Last Sunday, in the final weekend of the NFL's regular season, the Tennessee Titans scored touchdowns on four straight touches of the ball against the Jacksonville Jaguars, but none of these scores was accomplished by the Titans' offensive unit. I thank my father and brother for bringing this occurrence to my attention. As summarized in this
article:
The first touchdown was scored after [Tennessee's] Zach Brown intercepted [Jacksonville's] Chad Henne and ran the ball back 79 yards for a defensive touchdown. After the Jaguars offense was forced to punt on their very next possession, the punt was returned 69 yards for another touchdown by Darius Reynaud. The Jaguars offense once again took the field without a break and once again they were forced to punt and, once again, Darius Reynaud returned the punt for a touchdown, this time for 81 yards. Finally, the last touchdown was scored on another Zach Brown pick six, this time from 30 yards out.In the remainder of this posting, I will estimate the probability of the Titans' achievement. First, what is the likelihood of two interception returns for touchdowns (also known as "pick sixes")
and two punt returns for touchdowns occurring in the same game? (At this stage, we're not taking into account that all four touchdowns were scored by the same team and done consecutively.)
The "Fifth Down" blog in the
New York Times just happened to have a
write-up on pick sixes the other day, which provided the following statistics on this past NFL season:
- 17,788 passes were attempted.
- 71 of these passes were intercepted and brought back for touchdowns (an NFL record for one season).
Based on these figures, the ratio of attempted passes that ended up as pick sixes was .004 (4-in-1,000 or 1-in-250).
We need the same kind of statistics on punt returns and, fortunately, they are
readily available:
- There were 1,133 punt returns in the NFL this past seasons (excluding fair catches, in which the returner raises his hand to signal that he will not run with the ball after catching it, in exchange for not being clobbered).
- 18 punt returns resulted in touchdowns.
The TD ratio is thus .016, a little over 1-per-100.
To estimate the probability of two pick sixes and two punt-return TDs in one game, we multiply together:
.004 X .004 X .016 X .016 = .000000004 or 4-in-1 billion.
(This calculation assumes
independence of observations, that the outcome of any one play had no effect on the other plays, akin to coin tosses. However, there could be common causes amongst the plays, such as the Jaguars' offensive and special-teams players being poor tacklers.)
Next, let's tackle the issue of all four of these touchdowns being scored by Tennessee, instead of say, two by Tennessee and two by Jacksonville, or three by one team and one by the other. Let's assume, for each of the touchdowns, that it was just as likely Jacksonville could have scored it as Tennessee (i.e., a coin flip). The probability of the Titans winning all four coin flips, if you will, is 1/16. Multiplying our previous value of .000000004 by .0625 yields .00000000025 or 2.5-in-10 billion.
Lastly, four touchdowns by the same team, via two interceptions and two punt returns, would not necessarily have to occur in four consecutive possessions by the victimized team. Here is a screen-capture of ESPN.com's
drive chart for the Jaguars vs. the Titans.
The four drives leading to Tennessee touchdowns are denoted with blue arrows. Note also that Jacksonville's drive beginning with 33 seconds left in the first half is excluded, as there was little remaining time for anything major to happen. We are thus left with 12 Jacksonville possessions, with Tennessee's non-offensive touchdowns occurring on drives 5, 6, 7, and 8. There are many possible combinations of drives on which the four Titans' TDs could have occurred, nine of which are consecutive (e.g., 1-2-3-4, 2-3-4-5, ..., 9-10-11-12), and many additional others that are non-consecutive (e.g., 1-4-9-10, 4-6-9-11).
The total number of possible combinations can be obtained via the
n-choose-k principle, in this case raising the question of how many ways can four drives (
k) be selected from 12 total drives (
n). Using this
online calculator, we learn that there are 495 possible ways to select four objects from a total of 12. Nine of these sets are consecutive, as noted above. The probability of the four TDs occurring on four consecutive drives is thus 9/495 or .018.
We thus multiply our previous interim probability of .00000000025 by .018, yielding .0000000000045. This translates to roughly
4.5-in-1 trillion. Suffice it to say, we are unlikely to see anything like this ever again. The funny thing is, as shown in the above drive chart, Tennessee intercepted a Jacksonville pass on the Jaguars' next drive following the four touchdowns, but the interceptor didn't make a touchdown on the return. Therefore, we possibly could have had
five consecutive non-offensive TDs by Tennessee.
The "Game notes" at the bottom of this
ESPN.com article on the Jacksonville-Tennessee game cited two similar occurrences to what Tennessee pulled off: "The Titans also had three return touchdowns Sept. 23 in an overtime win over Detroit. Reynaud had a 105-yard kickoff return in that game [
link]. ... The Seahawks had four interception returns for TDs on Nov. 4, 1984, against Kansas City in a 45-0 rout, and three different players scored those TDs [
link]." I checked the records from these games and in neither case were the return touchdowns all consecutive.
As noted throughout, the above calculations have required several simplifying assumptions. Also, it's certainly possible that I made a calculation or logical error somewhere along the way. If you spot something, please let the world know in the Comments section.
