The expected value of $n$ dependent coins? Given $n$ dependent coins $c_1, c_2, c_3, \dots, c_n,$ assume that
$$c_1 \textrm{ is head}\implies c_n\textrm{ is tail}$$
but $P(c_1\textrm{ is head})=P(c_n\textrm{ is tail})=1/2,$ so they look fair separately.
Let $X = \textrm{# heads after toss all coins}$, can be decomposed into $X_1+X_2+\dots+X_n$ where each $X_i$ represents the number of heads appear on i-th coin, then
$$E(X) = E(\sum_{i=1}^{n}X_i)=\sum_{i=1}^{n}E(X_i)=n\cdot1/2,$$
which is the same as the value when these coins are independent? Why this is possible?
 A: Yes if also $c_n$ must be head if $c_1$ is tail.
If that is not the case then coin $c_n$ does not have probability $\frac12$ to be head.
If so then in this situation $X_1+X_n=1$.
A: Simplify the problem. Consider just two coins which each behave as if unbiased (when looked at individually), yet the second is dependent on the first such that it always shows tails when the first shows heads. 
Let $X_1,X_2$ indicate when they show heads.   Then we have been told that: $$\mathsf P(X_1{=}1)=\tfrac 12\\ \mathsf P(X_2{=}1)=\tfrac 12\\ \mathsf P(X_2{=}0\mid X_1{=}1)=1$$
Now, by the Law of Total Probability we can show $\mathsf P(X_2{=}1\mid X_1{=}0)=1$ $$\begin{split}\mathsf P(X_2{=}1) &= \mathsf P(X_1{=}1, X_2{=}1)+\mathsf P(X_1{=}0, X_2{=}1) \\&=\mathsf P(X_1{=}1)\mathsf P(X_2{=}1\mid X_1{=}1)+\mathsf P(X_1{=}0)\mathsf P(X_2{=}1\mid X_1{=}0)\\\tfrac 12&= \tfrac 12\cdot 0+\tfrac 12\cdot\mathsf P(X_2{=}1\mid X_1{=}0)\end{split}$$
Therefore the expectation shall be:
$$\begin{align}\mathsf E(X_1+X_2) &= (0{+}0)\mathsf P(X_1{=}0,X_2{=}0)+(0{+}1)\mathsf P(X_1{=}0,X_2{=}1)+(1{+}0)\mathsf P(X_1{=}1,X_2{=}0)+(1{+}1)\mathsf P(X_1{=}1,X_2{=}1)\\&=0\cdot 0+1\cdot\tfrac 12+1\cdot\tfrac 12+2\cdot 0\\ & = 1\end{align}$$
You should be able to see why that is equal to $\mathsf E(X_1)+\mathsf E(X_2)$
A: You need to take into account the dependance between $c_1$ and $c_n$:
$\mathbb{E}[X_n]=\mathbb{E}[X_n|X_1=1]\mathbb{P}[X_1=1]+\mathbb{E}[X_n|X_1=0]\mathbb{P}[X_1=0]=1\cdot\frac{1}{2}+\frac{1}{2}\cdot \frac{1}{2}=\frac{3}{4} \neq \frac{1}{2}$
where I used the law of total probability
