@JeremyRickard, thank you very much. Are there some references about this fact? I need to cite this fact in a paper.

@user64494, I think it is $\overline{L_5}$ in Adam P. Goucher's comment.

thank you very much!

sorry, I mean when $k \ge 1$, $D_k \le 3/4$.

thank you very much for your great answer! I am trying to understand every step of your proof. In the formula before (3), it is said that $\sum_z P(X_{J \cap K}=z) \sum_y P(X_{K\backslash J}=y-z)=1$. I am trying to understand this step. I think that $$\sum_z P(X_{J \cap K}=z) \sum_y P(X_{K\backslash J}=y-z) \\=\sum_z P(X_{J \cap K}=z) \sum_y P(X_{K\backslash J}=y) = \sum_{y,z} P(X_{J \cap K}=z) P(X_{K\backslash J}=y) \\=\sum_{y,z}P(X_{K}=y+z)=\sum_y P(X_{K}=y)=1$$. Is my understanding correct? I checked that when $k=2$, $D_k=3/4$. So maybe for $k \ge 1$, $D_k \ge 3/4$?

@Iosif Pinelis, thank you very much! Yes, the $X_i$'s are independent.

@Johannes Trost, thank you very much for your help last time. If $v_1, v_2$ have large overlap, how do you compute asymptotic of ${|v_1| - O(\min(|v_1|, |v_2|)^v) \choose x_1 - d}$. Is it possible to show that $f(v_1, v_2)<1$ when $|v_1|, |v_2|$ are large and $v_1, v_2$ have large overlap?