### An Enormous Number of Kilograms

For years the kilogram has been defined with respect to a platinum and iridium cylinder, but this is now no longer the case. Here's a puzzle about kilograms that's easy to state and understand, but the answer is very, very surprising.

I've always had a fascination with really large numbers. First 100 when I was really little, and as I got older and more sophisticated numbers like a googol and the smallest number that satisfies the conditions of the Archimedes cattle problem.

When I was an undergraduate I interviewed for a summer internship with an insurance company as an actuarial student. They gave me the following puzzle - what's the smallest number that when you move the last digit to the front it multiplies by 2? I calculated for a little while, then said "This can't be right, my answer has 18 digits!". It turns out that the smallest solution does, indeed, have 18 digits.

We can see this by letting our $$(n+1)$$-digit number $$x = 10 m + a$$, where $$m$$ is an $$n$$-digit number and $$0\leq a < 10$$. Moving $$a$$ to the front we get $$y = 10^n a + m$$, and our requirement is $$y = 2x$$. This gives: \begin{align} 20 m + 2 &= 10^n a + m \\ 19 m &= a(10^n-2) \\ m &= \frac{2a(5\cdot 10^{n-1} - 1)}{19} \end{align} The smallest $$m$$, if one exists, requires $$a,n$$ such that 19 divides $$5\cdot 10^{n-1}-1$$ (as 19 can't divide $$2 a$$) and the result has $$n$$-digits. It's easy to check that the smallest value of $$n$$ that satisfies the first condition is $$n=17$$. To get the smallest solution we try $$a=1$$, but this yields a value with only 16 digits. Setting $$a=2$$, however, yields the 17-digit $$m = 10526315789473684$$. The smallest solution to our puzzle is therefore the 18-digit number $$105263157894736842$$; that's surprisingly large.

Numbers with this type of property are known as parasitic numbers.

Later, I wondered if there were numbers with the slightly different, but equally interesting property, that moving the last digit to the front converted ("autoconverts") it from a value under one unit of measurement to an equivalent value under a different unit of measurement.

The first one I tried was moving the last digit to the front converts from Celsius to Fahrenheit. This is a fun puzzle that eventually made its way into the New York Times. The smallest such value is 275 C, which exactly equals 527 F. What's the next smallest temperature?

How about moving the first digit to the end? We'll need to use the little-known fact that, legally, a pound is exactly equal to 0.45359237 kilograms. Given this, does there exist a number such that moving the first digit to the end converts from pounds to kilograms? The answer is yes, but the smallest solution has 108,437,840 digits! The solution is similar to the above, but as it's computationally more involved I've written Sage code to solve it, which you can find in my GitHub puzzles repository.

The smallest number that autoconverts from gallons to liters, incidentally, is even bigger at 382,614,539 digits!

### 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…