This week the #puzzle is: Can You Gerrymander Like a Dummy? #probabilities #voting

Fiddler Nation has an upcoming election between its two major political parties: the Suns and the Moons. You are drawing a new electoral map, and you have been tasked with gerrymandering on behalf of your party, the Moons. Fortunately, you know exactly how every citizen will vote in the election.

Fiddler Nation has seven districts, each with an equal population. For simplicity, assume that the Moons win any district in which they receive at least half of the vote.

What is the smallest total fraction of Fiddler Nation’s population needed to vote Moon such that it’s possible for the Moons to win a majority of the seven districts?

And for extra credit:

In the next election cycle you’re back at it, gerrymandering on behalf of the Moon party once more. This time, you can draw a map that results in any set of probabilities P1, P2, P3, P4, P5, P6, and P7 that you’ll win each of the corresponding districts—subject to the constraint that the average of the probabilities is 1/2. In other words, P1 + P2 + P3 + P4 + P5 + P6 + P7 = 7/2.

The bosses of the Moon party want to go big. More specifically, they want to maximize the party’s chances of winning at least five of the seven districts. However, you are concerned that such a greedy attempt at gerrymandering may backfire, a phenomenon that has become known as dummymandering. When you draw a map that maximizes your chances of winning at least five districts, the probability the Moon party blows it and winds up losing most of the districts is LA.

Your personal preference is to simply maximize the party’s chances of winning at least four of the seven districts—a simple majority of them. When you draw a map that maximizes these chances, the probability the Moon party loses most of the districts is LB.

How much greater than LB is LA?

Can You Gerrymander Like a Dummy?

Solution, possibly incorrect:

Facts:

• There are 7 districts.
• A win = winning 4 districts.
• Winning a district = getting 50% of the votes or more.
• All 7 districts have the same size, say V voters.
• WLOG, win district 1-4.
• In order to win district 1, V/2 Moon votes are needed. Minimize the number of votes needed by having exactly V/2. (So V is even? Hm…)
• The same goes for districts 2-4.
• In order to minimize the number of votes needed for a total win, just have 0 Moon votes in district 5-7.
• In all this is 4 * V/2 Moon votes.
• There are 7 * V votes in all.

Examining the proportion p_M of Moon voters.

Result: 2/7 = 0.285714….

And for extra credit:

Program

One of the basic ideas of gerrymandering is, that you don’t waste votes in the districts, you don’t plan to win. Let’s use this idea.

Chance of actually losing, when going for a 5 district win is L_A, Same for a 4 district win, L_B.

Assumption for strategy: In order to maximize the chances of a 5 district win, give each of districts 1-5 the maximum probability p_A, and district 6-7 0.

This will be compared to going for a 4 district win.

Computing.

Result: 0.4181.

Just for the fun of it, I also wrote a program to simulate these 2 scenarios. Result confirmed.

---

And then I came back. Originally I had an idea, that spreading the voters over more districts might be better. I try it out in my program. And what do you know! This works for the 5 district case.

What happens when the voters are spread out?

Let’s say the voters are evenly distributed.

What are the chances 5 of the districts will be won?

And 6?

And all 7?

And all together?

This is a lower chance of losing, much better! (ETA: And then I forgot to count “a 4 districts win isn’t the intended win, but it’s still a win”.)

And my program says:

Go for 5 wins, 5 districts with p = 0.700, 2 districts with p = 0
Wins: 0.170090
Go for 5 wins, 6 districts with p = 0.583, 1 districts with p = 0
Wins: 0.208540
Go for 5 wins, 7 districts with p = 0.500, 0 districts with p = 0
Wins: 0.227930

Go for 4 wins, 4 districts with p = 0.875, 3 districts with p = 0
Wins: 0.584810
Go for 4 wins, 5 districts with p = 0.700, 2 districts with p = 0
Wins: 0.529170
Go for 4 wins, 6 districts with p = 0.583, 1 districts with p = 0
Wins: 0.510270
Go for 4 wins, 7 districts with p = 0.500, 0 districts with p = 0
Wins: 0.499410

Difference: 0.356880

So, for 5 districts the best strategy is to spread the voters evenly. (And it’s still not a very good strategy.) For 4 districts concentrate the votes in 4 districts. (Gives 58% chance of winning instead of the fair and just 50%.) And the difference is: 0.3596.

