The Squiddler, Extra Credit

There are $N$ contestants about to play Mingle, where $N$ is less than $500$. This time, in each round, the number for group size can be anywhere from $1$ to $10$, with each number equally likely to be called. Also, the numbers called in the rounds are all independent of each other. It appears this version of Mingle will continue until there are no more survivors.

Before the game starts, one of the contestants, Young-il, does some quick computation. He says to his companion, Gi-hun: “Oh no, it’s too bad we only have $N$ people.” (Of course, Young-il doesn’t actually say “$N$,” but rather the number that $N$ represents.)

Young-il continues: “If we had started with $N+1$ people instead, I’d expect the game to last about $10$ additional rounds, on average, than I’d expect from starting with $N$ people.”

What is the value of $N$?

In the extra-credit version of Mingle, we again assume that the contestants are all genial, cooperative mathematicians, and there will me no nefarious and/or inadvertent issues with people not being able to form the appropriate groups. In this version, it is even easier to believe this, since rather than unceremonious death, let's say that all contestants in Extra Credit Mingle get paid proportionally for the number of rounds lasted, so that everyone is incentivized for someone to last to the next round.

The extra credit wrinkle of only ever needing to make groups of size $1$ to $10$ introduces some very weird behavior. For instance, if we ever had $N = 2520 = \textsf{lcm} \{ 1, 2, \dots, 10 \}$ then the game will go on for an infinite number of rounds, since no matter what number is called, the contestants will be able to make appropriately round groups. Thus, while there is a nonzero expectation for the number of rounds in the Extra Credit Mingle for $N \lt 2520,$ for any $N \geq 2520$ the expected number of rounds is infinite.

Anyway, let's set up the conditional expectation chain. Let's say that for some $N,$ we know that $E_N$ is the expected number of rounds when starting with $N$ contestants. From the law of total probability, assuming that we know $E_k$ for all $k = 1, \dots, N,$ we can set up the following recursive formula: $$E_N = 1 + \sum_{i=1}^{10} \mathbb{P} \{ i \} E_{ i \lfloor N / i \rfloor } = 1 + \frac{1}{10} \sum_{i=1}^{10} E_{i \lfloor N / i \rfloor}, \,\, \text{for $N \lt 2520$}.$$ (Note, and last one, sorry, but I just think it's kinda neat ... that we see here in the case of $N=2520$ that since $i\lfloor N / i\rfloor = N$ for each $i = 1, \dots, 10,$ that we get the nonsensical $E_{2520} = 1 + E_{2520}.$)

Now there are a few ways to think about this. We can turn the recursion formula into a giant system of linear equations and then solve the $500 \times 500$ lower $10$-diagonal matrix inverse to solve the values of $E_N,$ $N = 1, \dots, 500.$ Instead, I opted for setting up a memoized recursive function in a few lines of Python:

from math import floor
exp_num_rounds = {}
def compute_exp_num_rounds(N):
    if N == 0: return 0
    if N not in exp_num_rounds.keys():
        transitions = []
        for i in range(10):
            transitions.append( int((i+1) * floor( N / ( i+ 1.0) )) )
        others = sum([ compute_exp_num_rounds(t) for t in transitions if t < N ])
        self = sum([ 1 for t in transitions if t == N])
        exp_num = (1.0 + 0.1 * others)/(1.0 - 0.1 * self)
        exp_num_rounds[N] = exp_num
    return exp_num_rounds[N]

Using this function I quickly calculated $E_N$ for $N = 1, \dots, 500,$ and in fact for all $N$ up to $2519.$ We find what looks from a distance as a relatively straight line with a few barely discernable jumps. If we plot the differences $E_{N+1} - E_N$ versus $N$ we see that there are only $9$ total values of the $N$ in $\{1, \dots, 2519\}$ such that the number of expected rounds would be more than $9$ more if they started with $N+1.$ Conveniently, enough, since there is only one such value of below $500$, we know that Young-il is upset that they are starting the round of Extra Credit Mingle with $N=359$ contestants.

