GATE DA 2024 | Question 62
Details of ten international cricket games between two teams “Green” and “Blue” are given in Table C. This table consists of matches played on different pitches, across formats along with their winners. The attribute Pitch can take one of two values: spin-friendly (represented as \(S\)) or pace-friendly (represented as \(F\)). The attribute Format can take one of two values: one-day match (represented as \(O\)) or test match (represented as \(T\)).
A cricket organization would like to use the information given in Table C to develop a decision-tree model to predict outcomes of future games between these two teams.
To develop such a model, the computed InformationGain(C, Pitch) with respect to the Target is ______ (rounded off to two decimal places).
| Match Number | Pitch | Format | Winner (Target) |
|---|---|---|---|
| 1 | \(S\) | \(T\) | Green |
| 2 | \(S\) | \(T\) | Blue |
| 3 | \(F\) | \(O\) | Blue |
| 4 | \(S\) | \(O\) | Blue |
| 5 | \(F\) | \(T\) | Green |
| 6 | \(F\) | \(O\) | Blue |
| 7 | \(S\) | \(O\) | Green |
| 8 | \(F\) | \(T\) | Blue |
| 9 | \(F\) | \(O\) | Blue |
| 10 | \(S\) | \(O\) | Green |
Explanation
Answer: 0.12 to 0.13
Information gain is the entropy of the Target minus the weighted entropy left after splitting on Pitch (logarithms to base 2).
- All 10 games: 4 Green and 6 Blue, so \(H(C) = -0.4\log_{2}0.4 - 0.6\log_{2}0.6 \approx 0.971\).
- \(S\) pitches (games 1, 2, 4, 7, 10): 3 Green and 2 Blue, the same 0.6 / 0.4 split with the colours swapped, so the entropy is again \(\approx 0.971\).
- \(F\) pitches (games 3, 5, 6, 8, 9): 1 Green and 4 Blue, so the entropy is \(-0.2\log_{2}0.2 - 0.8\log_{2}0.8 \approx 0.722\).
Each branch holds half of the games, so
\[ \text{Gain} = 0.971 - \left(\tfrac{1}{2} \times 0.971 + \tfrac{1}{2} \times 0.722\right) = \tfrac{1}{2}(0.971 - 0.722) \approx 0.1245 \]
which is 0.12 to two decimal places (the key accepts 0.12 to 0.13).
Answer: 0.12