This week the #puzzle is: How Fast Can You Surf? #trigonometry #maximum #differentiation

A surfer is riding along a crest of a sinusoidal wave given by the function y = 2·sin(x−t), where y is the height above sea level in meters, x is horizontal position in meters, and t is time in seconds. Based on this, the surfer knows that her speed—and the speed of the wave—is 1 meter per second.

But hold on—it appears that another wave is approaching from the opposite direction. This wave has half the amplitude but twice the speed: y = sin(x+2t).

The two waves, the forward and the backward, ”interfere,” meaning they combine. In other words, the height of the water at position x and time t is given by y = 2·sin(x−t) + sin(x+2t). Nevertheless, the surfer stays atop a crest of the combined wave.

As the wave moves along, what are the surfer’s slowest and fastest [horizontal] speeds, in meters per second?

And for extra credit:

The surfer rides the crest of the combined wave, y = 2·sin(x−t) + sin(x+2t), for a very long time.

What is her average [horizontal] speed, in meters per second?

How Fast Can You Surf?

Solution, possibly incorrect:

Desmos Program

Method 1: Code it up in Desmos, just to see what’s happening. Huh! The combined wave (purple) has a funny start and stop pattern, where horizontal movement seems to stop entirely, while the wave grows in size.

Method 2: Math. Everybody seems to agree, when $v$ = speed of wave (or in out case, speed of crest), $\lambda$ = wave length and $f$ = wave frequency:

This can’t quite fit our situation, because the speed obviously varies, while the wave length and the wave frequency stay constant.

Let’s just look at a single moment.

When does this function have a max or min?

And this max/min is:

Yeah, actually a max. So at time 0, the crest (or one crest) is at $(\frac{\pi}{2},3)$.

Let’s try this with a more general time instead.

Chain rule:

Looking for a max/min:

Identities:

I try out this function in Desmos. It actually follows the minimum, but the math is the same.

So far so good! A visual inspection shows these values for the speed of the crest.

Doing a test where I look at how much the crest moves over a short amount of time, $\epsilon$, 0 is confirmed. And the max value seems to be 4!

Method 3: Write a program. This simply supplies Desmos with data to show the path of the crest and a curve for the speed of the crest. As it uses the same math, it spits out the same results.

Result: $0 \le v \le 4$.

And for extra credit:

This is suspiciously easy.

Average speed = total distance / total time. From reading Desmos again, measuring between 2 maxima:

Result: 1.

This week the #puzzle is: Can You Swim With the Fishes? #probabilities #normaldistribution

Can You Swim With the Fishes?

Solution, possibly incorrect:

Program

Method 1:

Result 1: 1.666667

Method 2: Every time fish 1 goes by at speed 1, fish 2 goes by at speed 2 twice. The average of 1, 2 and 2 is 5/3 or 1.67.

Result: 1.67.

And for extra credit:

Method 1:

Result 1................:    1.2533140
Result 1................:    1.2533792
Result 1................:    1.2532510
Result 1................:    1.2533647
Result 1................:    1.2532415
======================================
Number of loops.... 5 * :    100980391
Target deviation........:    0.0001000
Deviation so far........:
................result 1:    0.0000632
======================================
Final result 1..........:    1.2533101
                         +/- 0.0000632

Result: 1.25.

Politiken benytter paywall på mange af sine artikler.

Jeg burde åbenbart ikke holde af at synge ’I Danmark er jeg født’, fordi det er jeg ikke. Men jeg synes sangen er skøn alligevel

(Sjovt nok tænker jeg primært på forvanskning af historien i 1984.)

Han slog igennem med blod dryppende ned ad sixpacken

(Ikke et ord om Murderbot. Suk. Men Wicker må godt komme med. Dobbelt suk.)

Tim Curry er død

Tommy Lee Jones har fødselsdag:

Og i øvrigt – inklusive engelske undertekster:

Bo er blind …

Star Trek!

This week the #puzzle is: Can You Frame the Triangle? #trigonometry #triangle #circumscribe #montecarlo #coding #minimum #average

Congratulations—you just won your school’s tabletop football championship! You flicked your lucky paper football to victory countless times, and would now like to frame it for posterity.

