Probability that a random voter voted for the same candidate twice Let's say there are only two candidates(Candidate A & Candidate B) in an election and both of them ran for office in both elections.  Candidate A received 48% of the vote in the first election and 51% of the vote in the second election.   We will assume that there is a fixed group of voters and all of them voted in both elections. So, if 100 people voted in first election, exact same 100 people also voted in second election.  Let's also assume a correlation coefficient of .8 between the voting habit of a voter(voting for the same or different candidate) in the first and second election. What is the probability that a randomly chosen voter voted for Candidate A in both elections?
N.B:- This is the same question as posted here https://www.quora.com/Bush-received-48-51-of-the-total-votes-in-two-elections-Whats-the-probability-that-a-randomly-chosen-voter-voted-for-Bush-in-both-the-elections . I am trying to better understand the answer posted by Aaron Brown as I dont understand the details of his answer. 
 A: You can express it as a contingency table:
$\begin{array}{cc}
\begin{array}{cc|cc}
&& \text{1st Bush} & \text{1st others} &  \\
&100& 48 & 52 \\\hline
\text{2nd Bush}& 51& a & b \\
\text{2nd others} & 49 & c & d \\
\end{array}
\end{array}$
Due to all the restrictions (everything needs to add up to get the margins) you can bring the $a,b,c,d$ down to a single parameter
$\begin{array}{cc}
\begin{array}{cc|cc}
&& \text{1st Bush} & \text{1st others} &  \\
&100& 48 & 52 \\\hline
\text{2nd Bush}& 51& a & 51-a \\
\text{2nd others} & 49 & 48-a & a+1 \\
\end{array}
\end{array}$
So in any case you will need some additional information (wheter or not this is some vaguely defined correlation or not) to express the size of the group that voted for Bush in both elections.

In terms of Aaron Brown's comment on quora you have 
$$a/100 = \frac{\rho+0.98}{4} \quad $$
I am not sure what type/definition of correlation relates to that. It is not corresponding to a later statement: 

If the events were independent then ρ would be zero and the fraction that voted for Bush both times would be 0.2448 which happens to equal 0.48×0.51. However, this would be very surprising.

because $\frac{0.98}{4} \neq 0.2448$. 
Instead, the case that the fraction that voted for Bush both times would be 0.48×0.51=0.2448 occurs when the phi coefficient is zero
A: I don't understand the answer provided in the link. So I simulated this scenario in R and I got a different result, which leads me to believe that the answer in the link might be wrong.
Note: I choose to interpret the statement "a correlation coefficient of 0.8" to really mean "a probability of 0.8".
mean(replicate(1e5,{
  tmp=sample(c("A","B"),48,prob=c(0.8,0.2),replace=T)
  sum(tmp==rep("A",48))/100
}))

[1] 0.3839619