Looking at the values with large jump sizes, we can start to build a theoretical argument for why these values exhibit this behavior. Note that in each of these cases, $$N \in \{ 359, 719, 839, 1079, 1259, 1439, 1679, 1799, 2159\}$$ that we have $i \not\mid N$ for each $i = 2, 3, \dots, 10$ while there is only one value of $j \in \{ 2, 3, \dots, 10 \}$ for which $j \not\mid N+1.$ In a sufficiently handwavy way, this relative $9:1$ ratio in probability of dropping below $N$ after round one, is what is driving the roughly $9$ to $10$ additional expected rounds.

The Squiddler

N.B. - Shout-out to colleague and fellow Fiddler enthusiast, Rohit Jawle, for the punny post name this week.

In Season 2 of Squid Game, contestants play a game of “Mingle” (spoiler alert!). For each round of this game, the contestants all wait in a common area until a number is called. They must then split up into groups of that number. Any contestants not in a group of the correct number after $30$ seconds … well, let’s just say bad things happen to them. For now, we’ll refer to contestants who make it through a round in a group of the correct size as having “survived” the round.

Suppose there are initially $N$ contestants. In the first round, contestants must form groups of 1. Okay, that’s not too hard; everyone survives. In the second round, contestants must form groups of $2.$ This continues (with groups of $k$ in round $k$) until the $38$th round, when contestants must form groups of $38.$ What is the smallest number $N$ such that there is at least one person who survives all $38$ rounds?

Let's firstly assume that the contestants are all genial, cooperative mathematicians, and there will me no nefarious and/or inadvertent issues with people not being able to form the appropriate groups of the appropriate sizes. This might not totally make sense in Squid Game context where the penalty for being the odd person out is so draconian as death, but perhaps everyone agreed on an inherent numbering ahead of time, and if your number just doesn't come up in one round or another, these astute and honorable mathematicians at least march to their demise knowing that they played the game optimally. Whatever that's worth.

Anyway ... Let's say that at the end of round $k$ there are $N_k$ contestants remaining. Since any stragglers who are not neatly aligned in a group of $k+1$ at the end of $30$ seconds will be eliminated at the end of round $k+1$, then we see that we must have $$N_{k+1} = (k+1) \left\lfloor \frac{N_k}{k+1} \right\rfloor,$$ where we start with let's say $N_0 = N$ contestants. In theory, we could start at the beginning and guess numbers, work $38$ times through this recursion formula and then see if we end up with anyone remaining after $38$ rounds. Then once we found some starting $N$ that works, for instance, starting $N=2025$ leaves $1634$ lucky remaining contestants after $38$ rounds, you can whittle that down to final a minimal number. Of course, we could also work backwards from the condition that we want to have $N_{38} \gt 0.$

In particular, since we need $N_{38} \gt 0$ and $N_{38} \in 38 \mathbb{Z},$ the minimal number we can have is $N_{38} = 38.$ From the recursion formula, we see that $$38 \left\lfloor \frac{N_{37}}{38} \right\rfloor = 38,$$ which implies that we are looking for $N_{37} \in \{ 38, 39, \dots, 75 \}.$ Since we must also have $N_{37} \in 37 \mathbb{Z},$ we get the minimal value of $N_{37} = 74.$ In turn, we see that $N_{37} = 74$ implies that we must have $N_{36} \in \{ 74, 75, \dots, 110 \} \cap 36 \mathbb{Z},$ that is $N_{36} = 108.$ We can follow all of these nested implications back up to the minimal number of contestants needed to have some number of survivors in each of the 38 rounds to be $N = N_1 = N_2 \geq 454$ contestants. For completeness, if we were to start with $N = 454$ contestants, then the sequence of contestants remaining after $k$ rounds would be as follows:

Round #	Contestants Remaining
0	454
1	454
2	454
3	453
4	452
5	450
6	450
7	448
8	448
9	441
10	440
11	440
12	432
13	429
14	420
15	420
16	416
17	408
18	396
19	380
20	380
21	378
22	374
23	368
24	360
25	350
26	338
27	324
28	308
29	290
30	270
31	248
32	224
33	198
34	170
35	140
36	108
37	74
38	38

Ruffles have maximally aligned ridges

Two large planar sheets have parallel semicircular cylindrical ridges with radius $1.$ Neighboring ridges are separated by a distance $L \geq 2$. The sheets are placed so that the ridges extrude toward each other, and so that the sheets cannot shift relative to each other in the horizontal direction, as shown in the cross-section below:

Which value of $L$ maximizes the empty space between the sheets?

