Take a random variable defined as

$$r=u_{11}v_{1}v_{1}+u_{12}v_{1}v_{2}+\dots+u_{n,n-1}v_{n}v_{n-1}+u_{nn}v_{n}v_{n}$$ where $v_{i}$ are independent uniform random variables from $\{0,\dots,b\}$, $u_{ij}$ are in $\{-T,\dots,0,\dots,T\}$ with the condition that probability that all $|u_{ij}|<\frac{T}2$ is at most some small $\epsilon\in(0,1)$, $u_{ij}$ are independent of $v_i,v_j$ and $u_{ij}$ might not be independent of $u_{i'j'}$ where either $i\neq i'$ or $j\neq j'$ holds.

Assume the mean of $r$ is $0$. Variance is at most $n^4b^4T^{2}$.

1. What is the probability that $|r|\leq\frac12\big(1+\frac1{n^c}\big)b^2T$ holds at any fixed $c\geq2$?

We know $P(|r|\leq k\sigma)\geq1-\frac1{k^2}$ and here $k=\frac 1{2n^2}\big(1+\frac1{n^c}\big)<1$ and so useless.

2. Under what broad conditions on distribution of $r$ can above probability be $>1-\frac1{n^2}$?

One possibility is for conditions on random variables that can force cancellations in summation defining $r$ leading to a smaller variance. If $u_{ij}$ are such that $\pm1$ signs occur equally likely with probability $>1-\frac1{n^2}$ then we can expect the variance close to $b^4T^2$ with probability $>1-\frac1{n^2}$ shaving a factor of $n^2$ which leads to desired probability for $|r|\leq\frac12\big(1+\frac1{n^c}\big)b^2T$ holds at any fixed $c\geq2$.

If the probability approaches $0$ then I am interested in how fast precisely it approaches $0$.

Take a random variable defined as

$$r=u_{11}v_{1}v_{1}+u_{12}v_{1}v_{2}+\dots+u_{n,n-1}v_{n}v_{n-1}+u_{nn}v_{n}v_{n}$$ where $v_{i}$ are independent uniform random variables from $\{0,\dots,b\}$, $u_{ij}$ are in $\{-T,\dots,0,\dots,T\}$ with the condition that probability that all $|u_{ij}|<\frac{T}2$ is at most some small $\epsilon\in(0,1)$, $u_{ij}$ are independent of $v_i,v_j$ and $u_{ij}$ might not be independent of $u_{i'j'}$ where either $i\neq i'$ or $j\neq j'$ holds.

Assume the mean of $r$ is $0$. Variance is at most $n^4b^4T^{2}$.

1. What is the probability that $|r|\leq\frac12\big(1+\frac1{n^c}\big)b^2T$ holds at any fixed $c\geq2$?

We know $P(|r|\leq k\sigma)\geq1-\frac1{k^2}$ and here $k=\frac 1{2n^2}\big(1+\frac1{n^c}\big)<1$ and so useless.

2. Under what broad conditions on distribution of $r$ can above probability be $>1-\frac1{n^2}$?

One possibility is for conditions on random variables that can force cancellations in summation defining $r$ leading to a smaller variance. If $u_{ij}$ are such that $\pm1$ signs occur equally likely with probability $>1-\frac1{n^2}$ then we can expect the variance close to $b^4T^2$ with probability $>1-\frac1{n^2}$ shaving a factor of $n^2$ which leads to desired probability for $|r|\leq\frac12\big(1+\frac1{n^c}\big)b^2T$ holds at any fixed $c\geq2$.

If the probability approaches $0$ on some conditions then I am interested in how fast precisely it approaches $0$ on those set conditions. This is implicit in query 1. where I have attempted the asymptotics using Chebyshev.

Take a random variable defined as

$$r=u_{11}v_{1}v_{1}+u_{12}v_{1}v_{2}+\dots+u_{n,n-1}v_{n}v_{n-1}+u_{nn}v_{n}v_{n}$$ where $v_{i}$ are independent uniform random variables from $\{0,\dots,b\}$, $u_{ij}$ are in $\{-T,\dots,0,\dots,T\}$ with the condition that probability that all $|u_{ij}|<\frac{T}2$ is at most some small $\epsilon\in(0,1)$, $u_{ij}$ are independent of $v_i,v_j$ and $u_{ij}$ may not be independent of $u_{i'j'}$ where either $i\neq i'$ or $j\neq j'$ holds.

Assume the mean of $r$ is $0$. Variance is at most $n^4b^4T^{2}$.

1. What is the probability that $|r|\leq\frac12\big(1+\frac1{n^c}\big)b^2T$ holds at any fixed $c\geq2$?

We know $P(|r|\leq k\sigma)\geq1-\frac1{k^2}$ and here $k=\frac 1{2n^2}\big(1+\frac1{n^c}\big)<1$ and so useless.

2. Under what broad conditions on distribution of $r$ can above probability be $>1-\frac1{n^2}$?

One possibility is for conditions on random variables that can force cancellations in summation defining $r$ leading to a smaller variance. If $u_{ij}$ are such that $\pm1$ signs occur equally likely with probability $>1-\frac1{n^2}$ then we can expect the variance close to $b^4T^2$ with probability $>1-\frac1{n^2}$ shaving a factor of $n^2$ which leads to desired probability for $|r|\leq\frac12\big(1+\frac1{n^c}\big)b^2T$ holds at any fixed $c\geq2$.

Take a random variable defined as

$$r=u_{11}v_{1}v_{1}+u_{12}v_{1}v_{2}+\dots+u_{n,n-1}v_{n}v_{n-1}+u_{nn}v_{n}v_{n}$$ where $v_{i}$ are independent uniform random variables from $\{0,\dots,b\}$, $u_{ij}$ are in $\{-T,\dots,0,\dots,T\}$ with the condition that probability that all $|u_{ij}|<\frac{T}2$ is at most some small $\epsilon\in(0,1)$, $u_{ij}$ are independent of $v_i,v_j$ and $u_{ij}$ might not be independent of $u_{i'j'}$ where either $i\neq i'$ or $j\neq j'$ holds.

Assume the mean of $r$ is $0$. Variance is at most $n^4b^4T^{2}$.

1. What is the probability that $|r|\leq\frac12\big(1+\frac1{n^c}\big)b^2T$ holds at any fixed $c\geq2$?

We know $P(|r|\leq k\sigma)\geq1-\frac1{k^2}$ and here $k=\frac 1{2n^2}\big(1+\frac1{n^c}\big)<1$ and so useless.

2. Under what broad conditions on distribution of $r$ can above probability be $>1-\frac1{n^2}$?

One possibility is for conditions on random variables that can force cancellations in summation defining $r$ leading to a smaller variance. If $u_{ij}$ are such that $\pm1$ signs occur equally likely with probability $>1-\frac1{n^2}$ then we can expect the variance close to $b^4T^2$ with probability $>1-\frac1{n^2}$ shaving a factor of $n^2$ which leads to desired probability for $|r|\leq\frac12\big(1+\frac1{n^c}\big)b^2T$ holds at any fixed $c\geq2$.

If the probability approaches $0$ then I am interested in how fast precisely it approaches $0$.