Uniquely Non-Genius Queen Bees

Let’s put the rank of “Genius” aside. Here are some other ranks you can attain in Spelling Bee:

• Amazing (if you get 50 percent of the maximum, rounded to the nearest whole number)
• Great (40 percent)
• Nice (25 percent)
• Solid (15 percent)
• Good (8 percent)
• Moving Up (5 percent)
• Good Start (2 percent)

Suppose a given round of Spelling Bee has some very large, randomly chosen point total. What is the probability that this total can be precisely determined from these cutoffs (i.e., from “Good Start” through “Amazing,” inclusive)?

The first thing that we see is that there are many, many, many levels that start with $G$, so let's forego the step of naming each function of $Q$ and instead have a mapping $F: \mathbb{N} \to \mathbb{N}^7$ with $$Q \mapsto F(Q) = \begin{pmatrix} [0.5Q], [0.4Q], [0.25Q], [0.15Q], [0.08Q], [0.05Q], [0.02Q] \end{pmatrix} \in \mathbb{N}^7.$$ Firstly, we see that we have the recursion formula $$F(100d+k) = (50d, 40d, 25d, 15d, 8d, 5d, 2d) + F(k),$$ for all $d, k \in \mathbb{N},$ so we need only really calculate the ratio for any sufficiently chosen $100$ values of $Q$ in order to determine how many of those are

Let's start with $Q=101$ and start enumerating:

q	F(q)	q	F(q)
101	(50,40,25,15,8,5,2)	151	(75,60,38,23,12,8,3)
102	(51,41,25,15,8,5,2)	152	(76,61,38,23,12,8,3)
103	(51,41,26,15,8,5,2)	153	(76,61,38,23,12,8,3)
104	(52,42,26,16,8,5,2)	154	(77,62,38,23,12,8,3)
105	(52,42,26,16,8,5,2)	155	(77,62,39,23,12,8,3)
106	(53,42,26,16,8,5,2)	156	(78,62,39,23,12,8,3)
107	(53,43,27,16,9,5,2)	157	(78,63,39,24,13,8,3)
108	(54,43,27,16,9,5,2)	158	(79,63,39,24,13,8,3)
109	(54,44,27,16,9,5,2)	159	(79,64,40,24,13,8,3)
110	(55,44,27,16,9,5,2)	160	(80,64,40,24,13,8,3)
111	(55,44,28,17,9,6,2)	161	(80,64,40,24,13,8,3)
112	(56,45,28,17,9,6,2)	162	(81,65,40,24,13,8,3)
113	(56,45,28,17,9,6,2)	163	(81,65,41,24,13,8,3)
114	(57,46,28,17,9,6,2)	164	(82,66,41,25,13,8,3)
115	(57,46,29,17,9,6,2)	165	(82,66,41,25,13,8,3)
116	(58,46,29,17,9,6,2)	166	(83,66,41,25,13,8,3)
117	(58,47,29,18,9,6,2)	167	(83,67,42,25,13,8,3)
118	(59,47,29,18,9,6,2)	168	(84,67,42,25,13,8,3)
119	(59,48,30,18,10,6,2)	169	(84,68,42,25,14,8,3)
120	(60,48,30,18,10,6,2)	170	(85,68,42,25,14,8,3)
121	(60,48,30,18,10,6,2)	171	(85,68,43,26,14,9,3)
122	(61,49,30,18,10,6,2)	172	(86,69,43,26,14,9,3)
123	(61,49,31,18,10,6,2)	173	(86,69,43,26,14,9,3)
124	(62,50,31,19,10,6,2)	174	(87,70,43,26,14,9,3)
125	(62,50,31,19,10,6,2)	175	(87,70,44,26,14,9,3)
126	(63,50,31,19,10,6,3)	176	(88,70,44,26,14,9,4)
127	(63,51,32,19,10,6,3)	177	(88,71,44,27,14,9,4)
128	(64,51,32,19,10,6,3)	178	(89,71,44,27,14,9,4)
129	(64,52,32,19,10,6,3)	179	(89,72,45,27,14,9,4)
130	(65,52,32,19,10,6,3)	180	(90,72,45,27,14,9,4)
131	(65,52,33,20,10,7,3)	181	(90,72,45,27,14,9,4)
132	(66,53,33,20,11,7,3)	182	(91,73,45,27,15,9,4)
133	(66,53,33,20,11,7,3)	183	(91,73,46,27,15,9,4)
134	(67,54,33,20,11,7,3)	184	(92,74,46,28,15,9,4)
135	(67,54,34,20,11,7,3)	185	(92,74,46,28,15,9,4)
136	(68,54,34,20,11,7,3)	186	(93,74,46,28,15,9,4)
137	(68,55,34,21,11,7,3)	187	(93,75,47,28,15,9,4)
138	(69,55,34,21,11,7,3)	188	(94,75,47,28,15,9,4)
139	(69,56,35,21,11,7,3)	189	(94,76,47,28,15,9,4)
140	(70,56,35,21,11,7,3)	190	(95,76,47,28,15,9,4)
141	(70,56,35,21,11,7,3)	191	(95,76,48,29,15,10,4)
142	(71,57,35,21,11,7,3)	192	(96,77,48,29,15,10,4)
143	(71,57,36,21,11,7,3)	193	(96,77,48,29,15,10,4)
144	(72,58,36,22,12,7,3)	194	(97,78,48,29,16,10,4)
145	(72,58,36,22,12,7,3)	195	(97,78,49,29,16,10,4)
146	(73,58,36,22,12,7,3)	196	(98,78,49,29,16,10,4)
147	(73,59,37,22,12,7,3)	197	(98,79,49,30,16,10,4)
148	(74,59,37,22,12,7,3)	198	(99,79,49,30,16,10,4)
149	(74,60,37,22,12,7,3)	199	(99,80,50,30,16,10,4)
150	(75,60,37,22,12,7,3)	200	(100,80,50,30,16,10,4)

