Let $X_i$, $i=1, \dots, n$, be identically distributed and suppose they satisfy $X_i\in \{0,1\}$. Without assuming any independence, is it true that $P(X_1=1|\sum_{i=1}^n X_i \neq 0)>P(X_1=1|X_i=1)$, for any $i\neq 1$.
If it is not true in this generality, what conditions would be needed for this to be true?
Any comment would be helpful.
This is not true:
Lets say that $P(X_1=1|X_2=1)=1$, whereas the less constrictive condition $\sum X_i >0$ could result in $P(X_1=1|\sum X_i >0)<1$
In general, you need to have $P(X_1=1\cap \sum_{i=1}^n X_i \neq 0)P(X_i=1)>P(X_1=1 \cap X_i=1)P(\sum_{i=1}^n X_i \neq 0)$ which reduces to:
$P(X_1=1)P(X_i=1)>P(X_1=1 \cap X_i=1)P(\sum_{i=1}^n X_i \neq 0)$ since $\{X_1=1\} \subset \{\sum_{i=1}^n X_i \neq 0\}$ The answer to this depends on the nature of the correlations.
If the $X_i$ are independent, then we get:
$P(X_1=1)P(X_i=1)>P(X_1=1)P(X_i=1)P(\sum_{i=1}^n X_i \neq 0)$ which will be true as long as $P(X_i=1)\neq 1$
If they are perfectly correlated, then we get:
$P(X_1=1)^2>P(X_1=1)P(\sum_{i=1}^n X_i \neq 0) \implies P(X_1=1)>P(\sum_{i=1}^n X_i \neq 0)$, which is a contradiction due to the fact that probabilities are always positive and $\{X_1=1\} \subset \{\sum_{i=1}^n X_i \neq 0\}$
Therefore, a sufficient general condition is that the variables are non-degenerate and pairwise independent or at least have sub-independent pairwise joint probabilities:$P(X_1=1 \cap X_i=1)\leq P(X_1=1)P(X_i=1)$