Is my understanding of CPA indistinguishability experiment correct? Following are steps to conduct Chosen Plaintext Attack (CPA) indistinguishibility experiment $PrivK_\mathcal{A,E}^{eav}(n)$.

$\mathcal{A}$ is the adversary $\mathcal{E}$ is the encryption scheme (Gen, Enc, Dec), $n$ is the length of the input.  


a secret key SK is generated using Gen($1^n)$
$\mathcal{A}$ is given input of $1^n$ and oracle access to $Enc_{sk}(.)$ and outputs a pair of messages $m_0$ and $m_1$ of the same length.
A random bit $b$ is chosen.  $Enc_{sk}(m_b)$ is computed to form $challenge$ ciphertext $C$ given to $\mathcal{A}$. 
$\mathcal{A}$ continues to have oracle access to $Enc_{sk}(.)$ and outputs another bit $b'$.
If $b'== b$ then $\mathcal{A}$ succeeded, otherwise $\mathcal{A}$ lost the game.


To better understand the experiment above, I decided to run it with simple values of my choosing: 


secret key SK 10101010 generated with length n=8 by defender (one who wants security)
$\mathcal{A}$ is given an input $1^n$ =  11101110 by defender.  Using the length of this input, $\mathcal{A}$ comes up with $m_0 =  11111111 $ and $m_1$ = 00000000
Defender randomly selects $b$ = 0 and encrypts $Enc_{sk}(m_b)$ = $Enc_{sk}(m_0)$ = $Enc_{sk}(11111111)$ = 10111011 = $C$ challenge and given to $\mathcal{A}$
Using $C$ = 10111011 $\mathcal{A}$ tries to guess if it came from $m_0$ or $m_1$.  Guesses $b'$ = 1.  This means $\mathcal{A}$ lost the game. 


My understanding is the adversary continues to pick different $m_0$ and $m_1$ values and has the defender encrypt them until he starts guessing correctly over 1/2 the time. 

Is this a correct understanding or am I missing something?
There are a surprisingly large number of subtly different definitions of CPA indistinguishability.  What you describe in points 1 through 5 is one that I have heard of, though at the end of your post you make it 'iterative', in the sense that $\mathcal{A}$ can play the game over and over with $\mathcal{E}$ using the same secret key but fresh, independently-random values for $b$ for each game.  I have not heard of the IND game being set up iteratively like that -- usually $\mathcal{A}$ only gets one bite at the apple in terms of guessing $b$. Are you sure you meant to define it that way?

You also don't really define what it means for $\mathcal{A}$ to "continue to have access to the oracle" in point 4.  Do you mean $\mathcal{A}$ can make more chosen plaintext queries before it has to guess $b$?  If so, do they have to be of the form $(m_0, m_1)$ where only the encryption of $m_b$ is returned to it, or can they be individual messages?  

Granting $\mathcal{A}$ up to $q_e$ queries of the form $(m_0, m_1)$ (i.e. so that every query is a 'challenge') before it has to make its guess for the value of $b$ is the more common Left Or Right (LOR) notion of IND-CPA.  BDJR'97 defines LOR IND-CPA (as well as several other notions of CPA security):
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.117.4734