From our recurrence formula we see that $F(100) = F(200) - (50,40,25,15,8,5,2) = (50,40,25,15,8,5,2) = F(101)$ and similarly that $F(201) = F(200).$ Therefore, we see that if $U \subset \mathbb{N}$ such that the values of $Q$ can be uniquely recovered from the mapping $F,$ then the only integers in \begin{align*}\{ 101, \dots, 200 \} \setminus U &= \{ 101, 104, 105, 112, 113, 120, \\ & \quad\quad 121, 124, 125, 132, 133, 140, \\ & \quad\quad 141, 144, 145, 152, 153, 160,\\ &\quad\quad 161, 164, 165, 172, 173, 180, \\ &\quad\quad 181, 184, 185, 192, 193, 200 \},\end{align*} which means that we are left with $$\left|U \cap \{101, \dots, 200\}\right| = 100 - 30 = 70.$$ Therefore, availing ourselves of the squeeze theorem and some of the other argumentation that we went through for the classic problem, we have that the probability that this total can be precisely determined from these cutoffs is $$\mathbb{P}(U) = 70\%.$$

Uniquely Genius Queen Bees

In Spelling Bee, a word game from The New York Times, you must create words using seven letters arranged in a honeycomb grid. Letters can be used more than once per word, but each word must be at least four letters long and include the central letter. You may recall a previous puzzle (RIP, FiveThirtyEight) based on Spelling Bee.

Each day, your goal is to score as many points as possible. Four-letter words are worth 1 point, longer words are worth the number of letters they contain, and “pangrams” (words containing every letter) provide 7 bonus points. But these specific details don’t matter for this week’s puzzle.

If you find all the words in a given day and therefore accrue the maximum number of points, you earn the “Queen Bee” ranking. Meanwhile, accruing smaller point totals earns you other rankings. In particular, the point cutoff for “Genius” is 70 percent of the maximum number of points, rounded to the nearest whole number. While the cutoff for “Genius” is clearly displayed in the puzzle, the “Queen Bee” point total is not readily shown.

Of course, this maximum total can be approximated by dividing the “Genius” cutoff by 0.7. Even so, the total may be ambiguous, since multiple “Queen Bee” values can result in the same “Genius” cutoff.

Suppose a given round of Spelling Bee has some very large, randomly chosen point total. What is the probability that this total can be precisely determined (i.e., without any ambiguity) from its point cutoff for “Genius”?

