Skip to main content

Why does Kaggle use Log-loss?

How Unfair are the NFL's Overtime Rules?

In 2010 the NFL amended its overtime rules, and in 2012 extended these to all regular season games. Previously, overtime was handled by sudden death - the first team to score won. The team winning a coin flip can elect to kick or receive (they invariably receive, as they should).

Assuming the game ends in the first overtime, the team with the first possession wins under the following scenarios:
  1. scores a touchdown on the first drive
  2. kicks a field goal on the first drive; other team fails to score on the second drive
  3. both teams kick a field goal on the first and second drives; win in sudden death
  4. doesn't score on the first drive; defensive score during second drive
  5. neither team scores on first or second drives; win in sudden death
Under this overtime procedure, roughly how often should be expect the team winning the coin flip to win the game?

For an average team the empirical probabilities of the above events per drive are:
  • \(\mathrm{defensiveTD} = \mathrm{Pr}(\text{defensive touchdown}) = 0.02\)
  • \(\mathrm{safety} = \mathrm{Pr}(\text{safety}) = 0.001\)
  • \(\mathrm{fieldGoal} = \mathrm{Pr}(\text{field goal}) = 0.118\)
  • \(\mathrm{offensiveTD} = \mathrm{Pr}(\text{offensive touchdown}) = 0.200\)
We'll also use the following:
  • \(\mathrm{defensiveScore} = \mathrm{Pr}(\text{any defensive score}) = \mathrm{defensiveTD} + \mathrm{safety}\)
  • \(\mathrm{offensiveScore} = \mathrm{Pr}(\text{any offensive score}) = \mathrm{fieldGoal} + \mathrm{offensiveTD}\)
  • \(\mathrm{noOFscore} = \mathrm{Pr}(\text{no offensive score}) = 1 - \mathrm{offensiveScore}\)
  • \(\mathrm{noScore} = \mathrm{Pr}(\text{no score}) = 1 - \mathrm{offensiveScore} - \mathrm{defensiveScore}\)
  • \(\mathrm{sdWin} = \mathrm{Pr}(\text{driving team winning under sudden death rules})\)
Then the probabilities of the above numbered outcomes is approximately:
  1. \(\mathrm{offensiveTD}\)
  2. \(\mathrm{fieldGoal}\times \mathrm{noOFscore}\)
  3. \(\mathrm{fieldGoal}\times \mathrm{fieldGoal}\times \mathrm{sdWin}\)
  4. \(\mathrm{noScore}\times \mathrm{defensiveScore}\)
  5. \(\mathrm{noScore}\times \mathrm{noScore}\times \mathrm{sdWin}\)
The last thing we need to work out is \(\text{sdWin}\). There are three ways for the team with possession to win:
  1. any offensive score on the first drive
  2. no offensive score; any defensive score on the second drive
  3. neither team scores on the first or second possessions; we're back to our original state
These three scenarios have values:
  1. \(\mathrm{offensiveScore}\)
  2. \(\mathrm{noOFscore}\times \mathrm{defensiveScore}\)
  3. \(\mathrm{noScore}\times \mathrm{noScore}\times \mathrm{sdWin}\)
Doing the math, we get that \begin{align*}
\mathrm{sdWin} &= \mathrm{offensiveScore} + \mathrm{noOFscore}\times \mathrm{defensiveScore} + {\mathrm{noScore}}^2\times\mathrm{sdWin};\\
\mathrm{sdWin} &=\frac{(\mathrm{offensiveScore} + \mathrm{noOFscore}\times \mathrm{defensiveScore})}{(1-{\mathrm{noScore}}^2)}.
Putting it all together we get \[
\text{win} = \mathrm{offensiveTD} + \mathrm{fieldGoal}\times \mathrm{noOFscore} + \mathrm{noScore}\times \mathrm{defensiveScore}\\
+ (\mathrm{fieldGoal}^2+ \mathrm{noScore}^2)\times \mathrm{sdWin}.\]
Plugging in our empirical values, we finally arrive at \[\mathrm{Pr}(\text{win coin flip, win game}) = 0.560.\] For comparison, under the original sudden death rules, \[\mathrm{Pr}(\text{win coin flip, win game}) = 0.589.\] So the NFL overtime rules are still ridiculously unfair in favor of the winner of the coin flip, but not as ridiculously unfair as they were under the original sudden death rules.

How do these numerical results compare to actual outcomes? Under the current overtime rules, there have been 51 overtime games. In 27 of these the team winning the coin toss won the game, in 21 the team losing the coin toss won the game and there have been 3 ties. That puts \(\mathrm{win} = \frac{27}{48} = 0.5625\) for games not ending in ties. Close enough!

If you'd like to tweak the probabilities for each event to see how the resulting probability for the winner of the coin flip changes, I have a simple Python script here.


Popular posts from this blog

A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written \( \Pr(A \mid B) \)
probability of B is written \( \Pr(B) \)

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: \[ \Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)} \]Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be \[ w=\frac{P^e}{P^e+1},\] where \(w\) is the win fraction expectation, \(P\) is runs/allowed (or similar) and \(e\) is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with \(g\) games is \[W = g\cdot w = g\cdot \frac{P^e}{P^e+1},\] so for one estimate we only need to compute the value of the partial derivative \(\frac{\partial W}{\partial P}\) at \(P=1\). Note that \[ W = g\left( 1-\frac{1}{P^e+1}\right), \] and so \[ \frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}\] and it follows \[ \frac{\partial W}{\partial P}(P=1) = …

Behind the Speadsheet

In the book "The Only Rule Is It Has to Work: Our Wild Experiment Building a New Kind of Baseball Team", Ben Lindbergh and Sam Miller recount a grand adventure to take command of an independent league baseball team, with the vision of trying every idea, sane or crazy, in an attempt to achieve a winning edge. Five infielders, four outfielders, defensive shifts, optimizing lineups - everything.

It was really an impossible task. Professional sports at every level are filled with highly accomplished and competitive athletes, with real lives and real egos. Now imagine walking in one day and suddenly trying to convince them that they should be doing things differently. Who do you think you are?

I was one of the analysts who helped Ben and Sam in this quest, and I wanted to write some thoughts down from my own perspective, not as one of the main characters, but as someone more behind the scenes. These are some very short initial thoughts only, but I'd like to followup with some…