OT: Something different, take the quiz!
by scstrato on Nov 21, 2010 1:05 PM EST
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PrintOk Braves fans, with the slew of rosterbation floating around on the site, I thought it might be nice to change things up a bit. A friend and I like to quiz each other about baseball topics every now and again and one day he came across a couple of questions that stumped both of us - at least initially. So let's see just how smart we all really are, huh?
Question 1: How can a fielder to turn an unassisted triple play without ever touching the baseball? (you must elaborate or give examples)
Question 2: Let's assume a pitcher throws 9 complete innings (regardless of score or who is winning and excluding errors and walks). How many total hits, actual base hits, could he have given up without allowing a single run (earned or otherwise)?
Now, I will preface this by saying the reality of either is slim to none but it is nonetheless a possibility. These are not trick questions! Also, please no cheating! The fun in this exercise will be in discussing all the possible, and possibly ridiculous, scenarios everyone comes up with.
Post your answers and I'll let you know if they are correct.
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The answer to question 2 would be 54.
Assuming that each runner can only advance one base on each hit, and that each hit with the bases loaded results in a runner being thrown out at home, the opposing team could theoretically have 6 hits per inning without scoring. Multiply that by 9 innings and you have 54 hits without a run scoring.
Wouldn't that be a fielder's choice?
I think the answer is 45.
Right but no enough info
The answer to question 2 is 54 but you didn’t detail how it is possible – which kinda makes me wonder if you didn’t look it up??? How can you get six hits in an inning without giving up a run?
Correction
You gave the wrong info. Look at it like this, in your scenario you would have 3 hits before the first out. On each of the next batted balls the runner is thrown out at home it would be considered a fielders choice, not a hit. So your total hits would be 3 for an inning. Does that help?
Yeah, that is true.
I think duxfan is right with 45.
Nope
The answer to question 2 is 54. A pitcher can give up 6 hits in an inning without giving up a single run (earned or unearned).
How about if
both of the first two hits result in the runner being thrown out attempting to stretcha single into a double, and the next 3 hits load the bases. With the bases loaded, a ball is hit to the outfield, but the runner on first thinks it’s caught so he stays at third to tag up, and the runner from second passes him. The runner from second is called out, but the batter is awarded a hit, which would make 6. If that happened once each inning (which is unlikely, but possible) there would be a total of 54 hits.
The easiest way...
…is to have the first five batters do what you say (the first two can also be picked off or caught stealing), the sixth hitter hits a base runner with the batted ball, and gets credit for the hit.
For the first question, it would have to be on interferance. I know of one case where a fielder was awarded an unassisted double play without touching the ball (on interference).
For the second, assuming no errors were made, it would probably be 45 (5 per inning). Outs would be made on runners taking extra bases on hits, but it is not likely that a runner out on a bases-loaded single w/o being ruled a fc. However, it could happen, and in this rare case, the answer would be 54.
"I always thought that record would stand until it was broken." -Yogi Berra
by iamthesgt on Nov 21, 2010 1:44 PM EST via mobile reply actions
clarification
A shortstop (Alvin Dark, if I’m not mistaken) was getting ready to field a ball when a runner on second ran into him. The umpire ruled the batter would have been out an of course the baserunner was out. So interference resulted in two outs.
"I always thought that record would stand until it was broken." -Yogi Berra
by iamthesgt on Nov 21, 2010 5:07 PM EST via mobile up reply actions
Answer to question #1
Runners on 2nd and 3rd no outs, pop fly (very fast runners and very high fly ball) the ball is going to land between 1st and second base and the second baseman is camped under the ball, the runner who hit the pop fly runs into the second baseman causing him to drop the ball, the other two runners have already crossed home plate and started to the dugout, unassisted triple play awarded to 2nd baseman and he never touches the ball
That's interesting
Though I have to admit I’m not sure what would come of the two runners who crossed the plate. I’ll dig into this one a little later.
Without looking at anyone else's posts...
…the answer to the second question is 54 hits. Each hit can result in an out if necessary, and the pitcher can face 6 batters (all hits) every inning without giving up a run.
Yes
It is 54 but I don’t understand your point of how each hit can result in an out. Here’s my point, if there are two outs and bases loaded (which is the only way you can have 5 hits to that point and no runs scored) how is it that the final out can be off a hit without the run scoring?
Oops.
I missed Willinn’s response above. The answer is the final AB has to result in hitting a runner with a batted ball.
without looking at above answers...
I’ll be surprised if #1 is possible.
#2-36 (1 single, 1 double play, 3 more singles to load the bases, 1 out = 4 hits x 9 innings, or 36 hits)
http://sportsandgrits.blogspot.com/
Status so far
So we have Question 2 answered. It is possible, though not likely, that a pitcher can give up 54 hits in a 9 inning game without surrendering a single run. An example: Single, caught stealing, single, pick-off, single, single, single. The sixth batter then hits a base-runner in fair territory resulting in the final out BUT getting credit for a hit.
Some pretty close guesses for Question 1 and I’ll go so far as to say this one is probably the least likely to occur. Any more guesses?
Couldn't resist looking it up,
So I won’t answer it. Real close though. Someone just has to put one more situation into what we’ve got.
question 1
eric bruntlett did it in 09. he caught the ball with his glove, stepped on the base, then tagged the runner from first coming towards him. The ball never touched him, it touched his glove.
by JohnRocker4CyYoung on Nov 21, 2010 6:03 PM EST reply actions
Nope
“Without ever touching the baseball” implies that it was not caught in his glove, sock, hat, shoe or otherwise.
Question # 1
How about this, runners on first and second no outs. Batter hits infield fly so he is out, the runner on first passes the runner on second so he is out and the ball hits the runner on second so he is out, the question is who gets credit for the triple play?
this is what I was thinking...
or some sort of strange scenario where a batted ball hits all three runners…not very likely either way. I think the pitcher would get credit though because he was the last player to touch the ball.
So, it would never happen?
Why would the runner on first pass the runner on second?
"If I have asthma, they won't let me scuba. And if I can’t scuba, then what’s this all been about?? What am I working toward??"
"You look like you should be married to one of the San Diego Padres."
This reminds me of a game I saw where the Cubs were playing.
The bases were loaded and someone got a hit, but it hit a runner…so it was ruled a base hit, but no runs scored and I remarked to my friend that only the Cubs could manage a base hit without getting a run.
The rarest play in baseball
Runners on first and second, no out. A fly ball is hit to the short stop, the infield fly rule is called and the batter is out. The runner on first advances to second, passes the lead runner, and is called out. When the fly ball comes down, it hits the runner who was on second base, but is now standing off base.
It has never happened in professional baseball, and, probably never will.
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