We consider the $n$-dimenstional finite vector space $\mathbb{F}\_2^{n}$ over the finite field of two elements. For a subset $A\subseteq \mathbb{F}\_2^{n}$ of even size $|A|=2m$ and a linear form $l\in(\mathbb{F}\_2^{n})^\*$ let us say that *$l$ bisects $A$*, if
$$|\{a\in A\ |\ l(a)=0\}|=|\{a\in A\ |\ l(a)=1\}|\,\,\,\,(=m).
$$
Equivalently, for a subspace $L\leq \mathbb{F}\_2^{n}$ of dimension $n-1$ (i.e. $|L|=2^{n-1}$), let us say that *$L$ bisects $A$*, if
$$|A\cap L|=|A\setminus L|\,\,\,\,(=m).
$$
**How many subsets $A\subseteq \mathbb{F}\_2^{n}$ of size $|A|=2m$ are there which are bisected by some suitable linear form (equivalently: by some subspace of dimension $n-1$)?**


An alternative problem which I'm interested in (and which is maybe easier to solve??) is:


**Determine the limit
$$\lim\_{n\rightarrow\infty}\frac{|\{A\subseteq \mathbb{F}\_2^{n}|\,|A| \text{ is even and $A$ is bisected}\}|}{|\{A\subseteq \mathbb{F}\_2^{n}|\,|A| \text{ is even}\}|}.
$$**
(Clearly, $|\{A\subseteq \mathbb{F}\_2^{n}|\,|A| \text{ is even}\}|=2^{2^n-1}$.)


(My calculations seem to support the conjecture that the limit equals 1.)



Number of linearly bisected subsets in finite vector space $F_2^n$
The probability that a random subset is bisected by a fixed linear subspace is


$$ \frac{ \sum\_{k=0}^{2^{n-1}}\binom{2^{n-1}}{k}^2}{2^{2^n-1}} = \frac{\binom{2^n}{2^{n-1}} }{2^{2^n-1} } $$


Using the asymptotic 


$$\binom{N}{N/2} \approx \frac{ C 2^{N}} { \sqrt N} $$


for some constant $C$, we obtain


$$\approx \frac{2 C}{ \sqrt{ 2^{n} } }$$


for some constant $C$. (By $\approx$ I mean they are equal up to multiplication by $1+o(1)$ )


There are $2^n-1$ linear subspaces, so the expected total number of bisecting subsets is large, roughly $2C 2^{n/2}$.


Given two transverse subspaces, they divide $\mathbb F\_2^n$ into four equal sets, and the probability that they both bisect is 


$$\frac{\sum\_{k=0}^{2^{n-2}} \sum\_{l=0}^{2^{n-2}}\binom{2^{n-2}}{k}^2\binom{2^{n-2}}{l}^2}{ 2^{2^n-1}} =\frac{\binom{2^{n-1}}{2^{n-2}}^2 }{2^{2^n-1} } $$


using the same asymptotic:


$$ \approx \frac{ \left( \frac{ C 2^{2^{n-1}}}{\sqrt{ 2^{n-1}}}\right)^2 } { 2^{2^n-1}} = \frac{ 4 C^2} { 2^n } $$


So the probabilities that two different linear subspaces bisect a set are approximately independent. This means that the expected value of the square of the number of linear subspaces that bisect is roughly $( 2 C 2^{n/2} )^2$.


Let $X$ be this number, viewed as a random variable.


Because $E[X^2] \approx E[X]^2$, the variance of $X$ is $o ( E[X]^2)$, so the standard deviation is $o(E[X])$ so by Chebyshev's inequality the probability that $X$ is $0$ is $o(1)$.


This implies that the limiting probability of at least one bisecting subspace exists approaches $1$, as you predict.


