So, the home team won yesterday, 6-4. Odd stat: their second baseman scored all six runs.
How many men did they leave on base in the 7th?
(Show your work.)
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Don’t I need to know where in the batting order the second baseman appears?Report
Intuitively, it’s an add-on requirement so there’s a unique solution to the question about the seventh inning. I have to hook up the new dryer and run other errands this afternoon, so won’t get a chance to lay this out until this evening. My initial interpretation of the requirement will be that it means all six runs are scored by the same position in the batting order.Report
SPOILER
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The pattern I come up with is this:
1. Run – Single – Single – Single – Out – Out – Out
2. Out – Out – Run – Single – Single – Single – Out
3. Out – Out – Single – Out
4. Run – Single – Single – Single – Out – Out – Out
5. Out – Out – Run – Single – Single – Single – Out
6. Out – Out – Single – Out
7. Run – Single – Single – Single – Out – Out – Out
8. Out – Out – Run . . . [various options for remainder of inning]
So, there have to have been three men left on base, because the maximum number of plate appearances need to be used between the 2nd baseman’s at-bats so that he comes to the plate the following inning with nobody on base.Report
Comments:
Home team only bats eight innings because it would be impossible for the second baseman to score two or more runs in the ninth inning. The maximum number of plate appearances per inning without scoring a run is six (three outs, three men left on base). The max. PA per inning and scoring only one run is seven, and two runs is eight.
In order to score six times in eight innings, I think he has to bat first, but I just started with the assumption that it would be easiest if he batted first.Report
I got a similar result, though I had the 2B batting first and sometimes being brought home by bases-loaded RBI singles. The puzzle doesn’t specify that this player gets the RBI associated with the run, only that she be the runner to cross home plate. She could get on base with a BB, also; that doesn’t matter.
Are we sure that there’s no way this could happen in nine innings? If Visitor and Home team are tied going in to the ninth, 2B’s sixth and final run might be a walk-off…
Anyway, I got distracted when considering if 2B batted second or third or even last in the order.Report
The final score is 6-4. If the home team batted in the 9th, the second baseman would have had to score twice i the 9th with no one else scoring; this is not possible.Report
Also, we know the 2nd baseman only came up 6 times, because 7 times would mean at least 55 plate appearances: 6 runs, 24 outs, and thus 25 LOB in 8 innings. That last is impossible.Report
This problem is taken from the article “Baseball Retrograde Analysis” by Jerry Butters and Jim Henle in the Mathematical Intelligencer, Vol.40 No.2 (2018), which contains several similar fun problems. I’m not sure whether it is behind a paywall.Report
Mr. Shaw has pretty much nailed it.Report
I was thinking ‘I need to do this with modulo arthimetic’, then I was thinking ‘I haven’t done modulo arthimetic in over 20 years, so yeah, no’
I started to brute force it, (but I had figured out the 8 inning thing and 5 full times thru the lineup plus the portions of the 6th time thru that are required) but started with the 2B later in the order, so gave up after getting lost in the weeds.
Now seeing the solution, I see I wasn’t completely on the wrong track, but…
…is there a ‘mathematically elegant’ way of solving this? Or just trying combos of batting order and run production until it works?
Eta – oh the other conclusion was that the 2b always batted in the same position in the lineup.
If that didn’t happen, it’s an appeal play, and if the away team failed to call out the illegal substitution, *that* would have been the headline.Report