Conditional probability of a zero inner product Consider a random $n$ by $n$ matrix $M$ chosen uniformly over all $n$ by $n$ $(0,1)$ matrices and a random vector $v \in \{-1,0,1\}^n$ chosen uniformly as well. Let $X = Mv$.  
What is $$P(X_i = 0 \mid \forall j < i \; X_j = 0 )\;?$$
Although the rows of $M$ are independent, this probability does not equal $P(X_i = 0)$ which is where I am stuck.
 A: Since the rows of $M$ are independent, you're basically conducting $n$ independent probes of $v$. If $v$ has $k$ entries $+1$ and $l$ entries $-1$, the probability of its product with a row $r$ of $M$ being $0$ is $2^{-(k+l)}$ times the number of ways of selecting $k$ of the $k+l$ non-zero entries in $v$ and choosing $r$ to be $1$ when (entry is selected) XOR (entry is $+1$). (If this seems a bit complicated and out of the blue, you can unpack it as choosing $j$ entries that are $+1$ in $v$ and $j$ entries that are $-1$ in $v$ to be $1$ in $r$, summing over $j$ and then applying Vandermonde's identity.) Taking into account the  a priori probability for $k$ entries to be $+1$ and $l$ entries to be $-1$, we get for the probability you're seeking
$$P(X_i = 0 \mid \forall j < i \; X_j = 0 )=\frac{\sigma_{i+1}}{\sigma_i}$$
with
$$
\sigma_i=\sum_{k+l\le n}\left(2^{-(k+l)}\binom{k+l}k\right)^i\binom n{k,l,n-k-l}\;.
$$
The quotient tends to $1$ for $i\to\infty$, as you grow more and more certain that $v$ consists of all $0$s.
