Given $s,t\in(0,1)$, $c>1$, $n\in\Bbb N$, pick ${n^t}$ random vectors $\{v\_i\}\_{i=1}^{{n^t}}$ such that each $v\_i\in\{x\in\{0,1\}^{2^n}:|x|\_{hamming}={2^{n-n^s}}\}$.


Denote $v\_j\cap v\_j$ to be vector that is $1$ at a location iff both $v\_i,v\_j$ are $1$s at that location. Denote $v=v\_1\cap v\_2\cap\dots\cap v\_{{n^t}-1}\cap v\_{{n^t}}$.


Is there is a $c>1$ such that $$\Bbb{Pr}(|v|\_{hamming}<2^{{n}/c})>\frac{1}2?$$



Probability of Hamming weight
The probability that a fixed entry of $v$ is 1 equals $2^{-n^{s+t}}$. Hence the expected Hamming weight of $v$ is $2^{n-n^{s+t}}$. If $s+t\geq 1$, this implies that with high probability the Hamming weight of $v$ is much smaller than $2^{n/c}$. If $s+t<1-\varepsilon$, use the second moment trick: If the first coordinate of $v\_1$ is 1, the conditional probability that the second is 1 is $\frac{2^{n−n^s}-1}{2^n-1}<2^{-n^s}$, hence the variance of the hamming weight of $v$ is slightly smaller than the variance of a sum of independent random variables would be. Hence $|v|$ is larger than $2^{(1-\varepsilon)n}$ with probability tending to 1. 


