Scenario: In a game that you are scoring, the visting team’s shortstop is involved in three plays during the game. In the first inning the shortstop drops an easy put-out, for which you charge an error. In the third inning, the player is involved in a standard 4-6-3 double-play. In the bottom of the ninth with two out, the player makes up for the earlier error with a spectacular diving catch to end the game.
As the official scorer, what would you calculate for the shortstop’s fielding average from the game?
A – 1/2 = .500.
BÂ – 2/3 = .667.
CÂ – 3/4 = .750.
D – 4/5 = .800.
Highlight the text below to reveal the answer:
Answer:Â C – 3/4 = .750.
Rule 10.21(d) states:
[To compute] Fielding average, divide the sum of putouts and assists by the sum of putouts, assists and errors (which shall be called chances).
2 comments
So far as I can see the answer should be .667
(1 assist + 1 putout) = 2 ÷ (1 error + 1 assist + 1 putout) = 3
2 ÷ 3 = 0.667
Or am I missing sonmething?
The trick here is that by being involved in the double play, the shortstop picks up a put-out and an assist, so he gets 1 error + 1 assist + 2 put-outs in total.