Let's assume that $Q$ is the Queen Bee point total and that $G$ is the Genius cutoff, and we have the functional definition $G: \mathbb{N} \to \mathbb{N}$ with $Q \mapsto G(Q) = [ 0.7 Q ],$ where $[\cdot]$ denotes the closest integer function, that is $[t] = \min \arg\!\min \{ |n - t | \mid n \in \mathbb{N} \},$ where we are rounding halves downward towards $0$. While the concept of uniform distribution on the natural numbers is untenable, let us assume that here we mean by some very large, randomly chosen point total of $Q$ that we have the following probabilistic definition. For any subset $A \subseteq \mathbb{N}$, let us define its probability as $$\mathbb{P} (A) = \lim_{n \to \infty} \frac{ \left| A \cap \{1, 2, \dots, n \} \right| }{n}.$$ This is loosely a limit of the sequence of uniform distributions on the sets $\{1, 2, \dots, n \}$ as $n \to \infty,$ but again I don't really want to get into the formal definitions of how that limit should be defined, let's just kinda run with it.

Anywhoozle, in this case, let's define $U = \{ n \in \mathbb{N} \mid |G^{-1}(n)| = 1 \}$ to be the desired set of all values of $Q$ that can be precisely and unambiguously determiend by the corresponding value of the $G$ function. It would probably be good to have some idea of what $U$ looks like before plowing ahead and getting to the calculation of the desired answer, that is, $\mathbb{P}(U).$ Let's start super simple and just give the first few values of the $G$ function:

q	0.7q	G(q)
1	0.7	1
2	1.4	1
3	2.1	2
4	2.8	3
5	3.5	3
6	4.2	4
7	4.9	5
8	5.6	6
9	6.3	6
10	7.0	7
11	7.7	8

Using these first 10 integers, we see that $3, 6, 7, 10 \in U.$ We also can see that for any $d \in \mathbb{N}$ that $$G(10d+k) = \left[0.7 \left(10d+k\right)\right] = 7d+G(k).$$ So we can surmise that in face for any $Q \in \mathbb{N}$ if $Q\! \mod\! 10 \in \{0, 3, 6, 7 \}$ then $Q \in U,$ then $$U = \left(10 \mathbb{N} \cup (3 + 10 \mathbb{N}) \cup (6 + 10\mathbb{N}) \cup (7 + 10\mathbb{N}) \right).$$ We see that $$\left| U \cap \{ 1,2, \dots, n\} \right| = \begin{cases} 4 \lfloor \frac{n}{10} \rfloor, &\text{if $n \!\mod\! 10 \in \{0,1,2\};$}\\ 4\lfloor \frac{n}{10} \rfloor+1, &\text{if $n \!\mod\!10 \in \{3,4,5\};$}\\ 4\lfloor \frac{n}{10} \rfloor+2, &\text{if $n \!\mod\!10 \equiv 6;$}\\ 4\lfloor \frac{n}{10} \rfloor+3, &\text{if $n \!\mod\!10 \in \{7,8,9\}.$}\end{cases}$$ Using the squeeze theorem and the fact that $$\lim_{n\to \infty} \frac{4 \lfloor \frac{n}{10} \rfloor}{n} \leq \mathbb{P} (U) \leq \lim_{n\to \infty} \frac{4 \lfloor \frac{n}{10} \rfloor + 3}{n},$$ we have the probability of getting a value of $Q$ that is precisely, unambiguously determined from its value of $G$ is $$\mathbb{P} (U) = \lim_{n \to \infty} \frac{1}{n} \left| U \cap \{1, 2, \dots, n \} \right| = \frac{2}{5} = 40\%.$$

Frederica's Trochoidal Film

For her next show, Frederica wants to mix things up. Instead of placing the light on the circumference of the wheel, she will pick a random point inside the circle. (Before you ask, let me clarify what “random” means here: Any two regions with the same area are equally likely to contain the point.)

As before, she will roll the wheel for one revolution along the ground and capture the motion with a single long exposure on her camera. On average, what can she expect the length of the path to be?

As before we will compute for any given starting point $$(x_0, y_0) \in \{ (x,y) \mid \sqrt{x^2 + (y-1)^2} \leq 1 \}$$ and calculate the path that it traces out as the wheel makes a full rotation. If the point $(x_0,y_0)$ is on the circumference of the circle, the path is known as a cycloid; however, if $(x_0,y_0)$ is in the interior of the wheel it is known as a curtate trochoid. Anywhoozle, let's again define our parametric equations ...

