Let the number of such configurations be n. Consider all possible configurations of chicks if chick A and the two adjacent chicks to A were removed. Let's focus our attention on one particular chick, say chick A. Consider the set of all possible orientations of the chicks, essentially, all the different ways the chicks can peck each other. I shall solve the problem generally, let the total number of chicks be k, such that k > 3 (the reason behind this shall become apparent soon). My explanation only works when the total number of chicks is positive and divisible by $4$. Each state is equally likely and this means that $25$ has to be the expected number of unpecked chicks. I believe this involution gives us a one to one correspondence between states with $x$ unpecked chicks and states with $50-x$ unpecked chicks. Change the directions of each chick that is divisible by $4$ and each chick that is congruent to $1 \mod 4$. So you number the chicks $1$ to $100$ such that consecutive numbers are adjacent to each other and $1$ is next to $100$. If we can show that having $x$ unpecked chicks is equally likely as having $50-x$ unpecked chicks, then the average of both scenarios gives you $25$ unpecked chicks. $51$ unpecked chicks is impossible because then there would be no way to have $100$ pecks totally. It's also possible to have $0$ unpecked chicks and any integer between $0$ and $50$ is possible. It is possible to have a state with $50$ unpecked chicks. But $E$ is just the probability that the $i$th chicken is unpecked, thus justifying our reasoning above. Then the total number of unpecked chickens is $\sum X_i$, so the expected number of unpecked chickens isīy linearity of expectation. What we're formally doing above is assigning a random variable $X_i$ to the $i$th chick that is equal to $0$ if the chick is pecked and $1$ if it's unpecked. This tells us that for any random variables $X$ and $Y$, independent or not, $E=E E$, where $E$ is the expected value. Surely we have to take this into account!"įortunately, there's a very useful theorem called linearity of expectation. "But wait," you might say, "the probabilities that chicks are unpecked aren't independent! If chick $n$ is unpecked, then chicks $n-2$ and $n 2$ will be pecked. Since this is true for every chick, we can add up $0.25(100)=25$ to get the number of unpecked chicks. So overall, it has a $0.25$ chance of not being pecked. For any individual chick, there is a $0.5$ chance that the one on its right won't peck it, and a $0.5$ chance that the one on its left won't peck it.
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