Originally Posted by AtlUmpSteve
JMO, but I think your request for a definition of an attempted putout is intentionally obtuse. It should be crystal clear that there must be an available putout, any possible play being made, to actually attempt a putout. Not think you are attempting a putout, there must be a putout available.
What part of the rule isn't clear? Is it the exception? Let me elaborate on the exception, as to how it can apply.
No runners on, batter gets ball 4. Catcher could legally throw to 1B, and you could consider that an attempted putout (even though BR is awarded first, can legally overrun first, and is not in jeopardy). Suppose catcher throws the ball to the shortstop (coach thinks shortstop is the best athlete that can make the best decisions); is that an attempted putout? Most people would say no, it is a defensive play made as a strategic move to keep the runner on first, but not an attempted putout. So, is the next batter awarded a ball? No, the exception applies to supercede the initial rule.
Another instance, R1 on third, R2 on first. After a pitch, catcher throws to shortstop, even though no runners were attempting an advance (same reason as first play). Again, not an attempted putout, so the initial rule would seem to apply; but with runners on base, the exception applies. No ball awarded.
If there are no runners, it isn't a strikeout or a put out, and the batter doesn't become a batter-runner, any throw by the catcher to anyone but the pitcher is a ball on the batter. Even if you think the offense tricked them, intentionally or unintentionally. The catcher simply must know the count to avoid being tricked, just as they must know when they must throw to first on a dropped third strike. That's the rule, and the intent of the rule, as well.
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