I have found out to my surprise that the list of who did or did not vote is publically available in Texas.
My question is, can information regarding my preference be ascertained by the fact that I decided not to vote.
If I vote: my options are (a) YES, (b) NO or possible (c) Register to vote and then not actually vote.
If I decide not to vote - apparently the list is available to registered interest groups : NOT VOTING can mean (a) Im not interested in voting, (b) Im voting with the 'majority' of the electorate.
Somehow, it seems that having a list of specific people who decided to vote or not vote mathematically allows interested groups to gather more more information about me personally than if I went to vote and chose a/b/c (if indeed c is really an option)
I am offended that the list of who voted and who did not vote, has been made available in the public domain. So, is there any basis in mathematic probability for me to be concerned?
Denote by $A$ that you support candidate $A$, and denote by $V$ that you voted.
Your question amounts to:
Mathematically, we can use the total probability law: $$\begin{align*} P(A) &= P(A \mid V) P(V) + P(A\mid \neg V)P(\neg V) \\ &= P(A \mid V) P(V) + P(A\mid \neg V)(1-P(V) )\end{align*}$$ $$P(A\mid \neg V) = \frac{P(A)-P(A\mid V)P(V)}{1-P(V)}.$$
Using the poll data, we can compute some of these quantities:
$$P(A\mid \neg V) = \frac{\overbrace{P(A)}^{\substack{\color{green}{\textrm{infer from}} \\ \color{green}{\textrm{phone polls}}}}-\overbrace{P(A\mid V)}^{\color{red}{\textrm{actual election tally}}} \overbrace{P(V)}^{\color{blue}{\substack{\textrm{% of voting} \\ \textrm{registered voters}}}}}{1-\underbrace{P(V)}_{\color{blue}{\substack{\textrm{% of voting}\\\textrm{registered voters} }}}}.$$
The big stretch is computing $P(A)$, since it's basically impossible to determine. However, pre-election phone polling does sample from both people who do and do not vote, so it may be sufficiently accurate.