Let's assume that the point of light is at $x(0) = x_0, y(0) = y_0$, where without loss of generality let's assume that $x_0 \gt 0,$ and of course $r = \sqrt{x_0^2 + (y_0-1)^2} \leq 1.$ Then we see that we can write $x(0) = r \sin \theta, y(0) = 1 + r \cos \theta,$ for $\theta = \tan^{-1} \frac{x_0}{y_0 - 1} \in [0, \pi].$ Since after a rotation of $t \in [0,2\pi)$ we will have the wheel centered at $(t,1),$ while the angle that the point of light now makes as measured with respect to the positive $y$-axis, will now be $\theta + t.$ So at time $t$ the point of light would be at \begin{align*} x(t) &= t + r \sin (\theta + t) \\ y(t) &= 1 + r \cos (\theta + t).\end{align*} Differentiating with respect to time we get \begin{align*} \frac{dx}{dt} &= 1 + r \cos (\theta + t) \\ \frac{dy}{dt} &= -r \sin (\theta + t), \end{align*} so we see that the arclength of the path the light takes is \begin{align*}\ell &= \int_0^{2\pi} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 }\,dt \\ &= \int_0^{2\pi} \sqrt{ \left( 1 + r \cos (\theta + t)\right)^2 + \left( - r \sin (\theta + t) \right)^2 } \,dt \\ &= \int_0^{2\pi} \sqrt{ 1 + r^2 + 2r \cos (\theta + t) } \,dt.\end{align*} We can easily set $u = \theta + t$ and then through periodicity recognize that we have \begin{align*}\ell = \ell(r) &= \int_0^{2\pi} \sqrt{ 1 + r^2 + 2r \cos(\theta+t) } \,dt\\ &= \int_\theta^{\theta + 2\pi} \sqrt{ 1 + r^2 + 2r \cos u} \,du\\& = \int_0^{2\pi} \sqrt{ 1 + r^2 + 2r \cos u } \,du,\end{align*} regardless of the the value of $\theta.$ Now at this point, if we wanted to we could do some more algebra and appeal to the complete elliptic integrals of the second kind to find that $$\ell(r) = 4(1+r)E\left(\frac{2 \sqrt{r}}{1+r}\right),$$ where $$E(k) = \int_0^{\pi/2} \sqrt{1-k^2\sin^2t}\,dt;$$ however, since we would still need to plug this into another integral, this doesn't really do us any good.

However, we can us the integral definition of $\ell(r)$ and the definition of the randomly chosen starting point to get an expression for the expected length of the path. If we want to get the probability that some starting point $(x_0,y_0)$ is in the annulus centered at $(0,1)$ between radii $r$ and $r + dr,$ we get that the area of the annulus is $2\pi r \,dr,$ while the total probability of the unit disk centered at $(0,1)$ is $\pi,$ then we see that the conditional probability of the starting point being in the annulus between radii $r$ and $r + dr$ is $2r\, dr.$ Therefore, we get from the law of total expectation that $$\mathcal{L} = \mathbb{E} \left[ \ell \right] = \int_0^1 \ell(r) 2r \,dr.$$ Plugging back in the integral form of $\ell(r),$ we get $$\mathcal{L} = \int_0^1 \ell(r) 2r\,dr = \int_0^1 \int_0^{2\pi} 2r \sqrt{ 1 + r^2 + 2r \cos u } \,du \,dr,$$ which we can also recognize as the polar area integral of the function $$f(x,y) = 2\sqrt{ (1-x)^2 + y^2 }$$ over the unit disk. That is, $$\mathcal{L} = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2\sqrt{(1-x)^2 + y^2} \,dy \,dx.$$

