The Yankees scored 13—count 'em--thirteen runs in the second inning tonight en route to a 19-8 shellacking of the D Rays. This was just one behind the AL record shared by the 2003 Red Sox (6/27 against the Marlins) and 1950 Indians (6/18 vs. the A's).

However, the Yanks weren't the only ones running up the score tonight. The Tigers beat the O's 13-3, the Cards beat the Pirates 11-1, and the Red Sox beat the Jays 12-7. Meanwhile the Braves beat the 'Stros 1-0 in 12. Go figure. In 15 slated games, three were won by at least ten runs, and two are still in progress. In one the Long Beach Angels lead the M's 6-0 in the sixth, so it may be four games before the night's through.

When you consider that in 1916 only 17 games in the entire season had a ten-run differential, it becomes more impressive. The most won by at least ten was 113 in 1996.

For fun let's figure out the odds of three ballgames by at least ten runs in a slate of 15. I won't explain how I derive it since it involves a lot of factorials and binomials. However, overall 3.27% of all games had a ten-run differential, meaning that there is slightly over one percent of a chance that 15 games will end up with a ten-run winner. Given that there are six solid months of games and let's say 63 nights of baseball per year, three 10+ victories in fifteen games shouldn't happen more than once every couple of years, less if you consider that until expansion, baseball never hosted more than eight games a night, and many nights throughout the season have much less than a 15-game slate anyway.

Here are the years with the highest odds:

Now the lowest:

2004 saw 3.83% of games won by at least 10 runs and the odds of our scenario were 1.60%.

By the way, I lied. Here's the formula: [15! / (3! * (15-3)!)] * P^3 * P^(15-3), where P is the probability in any given year. If you got a problem wid dat, tell it to Bernoulli, not me.