Crow probability question Twenty crows land randomly on a wire. Each crow is crowing at the nearest crow.
What is the expected number of crows that are not crowed at? 
I truly have no idea how to approach this problem.
I was thinking along the lines that the end crows each have probability $\frac12$ of not being crowed at. The penultimate crows on each end have $0$ probability of not being crowed at. For each "interior" crow, the probability of not being crowed at is $\frac14$. So the expected number of crows that are not crowed at is $2\cdot\frac12+16\cdot\frac14=5$.
 A: Put the first crow in the wire. Now the second, at distance $\delta_1$, taken from the random variable $\Delta$ (never found myself using $\Delta$ for a random variable before). As stated, this will be crowed at with probability $1$, so if $C_n$ is the event of the $n^{th}$ crow being crowed at, $$P(C_2)=1$$ 
Now take a $\delta_2$ for determining the distance between the second and the third crow. $P(C_1)=P(\Delta>\delta_1)$. Since $\delta1$ comes from $\Delta$, $$P(C_1)=P(\Delta>\Delta')=\frac{1}{2}$$ 
Now take a $\delta_3$ for determining the distance between the third and the fourth crow, and so on. For any interior crow, then, $$P(C_n)=P(\delta_n<\delta_{n-1}\cup\delta_n<\delta_{n+1})=\\
P(\delta_n<\delta_{n-1})+P(\delta_n<\delta_{n+1})-P(\delta_n<\delta_{n-1}\cap\delta_n<\delta_{n+1})=\\
\frac{1}{2}+\frac{1}{2}-P(\Delta<\Delta'\cap\Delta<\Delta'')=1-\frac{2}{6}=\frac{2}{3}
$$
For the last part of the calculus, you basicaly have three random variables following the same distribution, so you have six equiprobable ways to order $\Delta, \Delta', \Delta''$, and only in two of them $\Delta$ is the smallest.
And finally,
 $$P(C_{19})=P(C_2)=1\\
P(C_{20})=P(C_1)=\frac{1}{2}
$$
For the expected number, let's sum all this.
$$E[C]=2\frac{1}{2}+2+16\frac{2}{3}=\frac{41}{3}\\
E[C^C]=20-E[C]=\frac{19}{3}\approx 6.3
$$
And this is it, the expected number of uncrowed crows is $6.3$
