Average Cantor Distance

It’s possible to pick a random point in the Cantor set in the following way: Start with the entire number line from 0 to 1. Then, every time you remove a middle third, you give yourself a 50 percent chance of being on the left remaining third and a 50 percent chance of being on the right remaining third. Then, when you remove the middle third of that segment, you again give yourself a 50 percent chance of being on the left vs. the right, and so on.

Suppose I independently pick two random points in the Cantor set. On average, how far apart can I expect them to be?

Let's define $\Delta$ as the average distance between two different randomly chosen points in the Cantor set, say $a$ and $b \in \mathcal{C}.$ Based on the self-similarity of the Cantor set $\mathcal{C},$ we see that $\mathcal{C} = \frac{1}{3} \mathcal{C} \cup \left( \frac{2}{3} + \frac{1}{3} \mathcal{C} \right).$ Then from the law of total expectation and the fact that the probability of being in either the left or right half of the Cantor set is independently $\frac{1}{2}$ for each of $a$ and $b$, we get \begin{align*} \Delta &= \mathbb{E} \left[ |a-b| \mid a, b \in \frac{1}{3} \mathcal{C} \right] \mathbb{P} \left\{ a, b \in \frac{1}{3} \mathcal{C} \right\} \\ &\quad + \mathbb{E} \left[ |a-b| \mid a, b \in \frac{2}{3} + \frac{1}{3} \mathcal{C} \right] \mathbb{P} \left\{ a, b \in \frac{2}{3} + \frac{1}{3} \mathcal{C} \right\} \\ & \quad\quad + \mathbb{E} \left[ |a-b| \mid a \in \frac{1}{3} \mathcal{C}, b \in \frac{2}{3} + \frac{1}{3} \mathcal{C} \right] \mathbb{P} \left\{ a \in \frac{1}{3} \mathcal{C}, b \in \frac{2}{3} + \frac{1}{3} \mathcal{C} \right\} \\ &\quad\quad\quad + \mathbb{E} \left[ |a-b| \mid a \in \frac{2}{3} + \frac{1}{3} \mathcal{C}, b \in \frac{1}{3} \mathcal{C} \right] \mathbb{P} \left\{ a \in \frac{2}{3} + \frac{1}{3} \mathcal{C}, b \in \frac{1}{3} \mathcal{C} \right\} \\ &= \frac{ \Delta_1 + \Delta_2 + \Delta_3 + \Delta_4}{4}.\end{align*}

Since they are simply scaled copies of the original problem, we see that $\Delta_1 = \Delta_2 = \frac{1}{3} \Delta.$ However, let's take a look at $\Delta_3$ and $\Delta_4.$ For any $a \in \frac{2}{3} + \frac{1}{3} \mathcal{C},$ there is some $a^\prime \in \mathcal{C}$ such that $a = \frac{2}{3} + \frac{1}{3} a^\prime.$ So if $b \in \frac{1}{3} \mathcal{C}$ then there is some $b^\prime \in \mathcal{C}$ with $b=\frac{1}{3} b^\prime$. Therefore, we then have $|a-b| = \frac{2}{3} + \frac{1}{3}(a^\prime - b^\prime) \geq \frac{1}{3} \geq 0.$ So taking the expectation over all values of $a^\prime, b^\prime \in \mathcal{C}$ we get $$\Delta_3 = \Delta_4 = \mathbb{E} \left[ |a-b| \mid a \in \frac{2}{3} + \frac{1}{3} \mathcal{C}, b \in \frac{1}{3} \mathcal{C} \right] = \frac{2}{3} + \frac{1}{3} \mathbb{E} \left[ a^\prime - b^\prime \right] = \frac{2}{3}.$$

Thus putting everything together we get $$\Delta = \frac{1}{4} \sum_{i=1}^4 \Delta_i = \frac{ \frac{2}{3} \Delta + \frac{4}{3} }{4} = \frac{1}{6} \Delta + \frac{1}{3},$$ which means that $\frac{5}{6} \Delta = \frac{1}{3}$ or finally that the average distance between two randomly chosen points in the Cantor set is $$\Delta = \frac{6}{5}\cdot \frac{1}{3} = \frac{2}{5}.$$

Team Centroid is slacking

Two teams of shovelers plan to remove all the snow from a parking lot that’s shaped like a regular hexagon. Team Vertex initially places each of its six shovelers at the six corners of the lot. Meanwhile, Team Centroid initially places all its shovelers at the very center of the lot.

Each team is responsible for shoveling the snow that is initially closer to someone on their own team than anyone on the other team. What fraction of the lot’s snow is Team Centroid responsible for shoveling?

