### Instruction: The vertices $V$ of a centrally symmetric hexagon in the complex plane are given by \[V=\left\{   \sqrt{2}i,-\sqrt{2}i, \frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}.\]For each $j$, $1\leq j\leq 12$, an element $z_j$ is chosen from $V$ at random, independently of the other choices. Let $P={\prod}_{j=1}^{12}z_j$ be the product of the $12$ numbers selected.

The probability that $P=-1$ can be expressed in the form
\[\frac{a}{p^b},\]where $a,$ $b,$ $p$ are positive integers, $p$ is prime, and $a$ is not divisible by $p.$  Find $a + b + p.$.### Response: To find the probability that $P=-1$, I need to count how many ways I can choose $12$ numbers from $V$ whose product is $-1$, and divide that by the total number of ways to choose $12$ numbers from $V$.
The total number of ways to choose $12$ numbers from $V$ is just $6^{12}$, since there are $6$ choices for each number and they are independent.
To find how many ways I can choose $12$ numbers whose product is $-1$, I need to think about the possible values of the product and how they change as I multiply by different numbers from $V$.
I notice that the numbers in $V$ are all of the form $\pm\frac{1}{\sqrt{2}}(\cos\theta + i\sin\theta)$ for some angle $\theta$.
This means that when I multiply two numbers from $V$, their product is also of the same form, but the angle is the sum of their angles (modulo $2\pi$).
In fact, the angle of the product is just the sum of the angles of the factors, since adding multiples of $2\pi$ does not change the value of the product.
So, to get $-1$ as the product, I need the sum of the angles of the factors to be equal to $\pi$, since $-1 = \frac{1}{\sqrt{2}}(\cos\pi + i\sin\pi)$.
I also notice that the angles of the numbers in $V$ are multiples of $\frac{\pi}{4}$, so the sum of the angles of the factors will also be a multiple of $\frac{\pi}{4}$.
This means that I need to have exactly $4$ factors whose angle is $\pi$, and the rest of the factors can have any angle.
There are only two numbers in $V$ whose angle is $\pi$, namely $\sqrt{2}i$ and $-\sqrt{2}i$.
So, I need to choose exactly $4$ of these two numbers, and the rest can be any of the other $4$ numbers in $V$.
The number of ways to choose $4$ numbers from $\{\sqrt{2}i, -\sqrt{2}i\}$ is $\binom{4}{0} + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 16$.
The number of ways to choose the remaining $8$ numbers from the other $4$ numbers in $V$ is $4^8$.
So, the total number of ways to choose $12$ numbers from $V$ whose product is $-1$ is $16\cdot 4^8$.
Therefore, the probability that $P=-1$ is $\frac{16\cdot 4^8}{6^{12}} = \frac{2^4\cdot 2^{16}}{2^{12}\cdot 3^{12}} = \frac{2^8}{3^{12}} = \frac{256}{531441}$.
This is already in the form $\frac{a}{p^b}$, where $a=256$, $p=3$, and $b=12$.
So, the answer is $a+b+p = 256 + 12 + 3 = 271$.