### Solving the TopSpin Puzzle using GAP

TopSpin is an oval-track permutation puzzle that was made by Binary Arts; similar puzzles are made and sold by other manufacturers. Here's the Binary Arts TopSpin.

It's not a difficult puzzle to solve if you play around with it for a few hours and figure out how to generate various permutations. It's more interesting (and difficult) if you observe that the turntable has a distinguishable top and bottom. This suggests an interesting question - can you invert the turntable while keeping the numbers in the track in the same order?

The answer is, perhaps surprisingly, yes. Here's one way to find a sequence of operations that produces precisely this outcome.

GAP (Groups, Algorithms and Programming) is a freely available programming language that specializes in computational group theory, and it's perfect for solving permutation puzzles. Here's my GAP code for TopSpin. Label the top of the turntable with 21 and the bottom with 22. Flipping the turntable generates the permutation $(1,4)(2,3)(21,22);$ rotating the oval to the left generates the permutation $(2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,1).$ Denoting these two permutations as $$x$$ and $$y$$, we can now simply ask GAP to find a sequence of operations on the free group generated by $$x,y$$ that results in the pre-images $$(1,2)$$, flipping two adjacent numbers on the track, leaving everything else the same (including turntable parity); and $$(21,22)$$, flipping turntable parity, leaving everything else the same. This corresponds to the operation of flipping the turntable while keeping the order of the numbers in the track the same.

Running the code, we get these lovely results:
(1,2) = y*x^-1*y^-1*x^-1*y*x^-1*y*x^-1*y^-1*x^-1*y^-1*x^-1*y^2*x^-1*y^-1*x^-1*y^-1*x^-1*y^4*x^-1*y^-1*x^-1*y^2*x^-1*y^-2*x^-1*y*x^-1*y^2*x^-1*y^-3*x^-1*y^-3*x*y^-4*x*y*x*y^-1*x*y^4*x^-1*y^-5*x^-1*y^-1*x^-1*y^5*x^-1*y^-6*x^-1*y^2*x^-1*y^-1*x^-1*y^5*x*y*x*y*x*y*x*y*x*y^5*x*y*x*y^-1*x*y^-5*x^-1*y^-1*x^-1*y^-1*x^-1*y^-2*x*y*x*y^-1*x*y^4*x*y^-1*x*y^3*x^-1*y^-1*x^-1*y^6*x^-1*y^-1*x^-2*y^-4*x^-3*y^-3*x^-2*y^-1*x^-1*y*x^-1*y^2*x^-1*y^-2*x^-1*y^-1*x^-1*y*x*y*x^-1*y*x^-1*y^-1*x^-1*y^-2
(21,22) = y*x^-1*y^2*x*y^-1*x*y*x*y^-1*x*y^-2*x*y^2*x*y^-1*x*y*x*y*x*y^-2*x^-1*y*x^-1*y^-2*x^-1
Since there are sequences of operations that allow us to flip any two adjacent numbers or the parity of the turntable, it follows that all possible configurations are both solvable and achievable.

1. This comment has been removed by the author.

### 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) = …

### Mixed Models in R - Bigger, Faster, Stronger

When you start doing more advanced sports analytics you'll eventually starting working with what are known as hierarchical, nested or mixed effects models. These are models that contain both fixed and random effects. There are multiple ways of defining fixed vs random random effects, but one way I find particularly useful is that random effects are being "predicted" rather than "estimated", and this in turn involves some "shrinkage" towards the mean.

Here's some R code for NCAA ice hockey power rankings using a nested Poisson model (which can be found in my hockey GitHub repository):
model <- gs ~ year+field+d_div+o_div+game_length+(1|offense)+(1|defense)+(1|game_id) fit <- glmer(model, data=g, verbose=TRUE, family=poisson(link=log) ) The fixed effects are year, field (home/away/neutral), d_div (NCAA division of the defense), o_div (NCAA division of the offense) and game_length (number of overtime periods); off…