Firstly, let's establish some parameters in order to make sure we can turn this into a nice(r) optimization problem. Let $h$ be the vertical distance between the two planar sheets for a given value of $L \geq 2$. Let's assume that one cylindrical ridge on the lower sheet is centered at the origin, that is, $x = 0.$ By construction, the next cyliindrical ridge on the lower sheet (to the right) is centered at $x = L$ and the center of the semicircular cylindrical ridges on the upper sheet that is tangent to these two lower ridges should be centered at $x = \frac{L}{2}.$ Adding in the vertical axis, we have semicircles of radius $1$ centered at $(0,0)$, $(L, 0)$ and $(\frac{L}{2}, h).$

Let's next determine the objective function for optimization, in terms of $h$ and $L$. That is, we are trying to determine the maximal value of the cross-sectional area of empty space per unit length of one sheet. That is, we have $f(h, L) = \frac{A(h,L)}{L},$ where $A(h,L)$ is the cross-sectional area of the empty space bounded by the rectangle $0 \leq x \leq L$ and $0 \leq y \leq h.$ Now, the non-empty space in this rectangle is half of the unit semicircle centered at (0,0), the entire unit semicircle centered at (L/2, h) and half of the unit semicricle centered at (L, 0), so we need to subtract out the total non-empty area of $\pi,$ that is, $A(h,L) = hL - \pi,$ so we arrive at $$f(h, L) = \frac{A(h, L)}{L} = \frac{hL - \pi}{L} = h - \frac{\pi}{L}.$$

Now, let's determine the height $h$ as a function of $L,$ that is, $h = h(L).$ Since there is no ability to shift the sheets horizontally, it must be that the semicircle centered at $(\frac{L}{2}, h(L))$ is tangent to both of the lower semicircles. Thus, we can draw a straight line from $(0,0)$ to $(\frac{L}{2}, h(L))$ which must have length $2$, since each of the semicircles have unit radii, thus we have the Pythagorean theorem $$2^2 = \left(\frac{L}{2}\right)^2 + h(L)^2,$$ or equivalently $$h(L) = \frac{1}{2} \sqrt{16 - L^2}.$$ In actuality, we should be careful to make sure that the space between the sheets is at least the unit radius of the semicircular ridges, that is, $$h(L) = \max \left\{ 1, \frac{1}{2} \sqrt{16 - L^2} \right\}.$$ However, we can disregard the case where $L \gt 2\sqrt{3}$ where $h(L) = 1 \gt \frac{1}{2} \sqrt{16-L^2}$, since otherwise, barring some extreme static friction, we could horizontally shift the sheets relative to one another which is a no-no of the setup.

Having decided to ignore the other case, we have $h(L) = \frac{1}{2} \sqrt{16-L^2}$ for all $2 \leq L \leq 2\sqrt{3},$ so we now have $$f(L) = f(h(L), L) = \frac{1}{2} \sqrt{16 - L^2} - \frac{\pi}{L}.$$ Taking derivatives, we $$f^\prime(L) = \frac{-L}{2 \sqrt{16-L^2}} + \frac{\pi}{L^2},$$ which if we set equal to zero and solve for $L$ results in the sextic polynomial $$L^6 - 4\pi^2 L^2 + 64 \pi^2 = 0.$$ If we set $\ell = L^2,$ then the condition for optimality becomes a depressed cubic in $\ell$ with a single real zero, that can be handled by Cardano's formula, that is, $$\ell = \sqrt[3]{ 32 \pi^2 + \sqrt{ 1024 \pi^4 + \frac{64}{27} \pi^6} } + \sqrt[3]{ 32\pi^2 - \sqrt{1024 \pi^4 + \frac{64}{27} \pi^6}} \approx 7.06550552638\dots$$ Therefore, the optimal spacing of the ridges is $$L^* = \sqrt{\ell} \approx 2.65810186531\dots.$$

This is New Years, not Prime Day!

The number $2025$ is not prime. As a matter of fact, it’s a perfect square: $2025 = 45^2$.

You cannot make $2025$ by adding two distinct primes. To do so, you’d have to add an even prime and an odd prime. The only even prime is $2,$ but $2025 − 2 = 2023$, which is not prime (it’s equal to $7∙172$). But you can make $2025$ by adding three distinct primes. For example, $661 + 673 + 691 = 2025.$ You can also make $2025$ by adding four distinct primes: $2 + 659 + 673 + 691 = 2025.$

What is the greatest number of distinct primes that add up to $2025$?

First let's find the set of all primes less than $2025,$ that is, \begin{align*}P = \{ & 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101,\\ & 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197,\\ & 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311,\\ & 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431,\\ & 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557,\\ & 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661,\\ & 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809,\\ & 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937,\\ & 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049,\\ & 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153,\\ & 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277,\\ & 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381,\\ & 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487,\\ & 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597,\\ & 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699,\\ & 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823,\\ & 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949,\\ & 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017 \}.\end{align*} One way to encode the problem is as the following binary linear knapsack program: \begin{align*} N = \max \,\, & \sum_{p \in P} x_p \\ \text{s.t.} \,\, & \sum_{p \in P} p x_p = 2025 \\ & x_p \in \{ 0, 1 \}, \, \forall p \in P\end{align*}

We can first loosen the binary restrictions a bit to get an upper bound. If as opposed to having $x_p \in \{0,1\},$ if we use $x_p \in [0,1],$ $\forall p \in P,$ then we get an upper bound of $\tilde{N} = \frac{4624}{139} = 33.266187\dots,$ by greedily filling our knapsack with all of the smallest primes until we have to resort to a fractional piece, that is, $$\tilde{x}_p = \max \left\{ 0, \min \left\{ \frac{2025 - \sum_{j \in P, j \lt p } j }{p}, 1 \right\} \right\}, \, \, \forall p \in P.$$ So we know that $N \leq \lfloor \tilde{N} \rfloor = 33.$ If we have $33$ primes and $2$ is one of them, then the sum will be even and so could never be $2025.$ The smallest sum of $33$ primes that does not include $2$ is $2125,$ which is too large. So we can never have a set of $33$ primes add up to $2025,$ and hence the upper bound must be $N \leq 32.$ From here, we can either hunt and peck for the right tradeoff of primes, ... or we can just program a quick binary program and your local solver to arrive at the optimal answer is that you can make $2025$ into the sum of $N = 32$ primes (so not that far off from the non-integral approximation), since \begin{align*}2025 &= 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 47 \\ & + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89 + 101 + 103 + 107\\ & + 109 + 113 + 127 + 139 + 149 + 157 \end{align*}

It's my particles in a box!

You have three particles inside a unit square that all repel one another. The energy between each pair of particles is $1/r,$ where $r$ is the distance between them. To be clear, the particles can be anywhere inside the square or on its perimeter. The total energy of the system is the sum of the three pairwise energies among the particles.

What is the minimum energy of this system, and what arrangement of the particles produces it?

If we only had two particles, then the furthest away that the particles could be (and hence the minimal total energy arrangment), would be opposite corners of the unit square, e.g., one at $(0,0)$ and the other at $(1,1).$ In this configuration, the total energy would be $E_2 = \frac{1}{\sqrt{2}}.$ So let's assume that we have two of the three particles there and then see where the other particle would end up.

Assume that the third particle is at the point $X=X(a,b)$ for some $0 \leq a, b \leq 1.$ Then the total energy of the system would be $$E = E(a,b) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{a^2+b^2}} + \frac{1}{\sqrt{(1-a)^2 + (1-b)^2}}.$$ Taking derivatives gives $$\nabla E = \begin{pmatrix} -\frac{a}{(a^2 + b^2)^{3/2}} + \frac{1-a}{\left( (1-a)^2 + (1-b)^2 \right)^{3/2}} \\ -\frac{b}{(a^2 + b^2)^{3/2}} + \frac{1-b}{\left( (1-a)^2 + (1-b)^2 \right)^{3/2}}\end{pmatrix}.$$ The only analytic root of the gradient mapping is $a=b=\frac{1}{2},$ which would give a total energy of $E(\frac{1}{2},\frac{1}{2}) = \frac{1}{\sqrt{2}} + 2 \frac{2}{\sqrt{2}} = \frac{5}{\sqrt{2}},$ but this is a saddle point and we can do better by checking what happens if $X$ were along the perimeter of the unit square.

Assume without loss of generality and by availing ourselves of symmetry, that $b = 0$ and we just have $0 \leq a \leq 1.$ In this case, we have $$\tilde{E}(a) = E(a,0) = \frac{1}{\sqrt{2}} + \frac{1}{a} + \frac{1}{\sqrt{1 + (1-a)^2}}$$ which has derivative $$\tilde{E}^\prime = -\frac{1}{a^2} + \frac{1-a}{\left(1 + (1-a)^2\right)^{3/2}} = \frac{a^2(1-a) - \left(a^2 - 2a+2\right)^{3/2}}{a^2\left(a^2 - 2a+2\right)^{3/2}}.$$ Since $$a^2(1-a) \leq \frac{4}{27} \lt 1 \leq \left(a^2 - 2a + 2\right)^{3/2},$$ for all $a \gt 0,$ we have $\tilde{E}^\prime \lt 0$ for all $a \gt 0,$ so the lowest possible value is when $a^* = 1,$ that is when $X^* = (1,0),$ and the total energy is at its lowest $$E^* = \tilde{E}(1) = \frac{1}{\sqrt{2}} + 2 \approx 2.70710678119....$$ There are other possible arrangements of the particles that obtain the same minimal energy. Namely, any of the four right isoceles triangle formed by choosing three out of the four vertices of the unit square will give this minimal energy configuration.

Snoring while sorting

You are waiting in line to be sorted into one of the four houses of Logwarts (a posh wizarding boarding school in the Scottish highlands) by an anthropomorphic sorting hat. The hat is a bit of a snob about the whole matter, and refuses to sort two students in a row into the same house. If a student requests a certain house, but the previously sorted student was already sorted into that same house, then the hat chooses randomly from among the three remaining houses.

You are standing 10th in line, and you make plans to request Graphindor house for yourself. As for the other students in line, you can assume that they have random preferences from among the four houses. The first student steps up, and has a brief, quiet conversation with the hat. After a few moments, the hat proclaims, “Graphindor!” At this point, what is the probability that you will be sorted into Graphindor?

Let's define our probability space such that it will be useful for the this problem. Of course, as stated there are three other houses with fanciful names (perhaps Hufflepath, Ravenndiagram and Slytheorem), but as far as you're concerned it is Graphindor or bust, so luckily all of these outcomes are indistinguishable to you. Let $p_{i,k}$ be the probability that the $i$th wizardling on line is sorted into Graphindor, subject to the fact that the $k$th wizardling on line was the last one to actually be sorted to Graphindor.

As far as you are concerned, you get to Graphindor as long as the 9th wizard is not sorted there, since you will always ask the hat to sort you into Graphindor. So, given that the 1st wizardling on line was just sorted into Graphindor, the probability that you will be sorted into Graphindor is $p=1-p_{9,1}.$ All that remains now is to divine what the formula for $p_{i,k}$ is anyway.

