Simpson's Paradox


Case 1

Consider two containers, one blue, one red. The blue container has 45 white balls and 55 black inside it. The red container has 43 white balls and 57 black ones. From which container would you pick out one ball if you wanted to have the best chance of getting a white one? If you answered "blue" that is correct.

Now there are two more containers, green and orange. The green one contains 67 white balls and 33 black, while the orange one contains 64 white and 36 black. From which would you have a better chance of getting a white ball? Yes, "green" is correct.

Let's now pour the balls from the blue container into the green one, so that it contains 112 white balls and 88 black. We also pour the contents of the red container into the orange one, so it has 107 white balls and 93 black. Obviously you would pick a ball from the green container to maximise the chances of getting a white ball.

Case 2

So far so good. Suppose instead that the blue container has 5 white and 6 black, the red container 3 white and 4 black. The green has 6 white and 3 black, the orange has 9 white and 5 black.

Where would you pick from? Since 5/11 = 45% and 3/7 = 43% blue looks good again. Likewise, 6/9 = 67% and 9/14 = 64%, so again green looks better than orange. Note that these percentages are the same in case 2 as they were in case 1.

What about if we pour the balls from the blue container into the green and those from the red one into the orange?

From where would you pick now?

Solution

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