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 Hey coach, that's unfair! (Posted on 2003-01-05)
A first baseman is fortunate enough for his team to be playing in the World Series. When game one is about to start, he asks the coach if he's going to play. The coach responds "despite the fact that you have a higher batting average than our rookie first baseman, we're up against a left handed pitcher today, and he has a better average against lefties than you, so I'm going to play him". Well, the fellow figures that this is fair enough, baseball being a game of averages and all, and happily sits out the first game, knowing that the team will come up against a right hander at some point, giving him a chance to play.

Sure enough, game 2 is set to start, and the opponents are starting a right handed pitcher. The fellow asks the coach if hes going to play today. The coach responds "well, I know that you have a better average overall, but today we're facing a rightie, and our rookie has a better average against righties than you do, so we're going to play him today".

So, the regular player, who has a better average against pitchers in general, has a lower average against BOTH left and right handers????? The player feels cheated. How did this happen?

For reference, the players batting average is calculated using the following formula:
average = safe hits/at bats, and is recorded to three decimal places (though announcers generally multiply this fraction by 1000 to give a integer value). A good players average will be between .300 and .350, with higher averages possible, but rare. For example a player gets 20 safe hits in 80 "at bats" then his average is .250

 See The Solution Submitted by Cory Taylor Rating: 3.1818 (11 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 First thoughts | Comment 2 of 8 |
Let v = veteren player
Let n = the rookie player
Let R(x) = "at bats" for player x against righties
Let L(x) = "at bats" for player x against lefties
Let r(x) = "safe hits" against righties
Let l(x) = "safe hits" against lefties
Then
r(x)/R(x) = batting average against righties = f(x)
l(x)/L(x) = batting average against lefties = g(x)
[r(x) + l(x)]/[R(x) + L(x)] = overall batting average = h(x)

f(n) > f(v)
g(n) > g(v)
h(n) < h(v)

There seems to be some echo of "A Birthday Gamble," so I suspect that the rookie's average against righties is close to, but less than half, and he's only had 1 at bat against lefties which he managed to hit, while the veteren has a slightly lower average against righties, and a high, but normal average against lefties.

 Posted by TomM on 2003-01-05 12:53:23

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