Coin flips and binomial distribution Consider a sequence of $100$ flips of a fair coin. Let $X$ denote the number of pairs of heads in this sequence, i.e. $HH$. Notice that we also consider the pair $X_{100}X_1$ (like a cyclic ordering). Am I correct that $X$ is described by a binomial distribution since the trials are independent and fixed, only two outcomes, ...?
 A: I don't think it's binomial because each flip is not the same Bernoulli.
I am here defining trial $n$ to be the pair of flips $n$ and $n+1$, (mod $100$ since if $n= 100$, then you loop back to $1$). Success here means "pair of heads" and failure is "anything else."
Suppose flip $n$ is heads. Then the probability of trial $n$ being a success is $1/2$, since it's all dependent on flip $n+1$. However, if flip $n$ is tails, then the probability of trial $n$ being a success is $0$, since flip $n+1$ cannot make a pair of heads.
For this reason, since the trials don't have a fixed probability of success, I think it's not binomial.
A: Suppose $N = 2.$ Then  $P(HH) = 1/4$
Suppose $N = 3$. Then of the $8$ possibilities, $HHH$ has $3$ pairs, $3$ options have $1$ pair, and $4$ options have no pairs.
Suppose $N = 4$. Then of the $16$ possibilities, $1$ has $4$ pairs, $4$ have $2$ pairs, of the $6$ ways to have $2H2T$, $4$ of them have $1$ pair, and none of the ways with $1$ or $0$ heads have any pairs.
This is not binomial.
