Synchronized Archery? That's Irrational!

Two logicians are trying to earn a fabulous prize as a team. There are three targets, and, to win the prize, each logician must fire a single arrow and hit the same target as the other. Two of the targets are closer but are otherwise indistinguishable; the logicians know they each have a 98 percent chance of hitting either of these targets. The third target is farther away; the logicians know they each have a 70 percent chance of hitting that target.

The logicians can’t cooperate or consult in advance, and they have no knowledge of which target their counterpart is aiming for or whether they are successful. What is the probability they will win the prize?

Here let's fill in some of the gaps in the prompt. Let's assume that the way that the closer targets are indistinguishable is some Rube Goldberg machine such that if an archer shoots an arrow into a tube then through some elaborate Rube Goldberg-esque machine, perhaps with a Plinko like randomizer, there is a fifty percent probability that the arrow will end up hitting target 1, while there is a fifty percent probability that the arrow will end up hitting target 2. So here the interpretation is that the probability of each logical archer getting an arrow into the tube is $98\%.$ So the probability that they both hit the same target it they both aim for the closer tube is is $$\mathbb{P} \{ \text{both arrows in tube} \} \mathbb{P} \{ \text{both arrows in same target} \} = 98 \% \cdot 98 \% \cdot \frac{1}{2} = 48.02\%.$$

On the other hand, the probability that they both hit the same target if they both aim for the further target is $$\mathbb{P} \{ \text{both hit the further target} \} = 70 \% \cdot 70 \% = 49 \%.$$ Since both of the archers are logicians, they both decide to aim for the target with the largest probability of winning, meaning they will both aim at the further target and have a probability of $49\%$ to win the prize.

As before, there are still three targets, but their respective probabilities of being struck have changed. That said, two of the targets remain indistinguishable from each other and have the same probability of being struck. Moreover, all three probabilities are rational.

After doing some mental arithmetic, the logicians realize that it doesn’t matter which target they aim for—their probability of winning the prize is the same no matter what. What is their probability of winning the prize?

Let's assume that the probability of hitting the closer tube is now $p \in \mathbb{Q} \cap [0,1],$ while the probability of hitting the further target is now $q \in \mathbb{Q} \cap [0,1],$ which means by analogy to the above calculations that the probability of hitting the same target if aiming for the closer and further targets are $\frac{1}{2}p^2$ and $q^2,$ respectively. Being excellent logicians the general probability of winning is thus $$\pi (p, q) = \max \left\{ \frac{1}{2} p^2, q^2 \right\}$$ for any $(p,q) \in (\mathbb{Q} \cap [0,1])^2.$ If as the logical archers surmised there is no different in which target they aim for, then we must have that $\frac{1}{2} p^2 = q^2.$ However, since $\sqrt{2}$ is well known to be irrational, we see that the only possible solution to $\frac{1}{2} p^2 - q^2 = 0$ in $( \mathbb{Q} \cap [0,1] )^2$ is $p = q = 0.$ This means, that rather unsportsmanlike the probability of winning the prize is $\pi(0,0)=0$ if (a) the probabilities $p$ and $q$ are rational and (b) it doesn't matter which target to aim at.

Of course, that's not very fun. So if we wanted to drop condition (a) that both probabilities are rational, then we can replace $\frac{1}{2}p^2 - q^2 = 0$ with the linear condition $p - \sqrt{2} q = 0,$ or $p = \sqrt{2} q.$ In this case, we get $\tilde{\pi}(q) = \pi( \sqrt{2} q, q ) = q^2,$ for all $q \in [0, \sqrt{2} / 2 ].$ This gives a maximum probability of winning the prize when it still doesn't matter which target to aim at irrational probabilities are available of $\tilde{\pi}^* = \frac{1}{2}.$

Roving all over the planet

A rover is dropped down on a spherical planet with a radius of 1000 miles. The rover has been programmed with a very specific set of motions:

• First, it moves straight forward a fixed distance $s$, and stops.
• Without moving forward, it turns left 60 degrees.
• Next, the rover moves straight forward in this new direction another distance $s,$ and stops.
• Without moving forward, it again turns left 60 degrees.
• Finally, the rover moves straight forward in this new direction another distance $s,$ and stops.

To be picked up, the rover must complete its journey in the same place it was dropped down. What is the minimum value of $s,$ with $s > 0,$ for which this works?

