Details of ten international cricket games between two teams “Green” and…
2024
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).
Table C
Match NumberPitchFormatWinner (Target)1STGreen2STBlue3FOBlue4SOBlue5FTGreen6FOBlue7SOGreen8FTBlue9FOBlue10SOGreen
Attempted by 1 students.
Show answer & explanation
Correct answer: 0 to 1
Solution:
Step 1 — Compute overall entropy of the target (Winner).
There are 10 matches: Green wins = 4, Blue wins = 6.
Entropy(C) = - (4/10) log2(4/10) - (6/10) log2(6/10) = -0.4×(−1.32193) - 0.6×(−0.73697) ≈ 0.97
Step 2 — Compute entropy conditioned on Pitch.
Pitch = spin-friendly: 5 matches, Green = 3, Blue = 2.
Entropy(spin-friendly) = -0.6 log2 0.6 - 0.4 log2 0.4 ≈ 0.97
Pitch = pace-friendly: 5 matches, Green = 1, Blue = 4.
Entropy(pace-friendly) = -0.2 log2 0.2 - 0.8 log2 0.8 ≈ 0.72
Step 3 — Compute expected conditional entropy given Pitch.
Info(C | Pitch) = (5/10)×0.97 + (5/10)×0.72 = 0.5×(0.97 + 0.72) ≈ 0.84644
Step 4 — Information gain for attribute Pitch.
InformationGain(C, Pitch) = Entropy(C) − Info(C | Pitch) ≈ 0.97095 − 0.84644 ≈ 0.12451
Answer: InformationGain(C, Pitch) ≈ 0.12 (rounded to two decimal places).