7 = 49 starting grids turn all black.
- There are 512 equally likely starting grids in all.
- The probability is 49 divided by 512.
- That matches choice (A).
💡 Favorable grids over all grids gives the probability straight away.
8.G.A.1 Picture turning the grid 90 degrees clockwise. The center stays put. The top-lef 7.SP.C.7 Follow one square's fate. After the turn, some square from its loop lands on it. 7.SP.C.8 Because a square and the one that rotates onto it always live in the same loop, 7.SP.C.8 Take the 4 corners in their loop; two squares are neighbors in the loop if one t 7.SP.C.5 Put the pieces together: 1 way for the center, 7 for the corners, 7 for the edge Review
Reasonableness: The answer 49/512 is a fraction over 2^9, exactly the shape you expect when every one of the 9 squares is a fair coin flip, and 49 = 7 x 7 is a whole number of favorable grids, so the count is self-consistent. The value is about 0.096, a small but not tiny chance, which fits: needing the center black plus safe patterns in both loops is demanding but far from impossible. Choices (C) 121/1024 and (E) 9/32 do not sit over 512 with an integer top, so they cannot be a whole count out of 512; (A) does.
Alternative: Instead of the compact 1 + 4 + 2 count, list all 16 colorings of a 4-square loop and cross out any with two adjacent whites; exactly 7 survive (the Lucas-number pattern for a cycle of length 4). Multiplying 7 by 7 by the single forced-black center again gives 49 favorable grids out of 512.
CCSS standards used (min grade 8)
8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Tracking a 90 degree rotation to see that it fixes the center and cycles the corners in one loop and the edges in another.)7.SP.C.7Develop probability models and use them to find probabilities of events (Restating the paint rule as a condition on outcomes: a spot ends white exactly when it and the square rotating onto it both started white.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Splitting the grid into independent groups so favorable counts multiply, and systematically listing safe colorings of a 4-square loop.)7.SP.C.5Understand that the probability of a chance event is between 0 and 1 (Forming the final probability as favorable grids (49) over all equally likely grids (512).)
⭐ A quarter turn sorts the squares into a fixed center and two rings of four, so count each ring's safe patterns on its own and multiply.
⭐ A quarter turn sorts the squares into a fixed center and two rings of four, so count each ring's safe patterns on its own and multiply.
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