Let's say that we know $p_{i,k}$ for some $1 \leq k \leq i.$ There are two ways for the $(i+1)$th wizardling to end up sorted into Graphindor: either (a) the $i$th wizard is sorted to some ``not Graphindor'' house and the $(i+1)$th wizardling requested Graphindor; or (b) the $i$th wizard is sorted to some ``not Graphindor'' house and the $(i+1)$th wizardling requested that same ``not Graphindor'' house. In both cases, the probability of event (a) is $1/4$ while the probability of event (b) is $1/12 = 1/4 \cdot 1/3.$ Putting these two together, we get the recursion relationship $$p_{i+1,k} = \frac{1}{4} (1-p_{i,k}) + \frac{1}{12} (1-p_{i,k}) = \frac{1-p_{i,k}}{3}, \,\,\text{for $i \geq k \geq 1.$}$$ By construction, we have $p_{k,k} = 1,$ for each $k \geq 1.$

This recursive formula leads to the explicit formula $$p_{i,k} = \frac{1}{4} \left(1 + 3 \cdot (-3)^{k-i}\right), \,\,\, \text{for all $i \geq k \geq 1.$}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\star$$ We see that for $i=k \geq 1$ that we recover $p_{k,k} = \frac{1}{4} (1 + 3 \cdot (-3)^0 ) = 1,$ as our base case. If we assume that the formula $\star$ holds for some $i \geq k \geq 1,$ then from the recursive formula we see that \begin{align*}p_{i+1,k} = \frac{1-p_{i,k}}{3} &= \frac{1 - \frac{1}{4} \left(1 + 3 \cdot (-3)^{k-i} \right)}{3} \\ &= \frac{\frac{3}{4} - \frac{3}{4} \cdot (-3)^{k-i}}{3} \\ &= \frac{1}{4} \left( 1 + \frac{3}{-3} (-3)^{k-i} \right) \\ &= \frac{1}{4} \left( 1 + 3 \cdot (-3)^{k-i-1} \right),\end{align*} which follows formula $\star$ for the case $i+1$. Thus by induction we have proven the formula. This therefore allows us to answer that if you are $10$th in line and have resolved to request Graphindor no matter what that your probability of getting sorted into Graphindor after the first wizard is sorted into Graphindor is $$p=1-p_9 = 1 - \frac{1}{4} \left(1 + 3 \cdot (-3)^{1-9}\right) = \frac{3-3^{-7}}{4} = \frac{6560}{8748} = \frac{1640}{2187} \approx 0.7498856882....$$

Now, instead of being 10th in line, suppose you are $N$th in line, where $N$ is some value much greater than 10. Because so many students are being sorted in front of you, you decide you’ll take a nap. You wake up without any idea of how long you were out—it could have been a second, or it could have been an hour, you’re just not sure. It’s still not your turn to be sorted yet, but you see a student wearing the hat. After a brief moment, the hat shouts, “Graphindor!”

What is the smallest value of $N$ such that your probability of being sorted into Graphindor is greater than $p$? (To be clear, when you wake up, the student being sorted is anywhere from first in line to immediately before you in line with equal probability.)

OK, so now it becomes a little clearer why I chose such obscure nomenclature for the earlier problem. However, in this case where you are sitting at $N$th in line and then dozed off only to wake up as the wizardling in uniformly randomly distributed position $M$ is sorted into Graphindor, so we have the expected probability of being sorted into Graphindor is still dependent on whether or not the person immediately in front of you is sorted into Graphindor, that is $q=1-p_{N-1,M}$. However, here since $M$ is random we have \begin{align*} q = 1 - p_{N-1,M} &= 1 - \sum_{i=1}^{N-1} \mathbb{P} \{ M = i \} p_{N-1,i} \\ &= 1 - \frac{1}{N-1} \sum_{i=1}^{N-1} \frac{1}{4} \left(1 + 3 \cdot (-3)^{i-N+1} \right) \\ &= \frac{3}{4} - \frac{3 \cdot (-3)^{1-N}}{4(N-1)} \sum_{i=1}^{N-1} (-3)^i \\ &= \frac{3}{4} - \frac{3 \cdot (-3)^{1-N}}{4(N-1)} \frac{(-3) - (-3)^N}{1-(-3)} \\ &= \frac{3}{4} \left( 1 - \frac{ 9 \cdot (-3)^{-N} + 3}{4(N-1)} \right) \\ &= \frac{3}{16} \frac{4N-7-9\cdot(-3)^{-N}}{N-1}\end{align*} If we then want to know what is the minimal $N$ such that $q = q(N) \gt p = \frac{1640}{2187},$ we need get a nonlinear, monotonically increasing equation that we can just as easily guess and check for a solution.

In particular, if we solve the approximation where we ignore that pesky exponential function and equally ignore the integrality of $N$, we get $$\tilde{q}(x) = \frac{3 (4x-7)}{16(x-1)}.$$ If we were to solve $$\tilde{q}(x) = \frac{3(4x-7)}{16(x-1)} = p$$ for a non-integer value of $x$, we get $$x^* = \frac{21-16p}{12-16p} = \frac{19687}{4} = 4921.75.$$ This seems like a very good place to start hunting and pecking with $N^* = \lceil x^* \rceil = 4922.$ Let's first check $$\tilde{q}(N^*) = \frac{3 \cdot (4 \cdot 4922 -7)}{16 \cdot 4921} = \frac{59043}{78736} \approx 0.7498856939.....,$$ which is roughly $6 \times 10^{-9}$ larger than $p.$ Since $9 \cdot (-3)^{-N^*} \ll 10^{-9},$ we can confirm that $N^* = 4922$ is the smallest integer value such that if you fell asleep and randomly woke up to some wizardling ahead of you in line is being sorted to Graphindor then your probability of also getting sorted to Graphindor is $q(N) \gt p = \frac{1640}{2187}.$

Is that what squaring the circle means?

A pseudo-square has the following properties:

1. It is a simple, closed curve.
2. It has four sides, all the same length.
3. Each side is either a straight line segment or the arc of a circle.
4. The four sides are joined at four corners, with each corner having an internal angle of 90 degrees or 270 degrees.

The pseudo-square pictured above has two straight sides, which run radially between arcs of two concentric circles. Assuming this is a unit pseudo-square (i.e., each side has length 1), what is its area?

As you can see, I've pre-doctored the image from the prompt with some parameters. Let $r$ be the radius of the smaller circle and let $\theta$ be the angle inscribed between the two straight line edges. From here, if the shape is a unit pseudo-square then the larger radius is $1+r,$ so we can find the formula for the area of the shape in terms of $r$ and $\theta,$ as $$A(r,\theta) = \pi r^2 + \frac{1}{2} \theta \left((1+r)^2 - r^2\right) = \pi r^2 + \frac{1}{2} \theta (1 + 2r).$$ That is all well and good, but we now have to find the particular value of $r$ and theta that allow for this shape to be unit pseudo-square. In particular, the length of the arc on the larger of the two concentric circles is $$\ell = \theta (1 + r).$$ The length of the other arc is $$\tilde{\ell} = (2\pi - \theta) r.$$ Since the shape is a unit pseudo-square we have the nonlinear system of equations \begin{align*} \theta ( 1 + r) &= 1 \\ (2\pi - \theta) r &= 1.\end{align*} By setting $\theta = (1 + r)^{-1}$ and plugging into the second equation we get $$\left(2\pi - \frac{1}{1+r}\right) r = 1,$$ which is equivalent to the quadratic equation $$2\pi r^2 + 2(\pi - 1) r - 1 = 0.$$ Thus, if the shape is a unit pseudo-square then the smaller radius $r$ is equal to $$r^* = \frac{1-\pi + \sqrt{1+\pi^2}}{2\pi}.$$

By plugging $\theta = (1+r)^{-1}$ into the area formula we get $A(r) = \pi r^2 + \frac{1+2r}{2(1+r)},$ so plugging in $r^*$ from above we get the area of the unit psuedo-square with two straight lines is \begin{align*}A^* = A(r^*) &= \pi \left( \frac{1-\pi + \sqrt{1+\pi^2}}{2\pi} \right)^2 + \frac{ 1 + 2 \frac{1-\pi + \sqrt{1+\pi^2}}{2\pi} }{2 \left( 1 + \frac{1-\pi + \sqrt{1+\pi^2}}{2\pi} \right)}\\ &= \frac{1+\sqrt{1+\pi^2}}{2\pi} \approx 0.683874197466......\end{align*}

Can you find a unit pseudo-square that has three curved sides and just one straight side? What is the area of your new unit pseudo-square?

