GATE 2024 DA – 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 |
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Show answer and explanation
Correct answer: 0.11 to 0.13
Explanation
Green wins 4 of the 10 matches, so the entropy of the target is $H = -0.4\log_2 0.4 - 0.6\log_2 0.6 = 0.971$. For $Pitch = S$ (matches 1, 2, 4, 7, 10) Green wins 3 of 5, so the entropy is $H(0.6, 0.4) = 0.971$. For $Pitch = F$ (matches 3, 5, 6, 8, 9) Green wins 1 of 5, so the entropy is $H(0.2, 0.8) = 0.722$. The weighted entropy after the split is $0.5 \times 0.971 + 0.5 \times 0.722 = 0.846$. The information gain is $0.971 - 0.846 = 0.125$, which is about 0.12.