Make up another example of Simpson’s paradox by changing the numbers in Example B of Sec...

Make up another example of Simpson’s paradox by changing the numbers in Example B of Section 1.4.

Reference

Simpson’s Paradox

A black urn contains 5 red and 6 green balls, and a white urn contains 3 red and 4 green balls. You are allowed to choose an urn and then choose a ball at random from the urn. If you choose a red ball, you get a prize. Which urn should you choose to draw from? If you draw from the black urn, the probability of choosing a red ball is 5/ 11 = .455 (the number of ways you can draw a red ball divided by the total number of outcomes). If you choose to draw from the white urn, the probability of choosing a red ball is 3/ 7 = .429, so you should choose to draw from the black urn.

Now consider another game in which a second black urn has 6 red and 3 green balls, and a second white urn has 9 red and 5 green balls. If you draw from the black urn, the probability of a red ball is 6/ 9 = .667, whereas if you choose to draw from the white urn, the probability is 9/ 14 = .643. So, again you should choose to draw from the black urn.

In the final game, the contents of the second black urn are added to the first black urn, and the contents of the second white urn are added to the first white urn. Again, you can choose which urn to draw from. Which should you choose? Intuition says choose the black urn, but let’s calculate the probabilities. The black urn now contains 11 red and 9 green balls, so the probability of drawing a red ball from it is 11/ 20 = .55. The white urn nowcontains 12 red and 9 green balls, so the probability of drawing a red ball from it is 12/ 21 = .571. So, you should choose the white urn. This counterintuitive result is an example of Simpson’s paradox. For an example that occurred in real life, see Section 11.4.7. For more amusing examples, see Gardner (1976).