Since $$\int \sqrt{a^2 + t^2} \,dt = \frac{t}{2} \sqrt{a^2 + t^2} + \frac{a^2}{2} \ln \left| t + \sqrt{a^2 + t^2} \right| + C,$$ we can do this integration directly. We first get that \begin{align*}g(x) &= \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2 \sqrt{ (1-x)^2 + y^2} \,dy \\ &= 4 \int_0^{\sqrt{1-x^2}} \sqrt{ (1-x)^2 + y^2} \,dy \\ &= 4 \left[ \frac{y}{2} \sqrt{ (1-x)^2 + y^2 } + \frac{(1-x)^2}{2} \ln \left| y + \sqrt{ (1-x)^2 + y^2 } \right| \right]_0^{\sqrt{1-x^2}} \\ &= 4 \left[ \frac{\sqrt{1-x^2} \sqrt{2 - 2x}}{2} + \frac{(1-x)^2}{2} \ln \left| \sqrt{1-x^2} + \sqrt{2-2x} \right| - \frac{(1-x)^2}{2} \ln |1-x| \right] \\ &= 2\sqrt{2} (1-x) \sqrt{1+x} + 2(1-x)^2 \ln \left( \frac{\sqrt{1+x} + \sqrt{2}}{\sqrt{1-x}} \right).\end{align*} So we now have \begin{align*}\mathcal{L} &= \int_{-1}^1 g(x) \,dx \\ &= \int_{-1}^1 2\sqrt{2} (1-x)\sqrt{1+x} \,dx + \int_{-1}^1 2(1-x)^2 \ln \left( \frac{\sqrt{1+x} + \sqrt{2}}{\sqrt{1-x}} \right) \,dx.\end{align*} Using integration by parts, we have first integral is \begin{align*}I_1 &= \int_{-1}^1 2\sqrt{2}(1-x) \sqrt{1+x} \,dx \\ &= 2 \sqrt{2} \left( \left[ \frac{2}{3} (1-x) (1+x)^{3/2} \right]_{-1}^1 + \frac{2}{3} \int_{-1}^1 (1+x)^{3/2} \,dx \right) \\ &= 2 \sqrt{2} \left.\frac{4}{15} (1+x)^{5/2} \right|_{-1}^1 = 2\sqrt{2} \frac{4}{15} 2^{5/2} = \frac{64}{15}.\end{align*} We can break the second integral further into two integrals, namely $$I_2 = \int_{-1}^1 2(1-x)^2\ln \left( \sqrt{1+x} + \sqrt{2} \right) \,dx$$ and $$I_3 = \int_{-1}^1 2(1-x)^2 \ln \sqrt{1-x} \,dx = \int_{-1}^1 (1-x)^2 \ln (1-x) \,dx.$$

If we set $u = \ln \left( \sqrt{1+x} + \sqrt{2} \right)$ then we see that $$du = \frac{dx}{2 \sqrt{1+ x} ( \sqrt{1+x} + \sqrt{2} )} = \frac{ \sqrt{2} - \sqrt{1+x} }{2 \sqrt{1+x} (1-x)} dx,$$ so then if we pursue integration by parts we get to obtain \begin{align*}I_2 &= \int_{-1}^1 2(1-x)^2 \ln \left(\sqrt{1+x} + \sqrt{2}\right) \,dx \\ &= 2 \left( \left[ -\frac{(1-x)^3}{3} \ln \left( \sqrt{1+x} + \sqrt{2} \right) \right]_{-1}^1 + \int_{-1}^1 \frac{(1-x)^3}{3} \frac{\sqrt{2} - \sqrt{1+x}}{2 \sqrt{1+x} (1-x) } \,dx \right) \\ &= 2 \left( -\frac{4}{3} \ln 2 + \int_{-1}^1 \frac{\sqrt{2}}{6} \frac{(1-x)^2}{\sqrt{1+x}} \,dx - \int_{-1}^1 \frac{(1-x)^2}{6} \,dx \right) \\ &= 2 \left( -\frac{4}{3} \ln 2 + \left[ \frac{\sqrt{2}}{3} (1-x)^2 \sqrt{1+x} \right]_{-1}^1 + \frac{2\sqrt{2}}{3} \int_{-1}^1 (1-x) \sqrt{1+x} \,dx + \left[ \frac{(1-x)^3}{18} \right]_{-1}^1 \right) \\ &= 2 \left( -\frac{4}{3} \ln 2 + \frac{4}{9} + \frac{1}{3} \left( \frac{64}{15} \right) \right) = \frac{56}{15} -\frac{8}{3} \ln 2, \end{align*} where we are using our precalculated knowledge of $I_1$ in the second to last step.

