BP Unfiltered: The Ultimate Showdown of Ultimate Rarities

On Wednesday's episode of Effectively Wild, sponsored by the Baseball Reference Play Index and featuring BP's own Editor-in-Chief Sam Miller and some other guy who now works for Disney and is therefore dead to us, a listener asked one of those questions that means nothing in the grand scheme of things, but… can't… resist… answering.

Which is more likely? That a batter would hit 5 home runs in the same game or that a pitcher would record 5 strikeouts in the course of an inning? In baseball history, there have been 16 players who have hit 4 home runs in a game and 71 innings where a pitcher recorded 4 strikeouts in an inning (Chuck Finley did it three times in his career!)

Just in case: Yes, it is possible to record more than three strikeouts in an inning. If a catcher drops the ball on a third strike, then the ball is live and the batter becomes a runner who can try to make it to first and must either be tagged or thrown out. It happens once in a while that a batter swings at and misses a two strike pitch, but it gets to the backstop and he takes first safely. The official scoring is a strikeout, but with the runner to first and no out recorded.

Warning! Gory Mathematical Details Ahead!

Let me flip over this envelope here and first look at the probabilities of 5 K's in one inning:

Despite the fact that we know that pitchers vary in their ability to strike batters out and that catchers vary in their ability to block balls in the dirt from getting away, we're just going to use league averages. For a five-strikeout inning to happen, two things need to be true. All five "outs" need to be of the strikeout variety. We don't necessarily need this to be a 1-2-3-4-5 inning, so we don't care what happens in any plate appearances that don't result in outs. However, 2 of the first 4 strikeouts need to be of the dropped third strike kind. The fifth one will have to necessarily be a clean K, because if more than three of the first four are clean, then the inning is already over.

(Technically, we also need the first guy who gets on base to move off of first, while not recording an out, and to keep other runners off first as well. The drop rule doesn't apply with a runner on first—unless there are two outs—but we're going to solve that problem by ignoring it.)

In 2013, when a batter made an out, he did so via strikeout 29.3 percent of the time. Thankfully, we live in a high strikeout era! Using a simple binary distribution, the chances of five consecutive outs happening via strikeout are 0.216%.

From 2003-2013, 97.6 percent of all strikeouts had the putout credited to the catcher. Either he caught the third strike cleanly or if the ball fell, he scooped it up and tapped the batter on the shin. That leaves 2.4 percent of strikeouts in which either a throw down to first was needed or where the batter made it safely, and the batter only made safe landfall at first in 1 out of 8 of those situations (0.3 percent of all strikeouts). The chances that at least 2 out of the first 4 of the strikeouts in our scenario would also feature a runner getting to first are 0.00000108%.

Multiply those numbers together and you get a 0.0000000023328% chance that an inning would happen like this. As some manner of scaling, that would be once every 42,866,941,015 half innings or roughly 980,039 seasons. We're almost literally looking at a 1 in a million chance that it will happen this year.

You could make the case that certain pitchers who strike more guys out or would be more likely to have a 5 K inning, but they also only account for a small percentage of all innings thrown. Alternately, there could be a bunching effect (strikeouts, like celebrity deaths, tend to come in groups of three?) but bunching effects —better known as "the hot hand"—have not generally appeared when we've gone looking for them.

On to five home runs in a game!

Figuring this one from a straight-up probability angle is easy enough, but we do need to talk about a couple issues. For one, if a player were to hit four home runs in a game, the opposing pitcher(s) probably didn't bring their A-game that day and it may be a blowout. If he gets a fifth at-bat, he's probably facing a sacrificial bullpen lamb who will be demoted tomorrow (so he's probably not good at HR prevention), but that guy may also not want to be remembered as an answer to the trivia question "Who gave up Duane Kuiper's fifth home run in that game?" On the flip side, the batter is most definitely swinging for the fences! If the game is somehow still close, the guy with four home runs will be intentionally walked. Yes, the fans will boo, but we gotta win.

Again, we will solve the problem by ignoring it.

In 2013, 2.5 percent of all plate appearances ended in a home run. Assuming that our player is league-average and that he had to go 5-for-5, he would do so in 0.0000009765625% of games, or roughly one in 102,400,000 games. However, there are 18 hitters in each of 2,430 games every year, so we would expect to see this once every 2,341 years. If we give him 6 chances to hit 5 HR, we would expect to see it once every 398 years.

There probably are bunching effects here too. Maybe he's having a "locked in" day, but even if we pretended that there was a hitter who was always Barry Bonds in 2001 (Bonds homered in 11.6 percent of the plate appearances where he wasn't intentionally walked) and gave him six chances to hit five home runs each game, we would expect him to do it once every 54 years.

It's not likely that a five home run game would happen, but it's much more likely to happen (by a couple orders of magnitude) than a five strikeout inning. The dropped-third-strike non-strikeout turns out to be so rare that expecting two to happen in the same inning is just too much to ask on top of a pitcher striking out five batters in the same inning. Then again, what sort of pitcher would have the stuff to throw two wild pitches in an inning, and yet get five hitters to strike out?

I guess it would be one who was (wait for it…)

Effectively Wild.