[math-fun] limits, batting averages
(2n choose n) /( Sum(k=0..n-1, (2k choose k) )
This is 3, beacuse lim (2n choose n)/(2n-2 choose n-1) is 4. My favorite form of Simpson's paradox, using baseball batting averages, comes to mind. In the first half of the baseball season, A has a better batting average than B (.275 vs. .250); and the same for the second half of the season (.325 vs. .300). But overall, B can have a better average, due to hidden bias: (25+100)/(100+300) > (55+65)/(200+200). Err, this example is wrong. Bs average for the second part of the season is 100/300 = .333, bigger than As .325. Rich rcs@cs.arizona.edu
Sorry; change 100/300 to 96/300. The principle still works. Nick Richard Schroeppel wrote:
(2n choose n) /( Sum(k=0..n-1, (2k choose k) )
This is 3, beacuse lim (2n choose n)/(2n-2 choose n-1) is 4.
My favorite form of Simpson's paradox, using baseball batting averages, comes to mind. In the first half of the baseball season, A has a better batting average than B (.275 vs. .250); and the same for the second half of the season (.325 vs. .300). But overall, B can have a better average, due to hidden bias: (25+100)/(100+300) > (55+65)/(200+200).
Err, this example is wrong. Bs average for the second part of the season is 100/300 = .333, bigger than As .325.
Rich rcs@cs.arizona.edu
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Richard Schroeppel