Skip to main content

Why does Kaggle use Log-loss?

Baseball, Chess, Psychology and Pychometrics: Everyone Uses the Same Damn Rating System

Here's a short summary of the relationship between common models used in baseball, chess, psychology and education. The starting point for examining the connections between various extended models in these areas. The next steps include multiple attempts, guessing, ordinal and multinomial outcomes, uncertainty and volatility, multiple categories and interactions. There are also connections to standard optimization algorithms (neural  nets, simulated annealing).


Common in baseball and other sports, the log5 method provides an estimate for the probability \( p \) of participant 1 beating participant 2 given respective success probabilities \( p_1, p_2 \). Also let \( q_* = 1 -p_* \) in the following. The log5 estimate of the outcome is then:

p &= \frac{p_1 q_2}{p_1 q_2+q_1 p_2}\\
  &= \frac{p_1/q_1}{p_1/q_1+p_2/q_2}\\
\frac{p}{q} &= \frac{p_1}{q_1} \cdot \frac{q_2}{p_2}

The final form uses the odds ratio, \( \frac{p}{q} \). Additional factors can be easily chained using this form to provide more complex estimates. For example, let \( p_e \) be an environmental factor, then:

\frac{p}{q} &= \frac{p_1}{q_1} \cdot \frac{q_2}{p_2} \cdot \frac{q_e}{p_e}


The most common rating system in chess is the Elo rating system. This has also been adopted for various other uses, e.g. ``hot or not'' websites. This system assigns ratings \( R_1, R_2 \) to players 1 and 2 such that the probability of player 1 beating player 2 is approximately:

p &= \frac{e^{R_1/C}}{e^{R_1/C}+e^{R_2/C}}

Here \( C \) is just a scaling factor (typically \( 400/\ln{10} \) ). The Elo rating is connected to log5 via setting \( e^{R/C} = p/q \). We then recover:

\frac{p}{q} &= e^{R/C}\\
p &= \frac{e^{R/C}}{1+e^{R/C}}\\
R &= C\cdot \ln(p/q)

Note that \( p \) is also the probability of this player beating another player with Elo rating 0. The Elo system generally includes enhancements accounting for ties, first-move advantage and also an online algorithm for updating ratings. We'll revisit these features later.


The Bradley-Terry-Luce (BTL) model is commonly used in psychology. Given two items, the probability \( p \) that item 1 is ranked over item 2 is approximately:

p &= \frac{Q_1}{Q_1+Q_2}

In this context \( Q_* \) typically reflects the amount of a certain quality. That this model is equivalent to the previous models is immediate:

Q &= e^{R/C} = p/q\\
R &= C\cdot \ln(Q) = C\cdot \ln(p/q)\\
p &= \frac{Q}{1+Q}


The dichotomous (two-response) Rasch and item response models are commonly used in psychometrics. For the Rasch model, let \( r_1 \) represent a measurement of ability and \( r_2 \) the difficulty of the test item. The Rasch model estimates the probability of correct response \( p \) as:

p &= \frac{e^{r_1-r_2}}{1+e^{r_1-r_2}}

The one-parameter item response model estimates:

p &= \frac{1}{1+e^{r_2-r_1}}

These are clearly equivalent to each other and to the previous models.


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…