Use information theory to estimate the size of the subtrees rooted at each child, for each possible attribute. That is, try each attribute, evaluate and pick the best one.

• How much (expected) work to guess if an element x in a set S of size |S|?
  log₂|S|
  That is, at each step we can ask a yes/no question that eliminates at most 1/2 of the elements remaining.

• Given S = P ∪ N, where P and N are two disjoint sets, how hard is it to guess if an element x is in P or N?

  if x ∈ P, then log₂|P| = log₂p questions needed, where p = |P|
  if x ∈ N, then log₂|N| = log₂n questions needed, where n = |N|

  So, the expected number of questions that have to be asked is:

  (Pr(x ∈ P) * log₂p) + (Pr(x ∈ N) * log₂n)

  or, equivalently,

  (p/(p+n)) log₂p + (n/(p+n)) log₂n