The fraction of k-juntas with low influences in all of the coordinates Let $f:\{-1,1\}^n\to\{-1,1\}$ be a boolean function.
Define the influence of the $i$'th coordinate of $f$ as follows:
$$\operatorname{Inf}_i(f)=\Pr_{x}[f(x)\neq f(\hat x_i)]$$
where $x$ is uniformly picked from $\{-1,1\}^n$, and $\hat x_i$ is $x$ with its $i$'th coordinate flipped (e.g., say $x=(1,1,1,1,-1)$, then $\hat x_3=(1,1,-1,1,-1)$).
Denote by $J_k$ the set of all the k-juntas for which the influencing variables are the first $k$ variables. That is, for each $f\in J_k$, the function $f$ is a boolean function that holds $Inf_i(f)>0$ for $1\leq i\leq k$ and $Inf_i(f)=0$ for $i>k$.
Let $0\leq \epsilon \leq 1$. What is the probability (over uniformly selecting such a k-junta) that the influence in each influencing coordinate of the junta will be less than $\epsilon$ ? Formally:
$$\Pr_{f\in J_k}[\forall i\quad Inf_i(f)< \epsilon]=?$$ 
 A: I'm afraid the marginal distribution for the influence of a single coordinate is the easier part :-) I gave up on this problem because the interrelations between the influences of different coordinates seem too complicated. But if you're interested in the marginal distribution, here's why it's binomial: A given influence for the $i$-th coordinate corresponds to a given number $m$ of pairs of points differing only in the $i$-th coordinate that have different function values. There are $2^{k-1}\choose m$ different ways of choosing the remaining $k-1$ coordinates for these $m$ pairs. For each pair with different function values, there are $2$ possibilities, $(-1,1)$ and $(1,-1)$, and for each pair with the same function values, there are also $2$ possibilities, $(-1,-1)$ and $(1,1)$. Thus the factor $2^{2^{k-1}}$ arising from these possibilities is independent of $m$ and cancels out in the probabilities, which are therefore completely determined by the number $2^{k-1}\choose m$ of choices for the coordinates for the pairs.
Here's another idea I tried: We can write the influence of the $i$-th coordinate as
$$\operatorname{Inf}_i(f)=\Pr_{x}[f(x)\neq f(\hat x_i)]=\frac{1-2^{-n}\sum_xf(x)f(\hat x_i)}2\;.$$
With the Walsh–Fourier transform of $f$, 
$$f(x)=\sum_j a_jf_j(x)\;,$$
this becomes 
$$
\begin{align}
\operatorname{Inf}_i(f)
&=\frac{1-2^{-n}\sum_x(\sum_j a_jf_j(x))(\sum_{j'} a_{j'}f_{j'}(\hat x_i))}2\\
&=\frac{1-2^{-n}\sum_j\sum_{j'}a_ja_{j'}\sum_xf_j(x)f_{j'}(\hat x_i)}2\\
&=\frac{1-2^{-n}\sum_j\sum_{j'}a_ja_{j'}\sum_xf_j(x)(\pm f_{j'}(x))}2\\
&=\frac{1-\sum_j(\pm a_j^2)}2\;,\\
\end{align}
$$
where the sign depends on whether $f_j$ changes sign with $x_i$.
Unfortunately I think this doesn't get us much closer to solving the problem, because although the $a_j$ are linearly uncorrelated, they're not independent and their squares are highly correlated.
