### Solving Sir Arthur Eddington's Zoo Puzzle using Dirac Matrices

Sir Arthur Eddington posed this difficult logic puzzle to the readers of Caliban's (Hubert Phillips) newspaper puzzle columns, and famously (or infamously) provided a solution using Dirac matrices.

Sir Eddington's zoo puzzle:

I took some nephews and nieces to the Zoo, and we halted at a cage marked
1. Tovus Slithius, male and female.
2. Beregovus Mimsius, male and female.
3. Rathus Momus, male and female.
4. Jabberwockius Vulgaris, male and female.
The eight animals were asleep in a row, and the children began to guess which was which. "That one at the end is Mr Tove." "No, no! It's Mrs Jabberwock," and so on. I suggested that they should each write down the names in order from left to right, and offered a prize to the one who got most names right.

As the four species were easily distinguished, no mistake would arise in pairing the animals; naturally a child who identified one animal as Mr Tove identified the other animal of the same species as Mrs Tove.

The keeper, who consented to judge the lists, scrutinised them carefully. "Here's a queer thing. I take two of the lists, say, John's and Mary's. The animal which John supposes to be the animal which Mary supposes to be Mr Tove is the animal which Mary supposes to be the animal which John supposes to be Mrs Tove. It is just the same for every pair of lists, and for all four species.

"Curiouser and curiouser! Each boy supposes Mr Tove to be the animal which he supposes to be Mr Tove; but each girl supposes Mr Tove to be the animal which she supposes to be Mrs Tove. And similarly for the other animals. I mean, for instance, that the animal Mary calls Mr Tove is really Mrs Rathe, but the animal she calls Mrs Rathe is really Mrs Tove."

"It seems a little involved," I said, "but I suppose it is a remarkable coincidence."

"Very remarkable," replied Mr Dodgson (whom I had supposed to be the keeper) "and it could not have happened if you had brought any more children."

How many nephews and nieces were there? Was the winner a boy or a girl? And how many names did the winner get right?

Solution:

Note that every child permutes the species, and we may indicate these as $$4\times 4$$ permutation matrices, where the row number indicates the actual species numbers and the column number indicates the species number guessed by the child. We can use a little trick to include gender. Consider the species permutation matrix for a given child, but use a -1 in the entry if the child additionally switches gender, too.

Here's an illustrative example. Assume a child transposes species 1 (Tovus Slithius) and species 3 (Rathus Momus), but doesn't switch genders; and transposes species 2 (Beregovus Mimsius) and species 4 (Jabberwockius Vulgaris), but in this case also switches genders. We'd represent this child's guesses with the signed permutation matrix $A = \left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{array} \right).$ Let $$A_i$$ be the guess matrix for child $$i$$. The keeper's first observation implies that these matrices satisfy $$A_i A_j = - A_j A_i$$ for all $$i\neq j$$. If the child is a boy, the keeper's second observation implies that $$A_i A_i = I$$, where $$I$$ is the $$4\times 4$$ identity matrix; if the child is a girl, the keeper's second observation implies that $$A_i A_i = -I$$. Also note that this observation implies children either get species right, or transpose two species. Thus, the species permutations can be written as the product of disjoint transpositions.

From the second observation, any child that gets the species right for an animal must be a boy. Furthermore, because the species guesses are the product of disjoint transpositions, every child must get an even number of species correct.

If a boy had gotten all the species and genders right, his matrix would commute with the matrix of any other child, but we know there is at least one boy and at least one girl. Thus no child could have gotten all of the animals correct. Furthermore, if any boy gets a species right, no other child could also get that species right, otherwise this would violate the keeper's first observation.

We could continue along these lines and work out the answer, but let's use Eddington's sledgehammer - the keeper's observations imply that these matrices will form a set of anticommuting Dirac matrices if we multiply the girl's matrices by the imaginary unit $$i$$. Such sets have size at most five; sets of size five have three boys and two girls, and the unique winner is a boy with four animals correct.

Here's one possible set. Note that I've multiplied two of the matrices given on the Wolfram site by the imaginary unit $$i$$ to convert them to signed permutation matrices. As I've mentioned, this doesn't affect the anticommutativity, but it does change the square of the matrix from $$I$$ to $$-I$$; thus, Dirac matrices with imaginary entries correspond to girls and Dirac matrices with real entries correspond to boys. \begin{eqnarray}
A_1 &= \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right)\\
A_2 &= \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0
\end{array} \right)\\
A_3 &= \left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0
\end{array} \right)\\
A_4 &= \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1
\end{array} \right)\\
A_5 &= \left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0
\end{array} \right)\\
\end{eqnarray} The boys are $$A_1, A_3, A_4$$ and the girls are $$A_2, A_5$$. The unique winner was $$A_4$$ with four animals correct (two correct species with correct gender, so four animals).

### 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…

### 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…

### Gambling to Optimize Expected Median Bankroll

Gambling to optimize your expected bankroll mean is extremely risky, as you wager your entire bankroll for any favorable gamble, making ruin almost inevitable. But what if, instead, we gambled not to maximize the expected bankroll mean, but the expected bankroll median?

Let the probability of winning a favorable bet be $$p$$, and the net odds be $$b$$. That is, if we wager $$1$$ unit and win, we get back $$b$$ units (in addition to our wager). Assume our betting strategy is to wager some fraction $$f$$ of our bankroll, hence $$0 \leq f \leq 1$$. By our assumption, our betting strategy is invariant with respect to the actual size of our bankroll, and so if we were to repeat this gamble $$n$$ times with the same $$p$$ and $$b$$, the strategy wouldn't change. It follows we may assume an initial bankroll of size $$1$$.

Let $$q = 1-p$$. Now, after $$n$$  such gambles our bankroll would have a binomial distribution with probability mass function \[ \Pr(k,n,p) = \binom{n}{k} p^k q^{n-k…