A baseball-playing friend stymied me with this question. Seven batters step up to the plate in one inning, all from the same team. The inning ends and no one has scored. How? We came up with bases loaded, two pop-outs, a grand slam, and one more pop-out, when the other team claims the grand slam hitter didn't tag third and the four runs are nullified. But we regard this answer as a copout, and the guy won't tell us the solution. Help!
Another one of those days when I not only have to give the answers, I have to clean up the questions, too. An essential condition you left out is that no substitutions can be made in the lineup. If you do allow substitutions, the number of players who can step up to the plate is limited only by the number of players you’ve got suited up. Even a casual baseball fan like me (I hate wearing ties) knows you can substitute batters freely during an inning.
As for your appealed grand slam stratagem, I wouldn’t call it a copout. I’m obliged to call it wrong. Still, you’re on the right track, as we shall see. Failing to tag a base is what we call an appeal play. The runs are nullified only if the opposing team appeals to the umpire before the next pitch. Since you throw in an intervening out, the runs stand. I learn this from rule 7.10 of my Official Baseball Rules, the holy writ of the sport.
Here’s the correct solution, helpfully supplied by loyal Straight Dope reader Kirk Miller of Richardson, Texas:
“Bases are loaded (three batters). Two outs (two more batters). The sixth batter hits a grand slam home run, but fails to touch first base. The umpire throws a new ball to the pitcher, and the seventh batter steps up to the plate. The pitcher successfully appeals that the home run hitter failed to touch first base. The umpire signals the third out, and no runs count. (Had the batter failed to touch second, third, or home, one or more runs would have counted.)
“Note: The defensive team cannot appeal the missed first base until the ball is put into play, which is done by the pitcher stepping up on the pitching rubber while holding the ball (when the seventh batter steps up to the plate). So they couldn’t appeal until the seventh batter got up; hence, the number of batters in the conundrum (seven instead of six).”
Nice, eh? Follow up with this next stumper and you’ll have ’em convinced you’re a baseball god. Nine players have been named most valuable player in two consecutive years. Coincidentally they constitute a complete lineup (all positions covered). Name the nine.
SURPRISE ANSWER: Hal Newhouser, pitcher, 1944-’45; Yogi Berra, catcher, 1954-’55; Jimmie Foxx, first base, 1932-’33; Joe Morgan, second, 1975-’76; Ernie Banks, shortstop, 1958-’59 (he played short in the 50s, so no snotty letters please); Mike Schmidt, third base, 1980-’81; Mickey Mantle, outfield, 1956-’57; Roger Maris, outfield, 1960-’61; Dale Murphy, outfield, 1982-’83. Hope you can turn this to profitable use on your next trip to the corner saloon.
The amateurs get into the act
Here’s another good trivia question. In one year, the MVPs in the National League, the American League, the National Football League and the American Football League (before they merged) all wore same number. Name the players. HINT: the number was 32. ANSWER: in 1963, Sandy Koufax of the Dodgers, Elston Howard of the Yankees, Jim Brown of the Browns, and Cookie Gilchrist of the Buffalo Bills were MVPs. I can’t tell you how much money this has netted me over the years.
Hope you didn’t spend it, Dave, because you’re going to have to give it back. Only three of the four MVPs in 1963 wore number 32–Koufax, Howard, and Brown. The fourth MVP wasn’t Cookie Gilchrist of the Bills, but Clem Daniels of the Raiders. He was number 36.
Send questions to Cecil via email@example.com.