Independent sampling of dependent random variables

Question

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

Answer 1

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.