The coordinate difference of a nearest neighbor in a random set Consider the following problem:
$r$ vectors of length $t$ are drawn randomly, where each coordinate is an i.i.d Bernoulli random variable with success probability $$p_i , i=1...t.$$
What is the probability that the closest vector to the origin will have 1 in coordinate i?
In case there are multiple closest vectors, pick one uniformly.
The exact expression, for the simplified case $$p_i=0.5$$  is as follows:
$$\frac{1}{r}\left(\frac{1}{2}\right)^{rt}\sum_{k=1}^{t}k\sum_{z=1}^{r}{r \choose z}{t \choose k}^{z}\left(\sum_{k'=k+1}^{t}{t \choose k'}\right)^{r-z}$$
I'm interested in either a simplified expression, or the large scale dependecy on $r,t$. 
 A: In the simplified case, since all the coordinates are symmetric, if the closest vector has $k$ coordinates equal to $1$, then the probability that the $i$th coordinate has value $1$ is equal to $\frac{k}{t}$.  Hence, if we let $X$ denote the number of $1$'s in the closest vector, the probability that the $i$th coordinate is equal to $1$ is given by $\sum_{k=0}^t \frac{k}{t} \mathbb{P}(X = k) = \frac{1}{t} \sum_{k=0}^t k \mathbb{P}(X = k) = \frac{1}{t} \mathbb{E}[X]$.
Now the closest vector is the one with the fewest coordinates equal to $1$.  Let $X_j$ denote the number of $1$'s in the $j$th vector.  Then $X_j \sim \mathrm{Bin}\left(t,\frac12 \right)$, and $X = \min (X_1, X_2, ..., X_r)$.
We can use the formula 
$$\mathbb{E}[X] = \sum_{k \ge 1} \mathbb{P}(X \ge k) = \sum_{k=1}^t \mathbb{P}( \min(X_1, X_2, ..., X_r) \ge k) = \sum_{k=1}^t \mathbb{P}(X_1 \ge k, X_2 \ge k, ..., X_r \ge k) = \sum_{k=1}^t \mathbb{P}(X_1 \ge k)^r.$$
Hence the probability that the $i$th coordinate of the closest vector is equal to $1$ is $\frac{1}{t} \sum_{k=1}^t \mathbb{P}(X_1 \ge k)^r$, where $X_1 \sim \mathrm{Bin}\left(t,\frac12 \right)$.
To simplify things a little, we may use the normal approximation for the binomial distribution.  Note that when $t$ is large, $\mathrm{Bin}\left(t, \frac12 \right)$ converges in distribution to $\mathrm{N} \left( \frac{t}{2}, \frac{t}{4} \right)$.  If $\Phi(z) = \mathbb{P}(Z \le z)$, where $Z \sim N(0,1)$ is a standard normal, then $\mathbb{P}(X_1 \ge k) \approx 1 - \Phi \left(2 \left(k - \frac{t}{2} \right) / \sqrt{t} \right)$.  If we let $x = \frac{k}{t} \in [0,1]$, then $\mathbb{P}(X_1 \ge k) \approx 1 - \Phi\left( (2x - 1) \sqrt{t} \right)$.
Hence the probability that the $i$th coordinate of the closest vector to the origin is equal to $1$ can be approximated by $\frac{1}{t} \sum_{k=1}^t \left(1 - \Phi \left( (2x - 1) \sqrt{t} \right) \right)^r$, where $x = \frac{k}{t}$.  One could use this to get some sort of asymptotic estimates.  For example, if $r$ is fixed and $t \rightarrow \infty$, I think you would get a probability of the form $\frac12 - \Theta_r \left( t^{- \frac12} \right)$.
