Badenhop's numbers for the past two years, as provided by

**nvalvo**
2012 62.1 IP, 9.1 H/9, 0.9 HR/9, 1.7 BB/9, 6.1 SO/9

2013 62.1 IP, 9.0 H/9, 0.9 HR/9, 1.7 BB/9, 6.1 SO/9

PrometheusWakefield said:

Jesus, what are the chances of that happening?

Well, assuming that he pitches the same 63 innings in two consecutive years, about 1 in 9,000.

It's actually not so unusual for some of these numbers to be so consistent. For example, the home runs: Badenhop gave up 6 in 2012 and 6 in 2013. Home runs must be integers -- you can't give up half a home run. If a pitcher does pitch the same number of innings, and has roughly the same HR rates, he's very likely to give up somewhere between 3 and 8 HR. That means that in two consecutive years, the chances that the number of HR is identical are actually pretty good: about 1 in 6, or 16 percent.

Now, when the number of events is larger, the range of likely values increases -- it's much less probable that the number of strikeouts in two consecutive years should be 42 than that the number of home runs should be 6. Nonetheless, since the largest number of events here is only 63 per year (hits in 2012), and since the range of likely numbers scales as the square root of the number of events (so, for 63 hits per year, we expect a range of roughly +/- 8 hits), even matching 4 sets of events isn't that unlikely.

Running 13 simulations of one million instances each yielded probabilities ranging from 1 in 6897 to 1 in 10417. Code available upon request