Machine Learning

A cricketer has three attributes:

Batting ability, with values Amazing, OK, Useless
Bowling type, with values Fast, Spinner
Fielding position, with values Wickie, Close-in, Outfield

Given the following data that show whether or not the cricketer was selected for the team, use decision tree learning to build an optimal decision tree that predicts whether or not a player will be selected. Show all your working. Draw the final decision tree.

Player Batting Bowling Fielding Selected
----------------------------------------
1      Amazing Spinner Outfield Selected
2      OK      Fast    Outfield Left-out
3      Useless Spinner Outfield Selected
4      Amazing Spinner Wickie   Left-out
5      Amazing Fast    Outfield Selected
6      OK      Fast    Outfield Left-out
7      OK      Fast    Outfield Left-out
8      Amazing Spinner Close-in Selected
9      Amazing Fast    Close-in Selected
10     Useless Fast    Wickie   Left-out

Entropy = 1.0

Batting
Entropy(Amazing) = -0.8*log2(0.8) -0.2*log2(0.2) = 0.722
Entropy(OK) = 0.0
Entropy(Useless) = 1.0
Entropy(Batting) = 0.5 * 0.722 + 0.2 * 1= 0.561
InfoGain(Batting) = 1.0 - 0.561 = 0.439
Bowling
Entropy(Fast) = -0.333*log2(0.333) - 0.666*log2(0.666) = 0.918
Entropy(Spinner) = -0.75*log2(0.75) -0.25*log2(0.25) = 0.811
Entropy(Bowling) = 0.6 * 0.918 + 0.4 + 0.811 = 0.875
InfoGain(Bowling) = 1.0 - 0.875 = 0.125
Fielding
Entropy(Wickie) = 0.0
Entropy(Close-in) = 0.0
Entropy(Outfield) = 1.0
Entropy(Fielding) = 0.6 * 1.0 = 0.6
InfoGain(Fielding) = 1.0 - 0.6 = 0.4
Batting is best with remaining entropy 0.561

Amazing:Bowling
Entropy(Fast) = 0.0
Entropy(Spinner) = -0.666*log2(0.666) -0.333*log2(0.333) = 0.918
Entropy(Amazing:Bowling) = 0.6 * 0.918 = 0.551
InfoGain(Amazing:Bowling) = 0.722 - 0.551 = 0.171
Amazing:Fielding
Entropy(Wickie) = 0.0
Entropy(Close-in) = 0.0
Entropy(Outfield) = 0.0
Entropy(Amazing:Fielding) = 0.0
InfoGain(Amazing:Fielding) = 0.722 - 0.0 = 0.722
Fielding is best for Amazing with remaining entropy 0.0

Useless:Bowling
Entropy(Fast) = 0.0
Entropy(Spinner) = 0.0
Entropy(Useless:Bowling) = 0.0
InfoGain(Useless:Bowling) = 1.0 - 0.0 = 1.0
Bowling is best for Useless with remaining entropy 0.0. (But Fielding would also work.)

No more needed. The decision tree is: