Skip to main content

Probability and Cumulative Dice Sums

Post a Comment

Closed Under Means

Here's a nice little problem from the 13th All Soviet Union Mathematics Olympiad.
Given a set of real numbers \(S\) containing 0 and 1 that's closed under finite means, show that it contains all rational numbers in the interval \(\left[0,1\right]\).
This isn't a difficult problem, but there's a particularly nice solution.

First observe that if \(x\in S\) then both \(\frac{x}{4}\) and \(\frac{3x}{4}\) are in \(S\); average \(\{0,x\}\) to get \(\frac{x}{2}\), average \(\{0, \frac{x}{2}\}\) to get \(\frac{x}{4}\), average \(\{\frac{x}{2}, x\}\) to get \(\frac{3x}{4}\).

We could show any rational number \(\frac{m}{n}\) with \(1\leq m < n\) is in \(S\) if we had \(n\) distinct elements from \(S\) that summed to \(m\). Let's exhibit one.

Start with an array with \(m\) 1s on the left, \(n-m\) 0s on the right. Repeatedly replace adjacent \(x,y\) values with \(\frac{3(x+y)}{4}, \frac{(x+y)}{4}\), where either \(x=1,y\neq1\) or \(x\neq 0, y=0\), until there is one 0 and one 1 left. We can do this in exactly \(n-2\) steps, each resulting number is guaranteed to be in \(S\) by the above note, and each number is guaranteed to be distinct by construction!

Examples:

\(\frac{1}{3}: \left[1,0,0\right] \to \left[\frac{3}{4},\frac{1}{4},0\right] \)

\(\frac{2}{5}: \left[1,1,0,0,0\right] \to \left[1,\frac{3}{4},\frac{1}{4},0,0\right] \to \left[1,\frac{3}{4},\frac{3}{16},\frac{1}{16},0\right] \)

Comments

Post a Comment

Popular posts from this blog

Mining the First 3.5 Million California Unclaimed Property Records

As I mentioned in my previous article  the state of California has over $6 billion in assets listed in its unclaimed property database .  The search interface that California provides is really too simplistic for this type of search, as misspelled names and addresses are both common and no doubt responsible for some of these assets going unclaimed. There is an alternative, however - scrape the entire database and mine it at your leisure using any tools you want. Here's a basic little scraper written in Ruby . It's a slow process, but I've managed to pull about 10% of the full database in the past 24 hours ( 3.5 million out of about 36 million). What does the distribution of these unclaimed assets look like?  Among those with non-zero cash reported amounts: Total value - $511 million Median value - $15 Mean value - $157 90th percentile - $182 95th percentile - $398 98th percentile - $1,000 99th percentile - $1,937 99.9th percentile - $14,203 99.99th perc...
11 comments
Read more

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 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); offense (strength of offense), defense (strength of defense) and game_id are all random effects. The reason for modeling team offenses and defenses as random vs fixed effec...
3 comments
Read more

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 an...
20 comments
Read more