Mathematical Vick

It’s difficult to live inside the beltway and not cheer for the Washington Redskins. The cashier at the grocery store, the teller at the bank, and even the soft pretzel vendor at the corner of 7th and Independence — all of them are just a little friendlier after a Redskins victory. So even though I’ve been a Steelers fan since birth, I still root for the burgundy and gold, knowing that my reward will be better customer service the following day.

But this past Monday night, I found myself cheering against the Redskins and for Michael Vick. Are you kidding me? Six touchdowns — two running and four passing — with 333 passing yards, 80 rushing yards, and a 20-for-28 performance with 0 interceptions. Wow. With his effort, he set the record for most fantasy points ever earned by a quarterback.

His performance earned him a passer rating of 150.7, just shy of the perfect quarterback ranking of 158.3. Which brings me to a question — WTF? Since when has 158.3 ever been considered perfect? Why not just multiply the result by 100/158.3 and convert it to a 0‑100 scale? Or multiply by 28/158.3 to convert it to a 0‑28 scale, in which case the top score would be truly perfect?

Have you ever looked at the formula for passer rating in the NFL? What a mess. Here’s how it works:

Calculate a, b, c, and d as follows:

• a = 5 × (completions/attempts – 0.3)
• b = 0.25 × (yards/attempts – 3)
• c = 20 × touchdowns/attempts
• d = 2.375 – (25 × interceptions/attempts)

Then, the value of each of a, b, c, and d must be between 0 and 2.375. If the value is negative, use 0 instead; if it’s greater than 2.375, use 2.375. Finally, once you have the four values, the final passer rating is equal to:

100/6 × (a + b + c + d).

For Vick’s performance on Monday night, the calculations for a, b, c, and d look like this:

• a = 5 × (20/28 – 0.3) = 2.071
• b = 0.25 × (333/28 – 3) = 2.223
• c = 20 × 4/28 = 2.857
• d = 2.375 – (25 × 0/28) = 2.375

Note that c = 2.857 above, but because a, b, c, and d cannot exceed 2.375, a value of c = 2.375 is used in the final step. Consequently, his final passer rating was:

100/6 × (2.071 + 2.223 + 2.375 + 2.375) = 150.733,

which the media reports to the nearest tenth, 150.7.

There are some interesting questions that can be asked, based on the formula. For instance, to garner a perfect rating:

• What percent of passes must be completed?
• How many yards, on average, must be gained per pass attempt?
• What percent of passes must result in a touchdown?
• How many interceptions can be thrown?

(In case you want to think about these questions, answers are included at the bottom of the post.)

All of this football talk reminds me of a math joke…

A college football coach walked into the locker room before a big game, looked at his star quarterback, and said, “You’re academically ineligible because you failed your math mid-term. But we really need you today. I talked to your math professor, and he said that if you can answer just one question correctly, then you can play today. So, pay attention. I really need you to concentrate on the question I’m about to ask you.”

“Okay, coach,” the player agreed. “I’ll do my best.”

“Good,” said the coach. Then he asked, “Okay, now really focus. What is 2 + 2?” All of his teammates watched quietly while the quarterback thought about the question.

The quarterback thought for a moment. Sheepishly, he answered, “Um, 4?”

“Really?” said the coach. “Did you really just say 4?”

To which his teammates shouted, “Oh, c’mon, coach! Give him another chance!”

For a perfect passer rating of 158.3, a quarterback must do the following: 

• Complete 31 of every 40 passes (77.5%).
• Maintain at least 12.5 yards per attempt.
• Score a touchdown on 19 of every 160 passes (11.875%).
• Throw 0 interceptions.