How many power series of the form $1+\sum_{k=1}^{\infty} a_{k}x^{k}$ with $a_{k}\in \{1,0,1 \}$, that have a double zero $f(x)=f'(x)=0$ in $(0,1)$, are there. Ok, there are many ways to understand the question: set theoretical, topological, measure theoretical. I would be especially interested in the Bernoulli measures of the coefficient space $C\subseteq \{1,0,1\}^{\mathbb{N}}$ of such series.

At least the settheoretical question can be answered: the are the cardinality of the continuum many such series, as can be deduced from the results in this paper (not all of them attributed by the authors to themselves): MR2293600 (2007k:30003) Reviewed Shmerkin, Pablo(FINJVSMS); Solomyak, Boris(1WA) Zeros of {−1,0,1} power series and connectedness loci for selfaffine sets. (English summary) Experiment. Math. 15 (2006), no. 4, 499–511. 


Some more examples with polynomials: $$\matrix{\left( {z}^{6}+{z}^{5}{z}^{3}+z+1 \right) \left( z+{z}^{4}1 \right) ^{2}\cr \left( {z}^{8}+{z}^{7}{z}^{5}{z}^{4}{z}^{3}+z+1 \right) \left( z+ {z}^{6}1 \right) ^{2}\cr \left( {z}^{9}+{z}^{8}{z}^{6}{z}^{5}{z}^{4}{z}^{3}+z+1 \right) \left( z+{z}^{7}1 \right) ^{2}\cr \left( {z}^{4}{z}^{3}+{z}^{2}z+1 \right) \left( {z}^{2}+{z}^{5}1 \right) ^{2}\cr \left( {z}^{6}{z}^{5}+{z}^{4}{z}^{3}+{z}^{2}z+1 \right) \left( {z }^{2}+{z}^{7}1 \right) ^{2}\cr \left( {z}^{6}{z}^{5}+{z}^{3}z+1 \right) \left( {z}^{3}+{z}^{4}1 \right) ^{2}\cr \left( {z}^{7}{z}^{5}+{z}^{4}+{z}^{3}{z}^{2}+1 \right) \left( {z}^ {3}+{z}^{5}1 \right) ^{2}\cr \left( {z}^{10}+{z}^{8}{z}^{7}+{z}^{6}+{z}^{5}2\;{z}^{4}+{z}^{3}z +1 \right) \left( {z}^{3}+{z}^{7}1 \right) ^{2}\cr \left( {z}^{4}{z}^{3}+{z}^{2}z+1 \right) \left( {z}^{4}+{z}^{5}1 \right) ^{2}\cr \left( {z}^{4}{z}^{2}+1 \right) \left( {z}^{4}+{z}^{6}1 \right) ^{ 2}\cr \left( {z}^{6}{z}^{5}+{z}^{4}{z}^{3}+{z}^{2}z+1 \right) \left( {z }^{4}+{z}^{7}1 \right) ^{2}\cr \left( {z}^{8}{z}^{7}+{z}^{5}{z}^{4}+{z}^{3}z+1 \right) \left( {z }^{5}+{z}^{6}1 \right) ^{2}\cr \left( {z}^{6}{z}^{5}+{z}^{4}{z}^{3}+{z}^{2}z+1 \right) \left( {z }^{6}+{z}^{7}1 \right) ^{2}\cr \left( {z}^{14}{z}^{13}2\;{z}^{12}{z}^{11}+{z}^{9}+2\;{z}^{8}2\;{z}^{5}+2\;{z}^{2}+z+1 \right) \left( {z}^{5}{z}^{3}+{z}^{2}+z1 \right) ^{2}\cr \left( {z}^{5}+{z}^{4}{z}^{3}{z}^{2}+z+1 \right) \left( {z}^{5}+{z }^{3}{z}^{2}+z1 \right) ^{2}\cr \left( {z}^{15}+{z}^{14}{z}^{11}{z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z }^{6}{z}^{5}{z}^{4}+z+1 \right) \left( {z}^{5}{z}^{4}+{z}^{3}+z1 \right) ^{2}\cr }$$ 


I am going to address the question for $\mathrm{Bernoulli}(1/2)$ measures, using probabilistic language. This is not a complete answer, but I am trying to relate your question to the properties of the distribution of $f(x)$. Clearly, for $x<1/2$ we never even reach zero, but my guess is that for $x>1/2$ this distribution is absolutely continuous, though I am unable to prove this at the moment. So formally, at least, $$\displaystyle \mathsf{E} \, \sum_{f(x)=0} \mathsf{1}\{f^\prime(x) < \epsilon\} = \intop_0^1 \mathsf{E} \, \delta(f(x)) \mathsf{1}\{f(x)<\epsilon\} f^\prime(x) dx \le \epsilon \intop_0^1 \mathsf{E} \, \delta(f(x)) dx.$$ $\mathsf{E} \, \delta$ is the density at zero, and it can be made perfect sense of, provided that the law of $f(x)$ has continuous density at zero. I don't know whether it has continuous density, but if we manage to prove that $f(x)$ has at least bounded density for $x>1/2$, then we can write inequalities with approximations of $\delta$ to get the same results... 