We can draw a straight line from Team Centroid's central spot to each of the vertices. Drawing a perpendicular bisector of each of these lines we obtain a new smaller, rotated hexagon, shown in blue in the figure below. We note that the apothem of the smaller hexagon is one half of the circumradius of the larger hexagon. Since the area of a regular $n$-gon is given by $$A_n = na^2 \tan \frac{\pi}{n} = \frac{1}{2} nR^2 \sin \frac{2\pi}{n},$$ we see that the apothem and circumradius are related by $a = R \cos \frac{\pi}{n},$ so in this case we have $a = R \cos \frac{\pi}{6} = \frac{R \sqrt{3}}{2},$ or equivalently $R = \frac{2a}{\sqrt{3}}.$

Since we have that the apothem of the Team Centroid hexagon is one half of the circumradius of the Team Vertex hexagon, we have $$R_C = \frac{2}{\sqrt{3}} a_C = \frac{1}{\sqrt{3}} R_V.$$ Since areas scale with the square of the circumradius, we have that the area shoveled by Team Centroid is $\left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3}$ of the entire hexagonal parking lot.

Fewest Firsts

As we just noted, in 2026, all seven days of the week appear as the first of the month at least once. But you know, I decided that I don’t like that at all. Instead, I want as few days of the week as possible to appear as the first of the month in a given year.

To accomplish this, I have been granted the authority to change the number of days in each of that year’s 12 months, provided that there are still 365 or 366 days in the year and each month has at least 28 days and at most 31 days.

What are the fewest days of the week that can appear as the first of the month in such a calendar year? (And for fun, rather than for credit: How many such calendars can you design with this property?)

Let $N^*$ be the minimal number of days of the week that can appear as the first of the month of any calendar year. First from the Classic problem we see that the furthest away that two consecutive months can start is 3 days of the week apart, so this means that the smallest that we could possibly hope for is to have a calendar with only three possible days of the week for firsts of the month. That is, $N^* \gt 2,$ or equivalently that $N^* \geq 3.$

Therefore, all we have to do is provide some calendar year where $N=3$ is attained. For instance, take the following setup where March, June, and September have 29 days and all other months have 31 days. This would give $3 \cdot 29 + 9 \cdot 31 = 87 + 279 = 366$ days in the year. Similarly, if January 1 occurs on $x \in \mathcal{D},$ then February 1 occurs on $x+3,$ March 1 occurs on $x+6,$ April 1 occurs on $x+7 \equiv x,$ May 1 occurs on $x+3,$ June 1 occurs on $x+6,$ July 1 occurs on $x+7\equiv x,$ August 1 occurs on $x+3,$ September 1 occurs on $x+6,$ October 1 occurs on $x+7 \equiv x,$ November 1 occurs on $x + 3$ and finally, December 1 occurs on $x+6.$ Putting these altogther gives a subset $\mathcal{N} = \{ x, x + 3, x + 6 \} \subsetneq \mathcal{D},$ with $N= |\mathcal{N}| = 3.$ Therefore, we have that $N^* \leq N = 3$ and hence the fewest days of the week that can appear as the first of the month is $N^*=3.$

Can every day be the first?

In 2026, every day of the week is the first day of the month at least once:

• Monday is June 1.
• Tuesday is September 1 and December 1.
• Wednesday is April 1 and July 1.
• Thursday is January 1 and October 1.
• Friday is May 1.
• Saturday is August 1.
• Sunday is February 1, March 1, and November 1.

Is 2026 special in this regard? If so, when is the next year when one of the days of the week is not represented among the firsts of the month? Otherwise, if 2026 is not special in this regard, then why not?

First let's note that if a month has $28$ days, looking at you February, then the next month will start on the same day of the week that it starts. If a month has $29$ days, again looking less frequently at you February, then then next month will start on the next day of the week. If a month has $30$ days then the next month will start two days after it, and finally for all of those $31$ day months, the following month will start three days of the week after it.

Let's further represent the days of the week as numerals, $\mathcal{D} = \mathbb{Z} / 7\mathbb{Z}.$ If January 1 occurs on some $x \in \mathcal{D}$ then February 1 occurs on $x+3.$ If the year is not a leap year then we have March 1 also on $x+3,$ followed by April 1 on $x+6,$ May 1 on $x+8 \equiv x+1,$ June 1 on $x+4$, July 1 on $x+6,$ August 1 on $x+9 \equiv x+2$, September 1 on $x+5,$ October 1 on $x+7 \equiv x,$ November 1 on $x+3$ and finally December 1 on $x+5.$ Gropuing these all together we see that we have all the days of the week covered $\{ x, x+1, x+2, x+3, x+4, x+5, x+6\} = \mathcal{D}.$

If on the other hand, this is a leap year, then we have January 1 and February 1 on $x$ and $x+3$ as before, but then March 1 on $x+4,$ April 1 on $x+7 \equiv x,$ May 1 on $x+2,$ June 1 on $x+5,$ July 1 on $x+7 \equiv x$, August 1 on $x+3,$ September 1 on $x+6,$ October 1 on $x+8 \equiv x+1,$ November 1 on $x+4,$ and finally December 1 on $x+6.$ Again, grouping these all togehter, we see that we have all the days of the week covered $\{ x, x+1, x+2, x+3, x+4, x+5, x+6 \} = \mathcal{D}.$

Therefore, there is nothing special about 2026. Each and every year under the Gregorian system has this property where each day of the week is represented among the firsts of the month.

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 $\mathbb{E}[S_\infty] = 2.$