Using integration by parts gives $$\int t^2 \ln t \,dt = \frac{t^3}{3} \ln t - \int \frac{t^2}{3} \,dt = \frac{t^3}{3} \ln t - \frac{t^3}{9} + C.$$ With the u = 1-x substituion we get $$I_3 = \int_{-1}^1 (1-x)^2 \ln (1-x) \,dx = \int_{0}^2 u^2 \ln u \,du = \left[\frac{u^3}{3} \ln u - \frac{u^3}{9} \right]_{0}^2 = \frac{8}{3} \ln 2 - \frac{8}{9}.$$ So putting it altogether we get the expected length of the uniformly random point of light to be \begin{align*}\mathcal{L} &= I_1 + I_2 + I_3 \\ &= \frac{64}{15} + \left( \frac{56}{15} - \frac{8}{3} \ln 2 \right) + \left( \frac{8}{3} \ln 2 - \frac{8}{9} \right) \\ &= 8 - \frac{8}{9} = \frac{64}{9} = 7.111111\dots.\end{align*}

Frederica's Cycloidal Film

For her photography show, Frederica Fiddleria attaches a light to a point on the circumference of a circular wheel with a radius of 1 meter. She points a camera at the wheel and, during a single long exposure, rolls the wheel for one revolution along the ground.

When she develops the film, she is curious about the path the light took as the wheel rolled. What is the length of this path?

Lo, and behold! Frederica's light source is traveling along a cycloidal path, which has some fairly well defined properties, but let's try to derive them from scratch. Let's assume that the center of the wheel starts at the point $(0,1).$ Since the circumference of the wheel is $2\pi,$ after one full revolution of the wheel, the center would end up at the point $(2\pi, 1).$ Without loss of generality, let's assume that the light was added to a point $P$ on the circumference of the circle such that the angle between $P$, the center of the circle and the positive $y$-axis is $\theta \in [0,\pi].$ In this case, we see that the point of light is at $x(0) = \sin \theta,$ and $y(0) = \cos \theta.$ After rotating for some time $t \in [0,2\pi),$ the center of the circle will be at the point $(t,1),$ while the angle that the point of light now makes as measured with respect to the positive $y$-axis, will now be $\theta + t.$ So the point of light will be given by the parametric equations \begin{align*} x(t) &= t + \sin(\theta + t),\\ y(t) & = 1 + \cos (\theta + t).\end{align*} In particular, taking derivatives with respect to time, we get \begin{align*} \frac{dx}{dt} &= 1 + \cos (\theta + t) \\ \frac{dy}{dt} &= - \sin (\theta + t).\end{align*}

Therefore, integrating the arclength formula and appealing to the half-angle formula, we see that no matter what the value of $\theta$ is, the length of the path the light took is given by \begin{align*}\ell &= \int_0^{2\pi} \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } \,dt \\ &= \int_0^{2\pi} \sqrt{ \left(1 + \cos (\theta + t)\right)^2 + \left( - \sin (\theta + t) \right)^2 } \,dt \\ &= \int_0^{2\pi} \sqrt{ 2 + 2 \cos (\theta + t) } \,dt \\ &= 2 \int_0^{2\pi} \sqrt{ \frac{ 1 + \cos (\theta + t) }{2} } \,dt \\ &= 2 \int_0^{2\pi} \left|\cos \left( \frac{\theta + t}{2}\right)\right| \,dt \\ &= 2 \int_0^{\pi - \theta} \cos \left(\frac{\theta + t}{2}\right) \,dt + \int_{\pi-\theta}^{2\pi} - \cos \left(\frac{\theta + t}{2}\right) \,dt \\ &= 4 \left. \sin \left(\frac{\theta + t}{2}\right) \right|_0^{\pi - \theta} - 4 \left. \sin \left(\frac{\theta + t}{2}\right) \right|_{\pi - \theta}^{2\pi} \\ &= 4 \left( \sin \frac{\pi}{2} - \sin \frac{\theta}{2} \right) - 4 \left( \sin \left(\pi + \frac{\theta}{2}\right) - \sin \frac{\pi}{2} \right) \\ &= 4 + 4 \sin \frac{\theta}{2} + 4 - 4 \sin \frac{\theta}{2} = 8\end{align*}

Fiddler: The Helical Ball?

