### 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:

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

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

### 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…

### What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be $w=\frac{P^e}{P^e+1},$ where $$w$$ is the win fraction expectation, $$P$$ is runs/allowed (or similar) and $$e$$ is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with $$g$$ games is $W = g\cdot w = g\cdot \frac{P^e}{P^e+1},$ so for one estimate we only need to compute the value of the partial derivative $$\frac{\partial W}{\partial P}$$ at $$P=1$$. Note that $W = g\left( 1-\frac{1}{P^e+1}\right),$ and so $\frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}$ and it follows $\frac{\partial W}{\partial P}(P=1) = … ### Solving a Math Puzzle using Physics The following math problem, which appeared on a Scottish maths paper, has been making the internet rounds. The first two parts require students to interpret the meaning of the components of the formula $$T(x) = 5 \sqrt{36+x^2} + 4(20-x)$$, and the final "challenge" component involves finding the minimum of $$T(x)$$ over $$0 \leq x \leq 20$$. Usually this would require a differentiation, but if you know Snell's law you can write down the solution almost immediately. People normally think of Snell's law in the context of light and optics, but it's really a statement about least time across media permitting different velocities. One way to phrase Snell's law is that least travel time is achieved when \[ \frac{\sin{\theta_1}}{\sin{\theta_2}} = \frac{v_1}{v_2},$ where $$\theta_1, \theta_2$$ are the angles to the normal and $$v_1, v_2$$ are the travel velocities in the two media.

In our puzzle the crocodile has an implied travel velocity of 1/5 in the water …