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### Why does Kaggle use Log-loss?

If you're not familiar with Kaggle, it's an organization dedicated to data science competitions to both provide ways for companies to potentially do analytics at less cost, as well as to identify talented data scientists.

Competitions are scored using a variety of functions, and the most common for binary classification tasks with confidence is something called log-loss, which is essentially $$\sum_{i=1}^{n} p_i\cdot\log(p_i)$$, where $$p_i$$ is your model's claimed confidence for test data point $$i$$'s correct label. Why does Kaggle use this scoring function? Here I'll follow Terry Tao's argument.

Ideally what we'd like is a scoring function $$f(x)$$ that yields the maximum expected score precisely when the claimed confidence $$x_i$$ in the correct label for $$i$$ is actually what the submitter believes is the true probability (or frequency) of that outcome. This means that we want $L(x)=p\cdot f(x) + (1-p)\cdot f(1-x)$ for fixed $$p$$ to be maximized when $$x=p$$. Differentiating, this means $L'(x) = p\cdot f'(x) - (1-p)\cdot f'(1-x) = 0$ when $$x=p$$, hence $$p\cdot f'(p) = (1-p)\cdot f'(1-p)$$ for all $$p$$. This will be satisfied by any admissible $$f(x)$$ with $$x\cdot f'(x)$$ symmetric around $$x=\frac{1}{2}$$, but if we extend our analysis to multinomial outcomes we get the stronger conclusion that in fact $$x\cdot f'(x) = c_0$$ for some constant $$c_0$$. This in turn implies $$f(x)=c_0\cdot \log(x)+c_1$$. If we want $$f(1/2)=0$$ and $$f(1)=1$$, we end up with $$f(x)={\log}_2(2x)$$ and the expected score is $L(x)=x\cdot {\log}_2(2x) + (1-x)\cdot {\log}_2(2(1-x)).$

### 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…