If there are 100 ducks arranged on the edge of a circle and each duck pecks either the duck directly to their left or their right what is the expected number of unpecked ducks

Let's label the ducks $1, 2, 3, \ldots, 100$ clockwise around the circle. For each duck $i$, let $X_i$ be the random variable that equals $1$ if duck $i$ is unpecked and equals $0$ otherwise. Then the total number of unpecked ducks is $X_1 + X_2 + \cdots + X_{100}$, and we want to find the expected value of this sum.

Notice that if a duck is unpecked, then the ducks immediately to its left and right must both peck the duck two spaces away in the opposite direction. For example, if duck $i$ is unpecked, then ducks $i-2$ and $i+2$ must both peck duck $i$. Conversely, if ducks $i-2$ and $i+2$ both peck duck $i$, then duck $i$ is unpecked.

Thus, for each $i$, we have $$ X_i = \begin{cases} 1 & \text{if ducks } i-2 \text{ and } i+2 \text{ both peck duck } i \ 0 & \text{otherwise} \end{cases} $$ Note that the indices $i-2$ and $i+2$ should be interpreted modulo $100$, so for example when $i=1$, we have $X_1 = 1$ if ducks $99$ and $3$ both peck duck $1$.

Let's compute the probability that $X_i = 1$. There are two cases to consider:

• If duck $i$ pecks the duck to its left, then the duck two spaces to the right of $i$ must peck duck $i$. Therefore, the only way that $X_i = 1$ in this case is if duck $i-2$ pecks duck $i+2$.
• If duck $i$ pecks the duck to its right, then the duck two spaces to the left of $i$ must peck duck $i$. Therefore, the only way that $X_i = 1$ in this case is if duck $i+2$ pecks duck $i-2$.

Thus, the probability that $X_i = 1$ is equal to the probability that either duck $i-2$ pecks duck $i+2$ or duck $i+2$ pecks duck $i-2$. Since there are $99$ other ducks besides $i$ in the circle, each of which is equally likely to peck $i$, the probability that a given one of these ducks pecks $i+2$ is $\frac{1}{2}$, and similarly for pecking $i-2$. Therefore, the probability that $X_i = 1$ is $\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$.

Finally, we can compute the expected value of the total number of unpecked ducks as follows: \begin{align*} \mathbb{E}[X_1 + X_2 + \cdots + X_{100}] &= \mathbb{E}[X_1] + \mathbb{E}[X_2] + \cdots + \mathbb{E}[X_{100}] \ &= 100 \cdot \frac{1}{2} \ &= \boxed{50}. \end{align*