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desertnaut
desertnaut
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Suppose I have a dataset

A B C D
1 1 1 0
1 1 0 0
1 1 0 1
1 0 1 1

Here A,B,C,D are my independent features and D is my dependent feature. Now if I make a decision tree here.

A -> Yes/No = 2/2 - > Based on value 1-> 2/2 So we go further for B. 
B -> Yes/No = 2/2 -> Based on value 1-> 1/2 So we go further for C
When B=1 & C = 1 then D = 0
When B=1 & C = 0 then D = 0 and as well as D = 1 too.

So here we have a 50/50 chance, Or we can see we can't go further to get unique value to predict. So how does decision tree solve this problem? May be I couldn't clear my question. Sorry I am a noob.

Suppose I have a dataset

A B C D
1 1 1 0
1 1 0 0
1 1 0 1
1 0 1 1

Here A,B,C,D are my independent features and D is my dependent feature. Now if I make a decision tree here.

A -> Yes/No = 2/2 - > Based on value 1-> 2/2 So we go further for B. 
B -> Yes/No = 2/2 -> Based on value 1-> 1/2 So we go further for C
When B=1 & C = 1 then D = 0
When B=1 & C = 0 then D = 0 and as well as D = 1 too.

So here we have a 50/50 chance, Or we can see we can't go further to get unique value to predict. So how does decision tree solve this problem? May be I couldn't clear my question. Sorry I am a noob.

Suppose I have a dataset

A B C D
1 1 1 0
1 1 0 0
1 1 0 1
1 0 1 1

Here A,B,C,D are my independent features and D is my dependent feature. Now if I make a decision tree here.

A -> Yes/No = 2/2 - > Based on value 1-> 2/2 So we go further for B. 
B -> Yes/No = 2/2 -> Based on value 1-> 1/2 So we go further for C
When B=1 & C = 1 then D = 0
When B=1 & C = 0 then D = 0 and as well as D = 1 too.

So here we have a 50/50 chance, Or we can see we can't go further to get unique value to predict. So how does decision tree solve this problem?

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Hasibul Islam Polok
Hasibul Islam Polok
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what happens when a decision tree can't be split into further unit values?

Suppose I have a dataset

A B C D
1 1 1 0
1 1 0 0
1 1 0 1
1 0 1 1

Here A,B,C,D are my independent features and D is my dependent feature. Now if I make a decision tree here.

A -> Yes/No = 2/2 - > Based on value 1-> 2/2 So we go further for B. 
B -> Yes/No = 2/2 -> Based on value 1-> 1/2 So we go further for C
When B=1 & C = 1 then D = 0
When B=1 & C = 0 then D = 0 and as well as D = 1 too.

So here we have a 50/50 chance, Or we can see we can't go further to get unique value to predict. So how does decision tree solve this problem? May be I couldn't clear my question. Sorry I am a noob.