Talk:Fielder's choice

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More on definition?

The article defines a fielder's choice as being when a fielder, upon fielding a batted ball, chooses to try to put out another runner (emphasis mine). But would it still be considered a fielder's choice if the fielder is unsuccessful in putting out that other runner? The article is ambiguous about that. ekedolphin 11:59, August 14, 2005 (UTC)

I clarified this. The fielder is charged with an error and the batter still does not get a hit. Ellsworth 23:50, 14 June 2006 (UTC)[reply]

Actually, a batter can be safe on a fielder's choice where there is no putout and no error. A common example involves the pitcher fielding a groundball with plenty to time to retire the batter. He chooses instead to make a riskier play and attempt to retire a runner going from first to second. That runner beats the throw, and no outs are recorded. This is usually scored as a fielder's choice for the batter (as he doesn't deserve a hit, but no error to the fielder (the pitcher) because it wasn't a misplay, just a bad decision. (Mental errors don't count in the box score.) When I learned this long ago, it bothered me, because the books don't quite balance, but it happens to be the case. I'll try to locate a specific example. WHPratt (talk) 17:46, 20 February 2009 (UTC)WHPratt There must be hundreds of examples, but here's a game I recall from long ago, Retrosheet ID is http://www.retrosheet.org/boxesetc/1962/B06260BAL1962.htm. June 26, 1962 (N) at Memorial Stadium. Baltimore Orioles 1, Chicago White Sox 0. In the Baltimore ninth, there's a fielder's choice, but no outs and no errors in the inning. WHPratt (talk) 02:50, 26 February 2009 (UTC)WHPratt[reply]

Batter-runners are not "charged" with outs

Batter-runners are not "charged" with outs, so it is nonsense to state that a batter hitting into a fielder's choice is "charged" with an out. Check the MLB scoring rules. [1] Batter-runners are not charged with outs for the purpose of scorekeeping and statistics. Fielders are credited with putouts, but there is no such notion in baseball scoring for outs being charged to offensive players. To think that this error has persisted in Wikipedia since 2003 is embarrassing. —Preceding unsigned comment added by 71.134.73.90 (talk) 04:09, 6 October 2007 (UTC)[reply]

Misleading

suppose there is a runner on first, no outs. A difficult grounder is fielded by the shortstop, and his ONLY play is at second, which is a putout. The scoring is 'fielders choice', not a hit. CorvetteZ51 (talk) 10:26, 18 June 2011 (UTC)[reply]

For the purposes of baseball statistics, it's scored a Force Out. I know because I am an MLB scorekeeper. It's a FC if there's nobody on the preceding base when the ball is hit into play. — Preceding unsigned comment added by 67.232.120.225 (talk) 06:53, 6 February 2013 (UTC)[reply]