The Simpson's paradox is one of the easily understandable and at the same time astonishing phenomena in statistics. It occurs whenever groups of data show a particular trend, but that trend is reversed when the groups are combined. With the help of a simple example, the paradox can be understood immediately.

We consider the two disjoint sets $$\#1$$ and \ (\#2\) as well as $$G = \#1 \cup \#2$$ and test within these sets the success rate of $$A$$ and $$B$$:

 $$A$$ $$B$$ $$win$$ $$\#1$$ $$\frac{1}{1}=100\%$$ $$\frac{3}{4}=75\%$$ $$A$$ $$\#2$$ $$\frac{2}{5}=40\%$$ $$\frac{1}{3}=33\%$$ $$A$$ $$\#1 \cup \#2$$ $$\frac{3}{6}=50\%$$ $$\frac{4}{7}=57\%$$ $$B$$

It turns out that $$A$$ is more successful in both \ (\#1\) and $$\#2$$ than $$B$$, but surprisingly $$B$$ is more successful in \ (G\) than $$A$$. This example is also among those with the smallest amount of $$G$$ with $$|G|=13$$. There is no $$G$$ with $$|G|<13$$ ( proof by brute force).

We now subdivide the set $$G$$ instead of $$2$$ into $$3$$ disjoint subsets $$\#1, \, \#2, \, \#3$$ with $$\#1 \cup \#2 \cup \#3 = G$$ . Then we construct the exciting case that for every element $$e_k \neq \emptyset$$ the power set $$P(G)$$ of $$G$$ following applies: $$\forall e_1, e_2 \in P(G): |e_1| \neq |e_2| \Rightarrow win(e_1) \neq win(e_2) \land |e_1| = |e_2| \Rightarrow win(e_1) = win(e_2)$$ $$\forall e_1, e_2 \in P(G): |e_1| \neq |e_2| \Rightarrow win(e_1) \neq win(e_2) \land |e_1| = |e_2| \Rightarrow win(e_1) = win(e_2)$$

After a few hours of brute force on a standard Core i7, the following example can be found:

 $$A$$ $$B$$ $$C$$ $$win$$ $$\#1$$ $$\frac{6}{7}=85,71\%$$ $$\frac{12}{15}=80,00\%$$ $$\frac{22}{37}=59,46\%$$ $$A$$ $$\#2$$ $$\frac{95}{167}=56,89\%$$ $$\frac{48}{88}=54,55\%$$ $$\frac{38}{67}=56,72\%$$ $$A$$ $$\#3$$ $$\frac{48}{144}=33,33\%$$ $$\frac{16}{50}=32,00\%$$ $$\frac{2}{20}=10,00\%$$ $$A$$ $$\#1 \cup \#2$$ $$\frac{101}{174}=58,05\%$$ $$\frac{60}{103}=58,25\%$$ $$\frac{60}{104}=57,69\%$$ $$B$$ $$\#1 \cup \#3$$ $$\frac{54}{151}=35,76\%$$ $$\frac{28}{65}=43,08\%$$ $$\frac{24}{57}=42,11\%$$ $$B$$ $$\#2 \cup \#3$$ $$\frac{143}{311}=45,98\%$$ $$\frac{64}{138}=46,38\%$$ $$\frac{40}{87}=45,98\%$$ $$B$$ $$\#1 \cup \#2\cup \#3$$ $$\frac{149}{318}=46,86\%$$ $$\frac{76}{153}=49,67\%$$ $$\frac{62}{124}=50,00\%$$ $$C$$

If such cases occur in reality, any conclusions on a recommendation of the success of a group are both meaningful and meaningless.

At this point, we recommend the exciting reading Causality: Models, Reasoning and Inference by Judea Pearl .

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