### Asteroids Algebra

Problem:

Waldo is playing Asteroids. This game starts with 3 ships and you earn an extra ship for every 10000 points you score. At the end of his first game, Waldo noticed that he had the lowest score possible while averaging exactly 9000 points per ship. At the end of his second game, Waldo noticed that he had the highest score possible while averaging exactly 9000 points per ship. What were his two scores?

Solution:

Let the number of starting ships be $$S$$, points for an extra ship be $$X$$ and the player's average scoring rate be $$A$$. Furthermore, let $$R = A/X$$ and the number of bonus ships earned by the player during the game be $$B$$. The number of bonus ships earned by the player during his game is the floor of his final score over the bonus value $$X$$. Algebraically, we have $B = \lfloor (S+B)\cdot R \rfloor.$ Now let $(S+B)\cdot R = \lfloor (S+B)\cdot R \rfloor + f,$ where $$0 \leq f < 1$$. This gives us
\begin{aligned}
(S+B)\cdot R &= B + f ,\\
SR + B(R-1) &= f,\\
B &= (SR-f)/(1-R).
\end{aligned}
Together with $$0 \leq f < 1$$ we get that $\frac{SR-1}{1-R} < B \leq \frac{SR}{1-R}.$ Thus, the smallest possible number of bonus ships is $\left\lfloor \frac{SR-1}{1-R} + 1 \right\rfloor = \left\lfloor \frac{SR-R}{1-R}\right\rfloor$ and the greatest possible number of bonus ships is $\left\lfloor \frac{SR}{1-R}\right\rfloor.$ It follows that the lowest possible score while averaging $$A$$ is  $A\cdot \left\lfloor \frac{SR-R}{1-R} + S \right\rfloor = A\cdot \left\lfloor \frac{S-R}{1-R}\right\rfloor$ and the highest possible score while averaging $$A$$ is $A\cdot \left\lfloor \frac{SR}{1-R} + S\right\rfloor = A\cdot \left\lfloor \frac{S}{1-R}\right\rfloor.$
For Waldo, $$S=3$$, $$A=9000$$ and $$R=9000/10000 = 0.9$$. Waldo therefore scored
$9000\cdot \left\lfloor \frac{3-0.9}{1-0.9} \right\rfloor = 189000$ points in his first game, $9000\cdot \left\lfloor \frac{3}{1-0.9} \right\rfloor = 270000$ points in his second game.

### 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 …