Without loss of generality, we can assume that the one flat side is positioned along the bottom. We would need to have two curved sides connected to this flat bottom that curve towards each other, with there being a third circle that is tangent to the circular arcs. The resulting shape is sort of like the shape of some pawns in chess. See the figure below. We again insert some parameters. Let the flat side be the line segment from $(-\frac{1}{2},0)$ to $(\frac{1}{2},0).$ Let the two symmetric sides attached to flat bottom be arcs of the circles centered at $(-a,0)$ and $(a,0),$ respectively, each with radius $a+\frac{1}{2},$ with subtended angle $\theta$. Let the final arc be from the circle centered $(0,b)$ with radius $r$ and subtended angle $2(\pi - \theta).$

Let's first try to figure out how to quantify the area of the pseudo-square with three curved sides. The region denoted by $B$ has area given by $$A_B = \frac{1}{2} (\pi - \theta) r^2.$$ The region denoted by $C$ has area given by $$A_C = \frac{1}{2} r^2 \tan \theta.$$ Finally, the region denoted by $D$ has aread given by $$A_D = \frac{1}{2} \theta \left(a + \frac{1}{2}\right)^2 - \frac{1}{2} a^2 \tan \theta.$$ The area of the entire pseudo-square with three curved sides is thus \begin{align*}A = A(a,r,\theta) &= 2 \left(A_B + A_C + A_D\right) \\ &= (\pi - \theta + \tan \theta) r^2 + \theta \left(a + \frac{1}{2}\right)^2 - a^2 \tan \theta.\end{align*}

With some creative trigonometry we can obtain $r = r(a,\theta),$ thus reducing the dimensions of the problem a smidge. The region denoted by $C$ is a triangle whose height is $r$ and base $\beta,$ which is some portion of line segment of total length $a + \frac{1}{2},$ since that line segment is a radius of the circle centered at $(-a,0)$ of radius $a+\frac{1}{2}.$ Since the line segment, negative $x$-axis and positive $y$-axis form a triangle, we can compute $\beta$ as $$\beta = a + \frac{1}{2} - a\sec \theta.$$ Since $r = \beta \cot \theta,$ we have $$r = r(a,\theta) = \left( \left(a+\frac{1}{2}\right) - a \sec \theta \right) \cot \theta = \frac{ (a + \frac{1}{2}) \cos \theta - a }{ \sin \theta }.$$ Therefore, we have $$A = A(a,\theta) = (\pi - \theta + \tan \theta) \left( \frac{ \left(a + \frac{1}{2}\right) \cos \theta - a }{\sin \theta} \right)^2 + \theta \left(a + \frac{1}{2} \right)^2 - a^2 \tan \theta.$$

OK, now that that significantly more intricate foundational work is out of the way, we need to ensure that each of these curves sides is unit length. Since the more vertical arcs are symmetric, we only need to quantify $\ell=(a+\frac{1}{2}) \theta$ and $\tilde{\ell} = 2(\pi - \theta) r,$ and set them both equal to $1$ to obtain the nonlinear system of equations \begin{align*} (a + \frac{1}{2}) \theta &= 1 \\ 2(\pi - \theta) r = (\pi - \theta) \frac{(a+1/2)\cos \theta - a}{\sin \theta} &= 1\end{align*} Solving for $a$ in the second equation and then plugging back into the first we get $$\frac{ \sin \theta - \pi + \theta }{ 2(\pi - \theta) ( \cos \theta - 1) } \theta = 1.$$ This non-linear equation is solved with $$\theta^* \approx 0.74960359...$$ which leads to $$a^* = \frac{2-\theta^*}{2\theta^*} \approx 0.8340377...$$ which ultimately leads to the area of a unit pseudo-square with three curved sides of $$A^* = A(a^*, \theta^*) \approx 0.8317044...$$