# What is the probability of having distinct grid squares?

Problem

Consider a pair of $2 \times 2$ grids. Each cell of each grid is colored either black or white, with equal probability. Any two colorings are considered non-distinct if they only differ by a rotation. For example, the two colorings shown are non-distinct. If the two grids of four squares are colored at random, what is the probability that the two resulting colorings are distinct?

Attempt

From here on I will use the word $n$-black square to refer to a square with exactly $n$ black squares.

I solved this question by considering all possible permutations. I will not consider the cases of having an $n$-black square and an $n$-black square be distinct until after I count the cases of having an $n$-black square and a non $n$-black square (not equal to $n$) distinct. We have for case $1$: $0$ black this means that the other square can have $1,2,3$ or $4$ black for a total of $\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}$. Case 2: $1$ black this means we can have by similar reasoning a total of $\binom{4}{0}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}$. Case 3: $2$ black we have a total of $\binom{4}{0}+\binom{4}{1}+\binom{4}{3}+\binom{4}{4}$. Case 4: $3$-black we have a total of $\binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{4}$. Case 5: $4$-black we have $\binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{3}$.

Now to count the cases of having an $n$-black square and an $n$-black square be distinct, note that it is impossible for $0-$black and $1$-black. For a $2$-black it is $2*4+4*2 = 16$. For a $3$-black it is $4$ and there exist none for a $4$-black square.

Therefore, the answer I get after adding is $\dfrac{87}{256}$. This was not one of the answer choices, though.

Edit: I solved the question using complementary counting and got the correct answer of $\dfrac{101}{128}$, but what did I do wrong here?

## 2 Answers

for 3-black there are no distinct pairs ( exactly the same as 1-black)

• But I counted $4$. If what you were saying were the case, that would just get me farther away from the correct answer. – Puzzled417 Dec 13 '15 at 15:36

There are a total of $256$ colourings for the $2$ grids. The only time you need to think about the same number of blacks is $n=2$ for which there are $4\times 2+2\times 4=16$ distinct colourings

$$\begin{array}{cr} \\ \text{ # of Blacks} & \text{ # of distinct ways} \\ 0 & 1 \times 15 = 15 \\1 & 4 \times 12 = 48 \\2 & 6 \times 10 + 16 = 76 \\3&48 \\4&15 \end{array}$$

Summing these possibilities gives $202$, which agrees with your correct answer.