Eight labeled cards with binary features A and B demonstrate how decision trees pick questions. Starting entropy is 1 bit for a 4-positive/4-negative set. Splitting on A or B each yields two children with entropy about 0.811 bits, giving the same information gain of 0.189. Since both features tie, an alphabetical rule picks A, illustrating that greedy ID3 selection can be arbitrary when gains match. Useful for first-year AI students learning entropy, conditional entropy, and how ID3 differs from gain ratio and Gini impurity.
Narrated · 16:9 · Preview before teaching · automatic layout checks do not establish subject accuracy
# Information gain measures how much a question helps Eight sample cards carry binary features A and B and a class label. Use these exact rows: (A,B,label) = (0,0,+), (0,0,+), (0,1,+), (0,1,-), (1,0,+), (1,0,-), (1,1,-), (1,1,-). The initial four-positive, four-negative group has entropy 1 bit. Split by A: one child contains three positives and one negative, the other one positive and three negatives. Each child has entropy approximately 0.811278, giving information gain 0.188722. Repeat with B and show the same gain. Choose A using an explicit alphabetical tie rule, not because it is better. Build the entropy and weighted conditional entropy expressions from the visible counts. Explain that ID3 greedily chooses information gain, which differs from gain ratio and Gini impurity. Use a black background, readable cards, and cyan and orange class markers. Audience: 一年级人工智能专业学生