### Learn One Weird Trick And Easily Solve The Product-Sum Problem

A tribute to Martin Gardner.

For which sets of positive integers does the sum equal the product? For example, when does $$x_1 + x_2 = x_1\cdot x_2$$? It's easy to see that this is only true when $$x_1 = x_2 = 2$$.

In the general case our equality is $$\sum_{i=1}^{n} x_i= \prod_{i=1}^{n} x_i$$. We can rearrange terms to give $x_1+x_2+\sum_{i=3}^{n} x_i= x_1\cdot x_2\cdot \prod_{i=3}^{n} x_i,$ and this in turn factors nicely to give us $\left( x_1\cdot \prod_{i=3}^{n} x_i - 1\right)\cdot \left( x_2\cdot \prod_{i=3}^{n} x_i - 1\right) = \left( \prod_{i=3}^{n} x_i \right)\cdot \left(\sum_{i=3}^{n} x_i \right) + 1.$ How does this help? Consider the case $$n=5$$, and without loss of generality assume $$x_i \ge x_{i+1}$$ for all $$i$$. If $$x_3=x_4=x_5=1$$ our factorized equation becomes $(x_1-1)\cdot(x_2-1)=4,$ with the obvious solutions $$x_1=5, x_2=2; x_1=3, x_2=3$$. The only remaining case to consider is $$x_3=2$$, as any other case forces  $$\sum_{i=1}^{n} x_i < \prod_{i=1}^{n} x_i$$. For this case our equality becomes $\left( 2 x_1 - 1\right)\cdot \left( 2 x_2 - 1\right) = 9.$ This gives us the remaining solution $$x_1 = x_2 = x_3 = 2$$ with the other $$x_i = 1$$.

This is quite efficient. For example, for the case $$n=100$$ the only equations we need to consider are \begin{aligned} (x_1-1)\cdot(x_2-1) &= 99\\
(2 x_1-1)\cdot(2 x_2-1) &= 199\\
(3 x_1-1)\cdot(3 x_2-1) &= 301\\
(4 x_1-1)\cdot(4 x_2-1) &= 405\\
(4 x_1-1)\cdot(4 x_2-1) &= 401\\
(6 x_1-1)\cdot(6 x_2-1) &= 607\\
(9 x_1-1)\cdot(9 x_2-1) &= 919\\
(8 x_1-1)\cdot(8 x_2-1) &= 809\\
(12 x_1-1)\cdot(12 x_2-1) &= 1225\\
(16 x_1-1)\cdot(16 x_2-1) &= 1633.
\end{aligned} Now 199, 401, 607, 919, 809 are all prime ruling them out immediately, and 301 and 1633 don't have factors of the right form. The other equations yield the five solutions $$(100,2), (34,4), (12,10), (7,4,4), (3,3,2,2,2)$$ with the other $$x_i = 1$$.

For $$n = 1000$$ you'd need to consider 52 cases, but most of these are eliminated immediately. I get the six solutions $$(1000,2), (334,4), (112,10), (38,28), (67,4,4), (16,4,4,4)$$.

Have fun!

### 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…