Firstly, let's assume without loss of generality that the dropoff point is the north pole of this planet, that is, the initial dropoff and final pickup are at the point $O = (0,0,r).$ Let's assume again, without loss of generality that the initial path of the rover is along the the intersection of the sphere and the $xz$-plane, so that the first stop is at some point $P = (r \sin \frac{s}{r}, 0, r \cos \frac{s}{r}),$ that is, we are traveling along the 0th longitude through an angle of $\frac{s}{r}$ radians, such that the total length of the arc is $d_{OP} = r \frac{s}{r} = s.$

Since we will want to have the path from the last stop to the final pickup point be along the longitude that makes an angle \frac{2\pi}{3} with the x-axis, we have that we will want the second stop to be at $$Q = (r \sin \frac{s}{r} \cos \frac{2\pi}{3}, r \sin \frac{s}{r} \sin \frac{2\pi}{3}, r \cos \frac{s}{r} ) = ( -\frac{r}{2} \sin \frac{s}{r}, \frac{r\sqrt{3}}{2} \sin \frac{s}{r}, \cos \frac{s}{r} ).$$ Here again, by design, we have $d_{QO} = r \frac{s}{r} = s,$ and furthermore we have the desired internal angle of $120^\circ = \frac{2\pi}{3}$ angle between the paths $OP$ and $QO.$

One additional step to ensure that this path fully satisfies the programming steps above is to calculate great circle distance $d_{PQ}$ and ensure that $d_{PQ} = s.$ In fact, this can in fact be the final step, since from symmetry, if all of the side lengths are equal to $s$ and one of the angles has measure $120^\circ$, then all of the exterior angles are $60^\circ$ as desired. To calculate the great circle distance between P and Q, we need to find the angle between the 3-dimensional vectors P and Q, which we can do by taking the dot products, that is, \begin{align*}\cos \frac{d_{PQ}}{r} &= \frac{1}{r^2} \left\langle (r \sin \frac{s}{r}, 0, r \cos \frac{s}{r} ), (-\frac{r}{2} \sin \frac{s}{r}, \frac{r\sqrt{3}}{2} \sin \frac{s}{r}, r\cos \frac{s}{r} \right\rangle \\ &= -\frac{1}{2} \sin^2 \frac{s}{r} + \cos^2 \frac{s}{r} \\ &= \frac{3}{2} \cos^2 \frac{s}{r} - \frac{1}{2}.\end{align*} If we want to insist on $d_{PQ} = s$ as well, then we are left with a quadratic polynomial in $\cos \frac{s}{r},$ that is $$\frac{3}{2} \cos^2 \frac{s}{r} - \cos \frac{s}{r} - \frac{1}{2} = \frac{ \left(3 \cos \frac{s}{r} + 1\right) \left( \cos \frac{s}{r} - 1 \right) }{2} = 0.$$ Therefore, in order for our roving robot friend to find his way back to the north pole of this planet, we must have either $\cos \frac{s}{r} = -\frac{1}{3}$ or $\cos \frac{s}{r} = 1.$

Let's handle the latter case first, which would give $s = 2\pi k r$ for any $k = 1, 2, \dots,$ which intuitively makes sense, since obviously our tireless rover would not have any troubles doing a complete lap of the planet, coming back to the north pole each time and then rotating any arbitrary number of degrees and still ending up at the north pole. However, thankfully, we get a more interesting answer for $\cos \frac{s}{r} = - \frac{1}{3}.$ Let $a = \cos^{-1} \left(-\frac{1}{3}\right) \approx 1.91063323625\dots,$ then we see that $s = (2\pi k \pm a)r,$ for any $k = 0, 1, 2, \dots$ will satisfy $\cos \frac{s}{r} = - \frac{1}{3}.$ So the shortest distance that will allow the rover to complete its journey is $$s_1 = ar = 1000 \cos^{-1} \left(-\frac{1}{3}\right) \approx 1910.63323625\dots \,\text{miles}.$$

There are other values of $s$ for which the rover will end its journey where it was dropped down. How many such positive values of $s$ (including the answer you just found in the Fiddler) are less than $100,000$ miles?

To solve this Extra Credit problem, we can set $r = 1$ for convenience and define set of all possible positive distances that would work for our rover friend to complete its journey on a unit sphere as $$\mathfrak{S} = \left( -a + 2\pi (\mathbb{N} \setminus \{0\}) \right) \cup \left( 2\pi (\mathbb{N} \setminus \{0\}) \right) \cup \left(a + 2 \pi \mathbb{N} \right).$$ Then the set of all possible distances on our $r=1000$ mile radius planet that would work for our rover to complete its journey in less than $100,000$ miles is equivalent to the set $$\mathfrak{S}_{100} = \{ s \in \mathfrak{S} \mid s \lt 100\}$$ on our unit sphere planet. In this case, since we have $32 \pi \approx 100.53 \gt 100,$ we see that $$\mathfrak{S}_{100} = \{ a, 2\pi - a, 2\pi, 2\pi + a, \dots, 30 \pi, 30\pi + a, 32 \pi - a \},$$ so there are $|\mathfrak{S}_{100}| = 3 \cdot 15 + 2 = 47$ possible positive values of $s$ that will allow the rover to complete its journey.

