Suppose we have 50 black balls and 50 white balls arranged randomly in a circle. Find the expected number of black balls such that at least one of its neighbors is white.
Proposed solution: Let $X_i$ be the random variable taking value $1$ if the i-th black ball has a white neighbor and $0$ otherwise. Then by linearity of expectation, our desired quantity is $50 E[X_i]$. Now \begin{align} E[X_i] &= 1 - P(\text{both neighbors are black})\\ &=1-\frac{49*48}{100^2}=0.7648$. \end{align} Thus, the desired quantity is equal to $38.24$.
Is this the correct approach? Could anyone point to the flaws?
Correction: the probability calculation is not right based on the comments. The correct one is \begin{align} 1 - P(\text{both neighbors are black})&=\\ &=1- \frac{49*48}{99*98} \\ &=0.7575(75) \end{align} Hence, the correct answer is indeed $37.87$ or $1250/33$.