  The Boolean lattice $2^{[n]}$ is the family of all subsets of
$[n]=\{1,\dots,n\}$ ordered by inclusion, and a chain is a family of pairwise
comparable elements of $2^{[n]}$. Let $s=2^{n}/\binom{n}{\lfloor n/2\rfloor}$,
which is the average size of a chain in a minimal chain decomposition of
$2^{[n]}$. We prove that $2^{[n]}$ can be partitioned into $\binom{n}{\lfloor
n/2\rfloor}$ chains such that all but at most $o(1)$ proportion of the chains
have size $s(1+o(1))$. This asymptotically proves a conjecture of F\"uredi from
1985. Our proof is based on probabilistic arguments. To analyze our random
partition we develop a weighted variant of the graph container method. Using
this result, we also answer a Kalai-type question raised recently by Das,
Lamaison and Tran. What is the minimum number of forbidden comparable pairs
forcing that the largest subfamily of $2^{[n]}$ not containing any of them has
size at most $\binom{n}{\lfloor n/2\rfloor}$? We show that the answer is
$(\sqrt{\frac{\pi}{8}}+o(1))2^{n}\sqrt{n}$. Finally, we discuss how these
uniform chain decompositions can be used to optimize and simplify various
results in extremal set theory.
