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On average, an NFL offense needs 3.7 first downs (including the score itself) to score a touchdown. Therefore, the estimated TD rate would be 0.65^3.7 = 0.20 TDs per drive. (Note: The actual share of drives that resulted in touchdowns over the past five years is very close--19%.)
One way to think of those 4 extra yards is that they would typically require 0.4 more first downs to score. The resulting effect on the probability of scoring is 0.65^4.1 = 0.17. The difference is 0.20-0.17 = 0.03.
A difference of only 3% in the chance of scoring a TD on a typical offensive drive may seem very small, but it has a large impact on points. Given a league average of 12.4 drives per game (according to KC Joyner), the effect on two teams with a 4-yd difference in starting field position would be:
0.17 * 12.4 = 2.1 TDs per game (14.7 points)
0.20 * 12.4 = 2.5 TDs per game (17.4 points)
The result is a 0.4 TD per game advantage to a team with a 4-yd field position edge, the equivalent of 2.8 points per game. But it wouldn't work out exactly that way, because there is obviously no such thing as 0.4 touchdowns. So sometimes a team would end up with an additional TD, sometimes not, but perhaps sometimes 2 additional TDs. In my view, this effect is very meaningful.
Here is perhaps a simpler way to conceptualize it. Instead of saying the team with lesser starting field position needs 0.4 more 1st downs per drive to score, we could say that they need a full additional 1st down in 40% of its drives.
The resulting probabilities of successful TD drives are somewhat simpler to understand. This time I'll say the average # of drives per game is 10, which I believe is closer to the actual number than KC's 12.4 number.
0.65^3.7 = 0.20 probability of TD drive
0.20 * 10 drives/game = 2.1 TD drives/game
0.65^3.7 = 0.20 probability of TD drive
0.65^4.7 = 0.13 probability of TD drive
0.20 * 6 drives/game + 0.13 * 4 drives/game = 1.7 TD drives/game
Again, the difference is 0.4 TDs/game.
One way to think of those 4 extra yards is that they would typically require 0.4 more first downs to score. The resulting effect on the probability of scoring is 0.65^4.1 = 0.17. The difference is 0.20-0.17 = 0.03.
A difference of only 3% in the chance of scoring a TD on a typical offensive drive may seem very small, but it has a large impact on points. Given a league average of 12.4 drives per game (according to KC Joyner), the effect on two teams with a 4-yd difference in starting field position would be:
0.17 * 12.4 = 2.1 TDs per game (14.7 points)
0.20 * 12.4 = 2.5 TDs per game (17.4 points)
The result is a 0.4 TD per game advantage to a team with a 4-yd field position edge, the equivalent of 2.8 points per game. But it wouldn't work out exactly that way, because there is obviously no such thing as 0.4 touchdowns. So sometimes a team would end up with an additional TD, sometimes not, but perhaps sometimes 2 additional TDs. In my view, this effect is very meaningful.
Here is perhaps a simpler way to conceptualize it. Instead of saying the team with lesser starting field position needs 0.4 more 1st downs per drive to score, we could say that they need a full additional 1st down in 40% of its drives.
The resulting probabilities of successful TD drives are somewhat simpler to understand. This time I'll say the average # of drives per game is 10, which I believe is closer to the actual number than KC's 12.4 number.
0.65^3.7 = 0.20 probability of TD drive
0.20 * 10 drives/game = 2.1 TD drives/game
0.65^3.7 = 0.20 probability of TD drive
0.65^4.7 = 0.13 probability of TD drive
0.20 * 6 drives/game + 0.13 * 4 drives/game = 1.7 TD drives/game
Again, the difference is 0.4 TDs/game.
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