### A Dicey Problem: Solution

Problem: Waldo rolls one standard (fair) 6-sided die repeatedly, keeping a running total of what he rolls. To three decimal places, what's the probability that he eventually reaches a total of exactly 1,000,000?

Solution: On average Waldo will be adding $$\mu = (1+2+3+4+5+6)/6 = 3.5$$ per roll, so we would "intuitively" or "naively" expect that the probability of hitting a particular total that's large enough would be close to $$1/3.5$$. This turns out to be correct.

More generally, if we have any finite set of positive integers with no common factor, then the probability of hitting a particular total for large enough numbers is $$1/\mu$$, where $$\mu$$ is the expected value for a single roll. Note that the probabilities associated with the positive integers in our set don't need to be the same, either, just greater than zero and summing to $$1$$.

Proof: I'll show how you can prove this rigorously for the case where the outcomes are $$1$$ and $$2$$, each with probability $$1/2$$. Let $$p(n)$$ be the probability of totaling the value $$n$$. The only possible ways to total $$n$$ are to reach $$n-2$$ and immediately roll a $$2$$, or reach $$n-1$$ and immediately roll a $$1$$. This implies that $$p(n) = 1/2\cdot p(n-1) + 1/2\cdot p(n-2)$$, which is a linear, second order, homogeneous difference equation (recurrence relation). I'm going to apply the theory here, but for more details see Wikipedia: Recurrence relation.

In this case the associated characteristic polynomial is $$r^2 = r/2 + 1/2$$ with roots $$r_1 = 1, r_2 = -1/2$$. So we have $$p(n) = c_1 + c_2\cdot (-1/2)^n$$ for constants $$c_1, c_2$$. The boundary conditions are $$p(0) = 1$$ and $$p(1) = 1/2$$; solving we get $$c_1 = 2/3$$ and $$c_2 = 1/3$$. This implies that $$p(n)$$ converges to $$2/3$$ very rapidly; exponentially, in fact.

1. Really interesting. I wonder if you have a book where I can I understand better the difference equations?

2. Esteban, I'll write a blog entry on difference equations. They're nice to know and a very useful framework for analyzing recursive problems. Algorithms are a common setting, but far from the only one.

### 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…