Let $x\_1, \ldots, x\_n$ be **possibly dependent** random variables, each taking values $x\_i \in \{0, 1, 2\}$. Suppose further that in every outcome the number of random variables that equal 2 is exactly 1. Now for each $i \in \{1, \ldots, n\}$ define
$$
f\_i = \begin{cases}
\Pr[x\_i = 2 \mid x\_i \geq 1] & \text{if } x\_i \geq 1\\
0 & \text{if } x\_i =0
\end{cases},
$$
and for each $i \in \{1, \ldots, n\}$ let $y\_i$ be a Bernoulli random variable that is 1 independently with probability $f\_i$ and 0 otherwise.


Is the following conjecture correct or is there a distribution on $x\_i$'s refuting it?


**Conjecture:** There is a fixed $\epsilon > 0$ (i.e. $\epsilon$ being independent of $n$) such that with probability at least $\epsilon$, there is exactly one index $i$ where $y\_i = 1$.


Related question:
[Bounds on variance of sum of dependent random variables](https://mathoverflow.net/questions/369336/bounds-on-variance-of-sum-of-dependent-random-variables)



Independent sampling of dependent random variables
The answer is "no" (if i understand the question correctly).


Consider the following exchangeble joint distribution of the $x\_i$s. In event $A$, which occur with probabiluty $1/\sqrt n$, all the $x\_i$s are 1, exept for one 2. In the complement event $B$, all the $x\_i$s are 0 exept for one 2.


Under this distribution, $f\_i$ is either 0 or $1/\sqrt n$. Let $Y=\sum y\_i$. Since $E[ Y|A]=\sqrt n$, and $E[Y|B]=1/\sqrt n$, in either event it is too far from 1; therefore the probability that there is exaxtly one positive $b\_i$ is vanishing.