Not much of an average streak ...

The Fiddler Basketball Association’s All-Star Game consists of two teams: “East” and “West.” Every year these two teams play a game, each with a 50 percent chance of winning that’s independent of the outcomes of previous years.

Many, many years into the future, you look at the most recent results of the All-Star Game. On average, what is the longest current winning streak that one of the teams is on?

Let's assume that we are looking $N$ years into the future. If the current winning streak is $S = 1$, then the last two games would have to either be EW or WE, where the first $N-2$ games could have had any possible outcome. This means there are $2 \cdot 2^{N-2} = 2^{N-1}$ possible configurations with current streak $S=1$ out of a total of $2^N$ possible configurations, so $$\mathbb{P} \{ S = 1 \} = \frac{2^{N-1}}{2^N} = \frac{1}{2}.$$ Similarly, for some generic $k \in \{ 2, \dots, N \},$ there are only two possible configurations of the last $k+1$ games that will yield $S = k$, that is, $W\overbrace{EE\cdots E}^k$ or $E\overbrace{WW \cdots W}^k.$ Again, the first $N-k-1$ games could have had any possible outcome. So we see that $$\mathbb{P} \{ S_N = k \} = \frac{2 \cdot 2^{N-k-1}}{2^N} = 2^{-k},$$

So we see that the for $N$ years into the future we get $$\mathbb{E} [ S_N ] = \sum_{k=1}^N k \mathbb{P} \{ S_N = k \} = \sum_{k=1}^N k 2^{-k}.$$ We will resort to the use of the following friendly partial sum formulae for $$f(t) = \sum_{k=0}^n t^k = \frac{1-t^{n+1}}{1-t}$$ and $$g(t) = f^\prime (t) = \sum_{k=1}^n kt^{k-1} = \frac{1 - (n+1)t^k +nt^{n+1}}{(1-x)^2}.$$ Here we see that the average current streak after $N$ years is $$\mathbb{E} [S_N] = \frac{1}{2} g\left( \frac{1}{2} \right) = \frac{1}{2} \frac{1 - (N+1) 2^{-N} + N 2^{-N-1} }{\left(1 - \frac{1}{2}\right)^2} = 2 \left( 1 - \frac{N+2}{2^{N+1}} \right),$$ which leads naturally when $N \to \infty$ to the long term average of $\mathbb{E}[S_\infty] = 2.$

Rapidly scattering food

Frankie has stored all of her food on lily pad A. However, her food has a tendency to “fly” away. Every second, the food that’s on every lily pad splits up into six equal portions that instantaneously relocate to the six neighboring pads.

At zero seconds, all the food is on lily pad A. After one second, there’s no food on pad A, and $1/6$ of the food is on each of the surrounding six pads. After two seconds, $1/6$ of the food is again on pad A, while the rest of the food is elsewhere.

After how many seconds $N$ (with $N > 2$) will pad A have less than $1$ percent of its original amount?

Here we can code up the six possible directions and let the probability distribution evolve through time. Let's say that at time $t$ we have $\pi(t, a, b) \in (0,1)$ proportion of all of Frankie's food at the lily pad with center at $a\zeta + b\xi.$ Then we see that this means that it recursively should have received 1/6 of the food on lily pads $(a\pm 1, b)$, $(a, b\pm 1),$ and $(a \pm 1, b\mp 1)$ at time $t-1,$ that is \begin{align*}\pi( t, a, b ) &= \frac{1}{6} \Biggl( \pi (t-1, a+1, b) + \pi( t-1, a-1, b) \\ & \quad\quad\quad +\pi( t-1, a, b+1) + \pi (t -1, a, b-1) \\ &\quad\quad\quad\quad + \pi( t -1, a+1, b-1) + \pi (t-1, a-1, b+1) \Biggr),\end{align*} for all $a, b \in \mathbb{Z}, t \in \mathbb{N},$ with initial condition $\pi(0,0,0) = 1$ and $\pi(0,a,b) = 0$ for $a\ne b \ne 0.$

Coding this up in Python and then solving for $T = \min \{ t \gt 2 \mid \pi(t, 0,0) \lt 0.01 \}$ yields that Frankie's food on lily pad A will be less than $1\%$ of the total after $T = 17$ seconds.

Slowly spiraling eastward

Frankie the frog is hopping on a large, packed grid of lily pads, shown below. The pads are circular and each is a distance 1 from its nearest neighbors. (More concretely: Each pad has a diameter of 1 and they are arranged in a hexagonal lattice.) Frankie starts at (0, 0), the center of the pad labeled A. Then she hops due east to pad B at (1, 0), and from there she hops to pad C at (1.5, √(3)/2).

