# Why decision tree method for lower bound on finding a minimum doesn’t work

(Motivated by this question. Also I suspect that my question is a bit too broad)

We know $$\Omega(n \log n)$$ lower bound for sorting: we can build a decision tree where each inner node is a comparison and each leaf is a permutation. Since there are $$n!$$ leaves, the minimum tree height is $$\Omega(\log (n!)) = \Omega (n \log n)$$.

However, it doesn’t work for the following problem: find a minimum in the array. For this problem, the results (the leaves) are just indices of the minimum element. There are $$n$$ of them, and therefore the reasoning above gives $$\Omega(\log n)$$ lower bound, which is obviously an understatement.

My question: why does this method works for sorting and doesn’t work for minimum? Is there some greater intuition or simply "it just happens" and we were "lucky" that sorting has so many possible answers?

I guess the lower bound from decision tree makes perfect sense: we do can ask yes/no questions so that we need $$O(\log n)$$ answers: namely, we can use binary search for the desired index. My question still remains.