Instead of requiring a vertical line down the cylinder’s surface, now any helix down the surface is allowed. An example of such a helix passing through all three gaps is shown below.

What is the probability that there exists at least one such helix that can pass through each ring’s gap?

Again, let's assume that we have cylinder with radius one, that the broken rings are embedded within the planes $z=2,$ $z=1$ and $z=0$ and finally that the gap on the ring at $z=2$ is from $\theta \in [0, \pi / 4],$ while the gaps at $z=1$ and $z=0$ are from $\theta \in [\alpha, \alpha+\pi/4]$ and $\theta \in [\beta, \beta + \pi / 4]$ for some $\alpha, \beta \sim U(0,2\pi),$ respectively. Here again we will play a little fast and loose with notation and say that $[\psi_1, \psi_2] \subseteq [0,2\pi]$ should mean the set $[0, \psi_2 - 2\pi] \cup [\psi_1, 2\pi]$ in the case that $\psi_1 \lt 2\pi \lt \psi_2.$

We also see that the helical path from any point $(2, \theta_2)$ to $(1, \theta_1)$ for any $\theta_1, \theta_2 \in [0,2\pi)$ is given by $z(t) = 2-t,$ $\theta(t) = \theta_2 + t(\theta_1 - \theta_2).$ All we need to do is understand where this helical path will hit the plane $z = 0,$ in this case at $\theta(2) = 2 \theta_1 - \theta_2.$ Since any helical path that gets through the upper and middle gaps will have $\theta_2 \in [0, \pi/4]$ and $\theta_1 \in [\alpha, \alpha + \pi/4],$ we see that $\theta_0 = \theta(2) = 2 \theta_1 - \theta_2 \in [2 \alpha - \pi/4, 2 \alpha + \pi/2].$ See the figure below, for instance for the choice of $\alpha = 4.613846199712232.$

So in order to fit through all of the gaps, then we would need to have $$[\beta, \beta + \pi/4] \cap [2 \alpha - \pi/4, 2 \alpha + \pi/2 ] \ne \emptyset,$$ that is, we would need to have $\beta \in [ 2 \alpha - \pi/2, 2\alpha + \pi/2].$ Therefore, the conditional probability of there being a helical path conditional on choice of $\alpha$ is $$p(\alpha) = \mathbb{P} \{ \beta \in [ 2 \alpha - \pi/2, 2 \alpha + \pi/2 ] \mid \alpha \} = \frac{\pi}{2\pi} \equiv \frac{1}{2}.$$ Therefore, the probability of there being at least one such helix that can pass through each of the gaps is $p = \int_0^{2\pi} p(\alpha) \,d\alpha = \frac{1}{2}.$

Ring around the cylinder

A long vertical cylinder has three narrow open rings, each of which wraps around seven-eighths of the cylinder (leaving a one-eighth “gap”). The rings are evenly spaced vertically, but are otherwise randomly rotated about the cylinder’s central axis. For some orientations of the rings, there exists at least one vertical line down the cylinder’s surface that passes through each ring’s gap, as illustrated below.

What is the probability that at least one such vertical line exists?

Let's define some terms. Let's assume we have a unit cylinder and that the broken rings are embedded within the planes $z=2,$ $z=1$ and $z=0.$ Let's, without loss of generality, assume that the opening at the topmost ring is one-eighth of the way around and oriented such that it is from $\theta \in [0, \pi/4].$ Let's further assume that for $\alpha, \beta \sim U(0,2\pi)$ that the middle and bottom rings at z=1 and z=0 are given by $\theta \in [\alpha, \alpha+\pi/4]$ and $\theta \in [\beta, \beta+\pi/4],$ respectively. Here we shall abuse notation slightly and assume that if $\psi_1 \lt 2\pi \lt \psi_2$ then by $[\psi_1, \psi_2] \subseteq [0,2\pi]$ we mean $[0, \psi_2 - 2\pi] \cup [ \psi_2, 2\pi].$