She wants to continue hopping in a counterclockwise, spiral-like pattern. Each of her jumps is to the center of a neighboring pad, a net distance of 1. But there are two rules her spiral must follow:

• Each next pad must be in a more counterclockwise direction (relative to spiral’s origin at pad A) than the previous pad.
• Each pad must be farther from A than the previous pad.

After a number of hops spiraling around, Frankie realizes she is, once again, due east from A. What is the closest to A she could possibly be? That is, what is the minimum possible distance between the center of the pad she’s currently on from the center of pad A?

There are only six possible directions for Frankie to jump in, namely $\pm 1$ and $( \pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2} ).$ If we let $\zeta = e^{i \pi/3}$ then we see that the directions are actually all $\zeta,$ $\zeta^2,$ $\zeta^3 = -1,$ $\zeta^4,$ $\zeta^5$ and $\zeta^6 = 1.$ Let's let define $\xi = \zeta^2$ and not that $1 = \zeta - \xi.$ In this way we see that the directions that Frankie can jump in are either $\pm \zeta$, $\pm \xi$ or $\pm 1 = \pm (\zeta - \xi),$ so all in terms of $\zeta, \xi$ coordinates. While we are still establishing preliminaries, let's also see that if we have some point $z=a\zeta + b\xi$ for any $a, b \in \mathbb{Z},$ that \begin{align*}|z|^2 &= (a\zeta + b\xi) \left(\frac{a}{\zeta} + \frac{b}{\xi}\right) \\ &= a^2 + b^2 + ab\left(\zeta + \frac{1}{\zeta}\right) \\ &= a^2 + b^2 + 2ab\cos \frac{\pi}{3} \\&= a^2 + b^2 + ab.\end{align*}

Since she wants to do make the tightest possible spiral, she will want to keep going in the same direction until the next possible clockwise motion, which will be dictated by her more counterclockwise and more distant from A requirements.

Let's see this in action. We have $C = 1 + \zeta = 2\zeta - \xi,$ with $$|C| = \sqrt{2^2 + 1^2 + 2\cdot (-1)} = \sqrt{4 + 1 - 2} = \sqrt{3}.$$ Since $|C + \xi| = |2\zeta| = 2 \gt |C|,$ we have $D = 2\zeta.$.

Now we will keep adding $\xi$ steps until $\xi - \zeta = -1$ gives us an acceptable step. So let's build up the cases here. We will have steps of $2\zeta + n\xi$ for each $n = 0, 1, \dots, N_1$ where we have \begin{align*}N_1 &= \min \{ k \mid |\zeta + (k+1)\xi| \gt |2\zeta + k\xi|\} \\ &= \min \{ k \mid 1 + (k+1)^2 + k+1 \gt 4 + k^2 + 2k \}\\ &= \min \{ k \mid k \gt 1 \} = 2.\end{align*} So after we are done with $3\xi$ steps we find ourselves at $G = 2\zeta + 2\xi = (0,2\sqrt{3})$ and ready to make steps in the $-1$ direction.

Again we will keep adding $-1=-\zeta + \xi$ steps until $-\zeta$ gives us an acceptable step. So we have steps $(2-n)\zeta + (2 + n)\xi$ for $n = 0, 1, \dots, N_2$ where \begin{align*}N_2 &= \min \{ k \mid | (1-k)\zeta + (2+k) \xi | \gt | (2-k) \zeta + (2+k) \xi | \} \\ &= \min \{ k \mid (1-k)^2 + (2+k)^2 + (1-k)(2+k)\gt (2-k)^2 + (2+k)^2 + (2-k)(2+k) \} \\ &= \min \{ k \mid k^2 +k + 7 \gt k^2 + 12 \} \\ &= \min \{ k \mid k \gt 5 \} = 6.\end{align*} So we will find ourselves at $M = -4\zeta + 8\xi = (-6,2\sqrt{3}).$

Noting that the number of $-1$ steps was exactly twice the number of $\xi$ steps, we continue on we see that we take 12 $(-\zeta)$ steps followed by 24 $(-\xi)$ steps, then 48 $+1 = \zeta - \xi$ steps to arrive at $32\zeta - 64 \xi,$ ready to take new $\zeta$ steps. Eventually we will arrive at the positive $x$-axis, which is due east of the center of A, whenever we have some number of $n (\zeta - \xi) = (n,0).$ Since we are starting with $-64\xi,$ Frankie will arrive at the point $64\zeta - 64\xi = (64,0),$ which is 64 units from the center of A.