Any vertical line on this cylinder can be written in cylindrical coordinates as $\theta = \theta_0$ for some $\theta_0 \in [0,2\pi).$ If a vertical line can pass through the top and middle gaps, then we must have $[0, \pi/4] \cap [\alpha, \alpha+\pi/4] \ne \emptyset,$ or equivalently, $\alpha \leq \pi/4$ and $\alpha + \pi/4 \geq 0,$ that is $\alpha \in [-\pi/4, \pi/4].$ In order for this line to pass through all three gaps, we must have $$[ \beta, \beta + \pi/4 ] \cap [ \alpha, \alpha + \pi/4 ] \cap [0, \pi/4] \ne \emptyset,$$ or equivalently $\beta \in [ \max \{ 0, \alpha \} - \pi/4, \min \{0, \alpha\} + \pi/4 ].$

Therefore, the probability that a vertical line can pass through all gaps is \begin{align*}p &= \int_{-\pi/4}^{\pi/4} \int_{\max \{0, \alpha\} - \pi/4}^{\min \{0, \alpha\} + \pi/4} \frac{d\beta}{2\pi} \, \frac{d\alpha}{2\pi} \\ &= \int_{-\pi/4}^{\pi/4} \left( \frac{\pi}{2} - |\alpha| \right) \,\frac{d\alpha}{4\pi^2}\\ &= \frac{1}{2\pi^2} \int_0^{\pi/4} \left( \frac{\pi}{2} - \alpha \right), d\alpha\\ &= \frac{1}{2\pi^2} \int_{\pi/4}^{\pi/2} u \,du = \frac{1}{4\pi^2} \left[ \left(\frac{\pi}{2}\right)^2 - \left(\frac{\pi}{4}\right)^2 \right] \\ &= \frac{1}{4\pi^2} \frac{3\pi^2}{16} = \frac{3}{64} = 4.6875\%\end{align*}

Fiddling for the win

Congratulations to Fiddler Nation for making it to the semifinals of the World Cup! All four teams that made it this far are equally matched in that they each possess the same total amount of “energy.” In advance of each semifinal game, teams must independently decide how much of their energy to allocate to the match; all remaining energy goes toward the finals. The team that spends more energy in any given game will win. The semifinals and finals occur so close in time that teams can’t recuperate any of their energy in between.

You’ve heard that the managers for the other three teams are abysmal and have no idea how to allocate their teams’ energy. Each of the other managers will independently pick a random percentage between 0 and 100 and allocate that portion of their team’s energy to the semifinal game; the rest of that team’s energy will go toward the final. Since you’re the cleverest manager of the bunch, you can choose an optimal strategy that will maximize Fiddler Nation’s probability of winning the World Cup. What is this optimal probability?

Let's define some terms, let's let $V$, $W$ and $X \sim U(0,1)$ be the i.i.d. random energy levels selected by the other three coaches to the first match. Without loss of generality, let's assume Fiddler Nation's first draw is against the team that devotes $V$ energy to the first match. Therefore, if we devote some level of energy $E \in [0,1]$ to the first match, then the probability of making the final is $\mathbb{P} \{ V \leq E \} = E.$

On the other hand since the other two coaches devoted $W$ and $X$ energies respectively to the first match, the coach that devoted $\max \{ W, X \}$ will have won the match, and the amount of energy remaining for the final is only $1 - \max \{W, X \}.$ Again, if our strategy was to devote $E$ to the first match that Fiddler Nation's fierce fiddling team would only have $1-E$ energy left for the final, which means that we will win the final only if $1 - E \geq 1 - \max \{W,X\},$ or equivalently, $\max \{W,X \} \geq E.$ So the probability of winning the final if you only have $1-E$ energy remaining is $$\mathbb{P} \{ \max \{W, X \} \geq E \} = 1 - \mathbb{P} \{ \max \{ W, X \} \leq E \} = 1 - E^2.$$

Putting these together we see that the probability of winning the final by devoting energy $E$ to the semifinal is $$f(E) = \mathbb{P} \{ V \leq E \} \mathbb{P} \{ \max \{W, X \} \geq E \} = E (1-E^2) = E - E^3.$$ In this case, we want to maximize this probability, which will occur when $f^\prime(E) = 1-3E^2 = 0,$ or at $E^* = \frac{1}{\sqrt{3}}.$ In this case, we see that the maximum probability of winning the World Cup is $$f(E^*) = \frac{1}{\sqrt{3}} - \left(\frac{1}{\sqrt{3}}\right)^3 = \frac{2\sqrt{3}}{9} \approx 0.38490017946\dots.$$