The football is an equilateral triangle with a side length of 1 inch. What is the side length of the smallest square frame that will contain the football?

And for extra credit:

To add a little excitement to the framing process, your new plan is to spin the triangular football by a random angle, and then frame it with a square that is not rotated. That is, the square consists of two perfectly horizontal sides and two perfectly vertical sides.

On average, what can you expect the side length of the smallest resulting square frame to be?

Can You Frame the Triangle?

Solution, possibly incorrect:

Desmos Program

Checking… Yeah, this is a triangle, where all 3 sides have the same length.

My sense is, that this square will have side length 1 inch. But let me just check. Oh, that’s not right. Messing around with Desmos gives me something like 0.966. This is achieved, when 1 side of the triangle has exactly a 45 deg angle on the horizontal and vertical axes. This was actually fun. To go from an angle to a triangle, to a circumscribed rectangle, to a circumscribed square. Visit Desmos to see an animation!

So that’s method 1. Method 2 is monte carlo (where it’s kind of overkill to have a variance check, I’m looking for a minimum). Result confirmed.

0.966 inches.

And for extra credit:

The easiest is to let the monte carlo register a lot of squares and take the average. (It was already registering a lot of them and taking the minimum.)

======================================
Number of loops.... 5 * :        32010
Target deviation........:    0.0001000
Deviation so far........:
................result 1:    0.0000029
................result 2:    0.0000886
======================================
Final result 1..........:    0.9659279
                         +/- 0.0000029
Final result 2..........:    0.9886172
                         +/- 0.0000886

0.989 inches. This makes sense, as being between the minimum and 1.

This week the #puzzle is: Can You Swap the Cups? #average #ExpectedValue #montecarlo #coding

My friend has three cups, labeled “A,” “B,” and “C” in a row. She randomly picks two different cups and swaps their positions. Then she does this again and again, picking a random pair each time, until all three cups are back in their original order. For example, here is one such sequence of swaps:

– A, B, C (original) – C, B, A (first and third were swapped) – B, C, A (first and second were swapped) – B, A, C (second and third were swapped) – A, B, C (first and second were swapped)

In this example, the cups returned to their original order after four swaps.

On average, how many swaps would you expect until the cups return to their original order?

And for extra credit:

In total, there are six possible orders of the cups. Without any swaps, one of those orders (A, B, C) has already been achieved.

On average, how many swaps would you expect until the remaining five orders (and thus, all six) are also achieved?

Can You Swap the Cups?

Intermission

So, last week was not good.

• I didn’t read the instructions correctly. I thought the 1st quiz had 4 questions, not 3.
• I did notice other answers using strategies involving “if the 1st correct answer was A, there’s a slightly better chance the 3rd answer is A”, but I didn’t understand why this could be until the official answer was out.

Solution, possibly incorrect:

Program

Method 1: Ye olde monte carlo. This time I play around with variance and deviation calculations, so that the program will keep going, until I get the number of decimals, I want.

======================================
Number of loops.... 3 * :    102970102
Target deviation........:    0.0010000
Deviations so far, res 1:    0.0003204
======================================
Final result 1..........:    5.9999515 +/-  0.0003204

Result: 6.00. Maybe simply 6?

Method 2: Look closely at what’s going on. This is a diagram of all states and all changes between states.

And that’s a little… Wait, that can actually be shown in a better way.

“abc” is the beginning state. Each line represents 1 swap. How many steps does it take to get from a state back to the beginning state?

If we’re already in the beginning state, 0 steps.

Here I’ve called “a state on the left that isn’t the beginning state” “left”. 1/3 of the time, take 1 step and get back to the beginning state. 2/3 of the time, take 1 step and reach a left state. And actually, let’s call all the states on the right “right”.

In a left state, I have to take 1 step to get to a right state. Let’s solve this.

There’s always a 1st step, away from the beginning state. And then it will take 5 further steps/swaps to get back. That’s 6. Result confirmed.

And for extra credit:

Same program, I just use the data slightly differently.

======================================
Number of loops.........:    102970102
Target deviation........:    0.0010000
Deviations so far, res 2:    0.0005949
======================================
Final result 2..........:   12.2144230 +/-  0.0005949

Result: 12.21 – 12.22.

Method 2. Let me see. First some new notation.

How many steps left on average, when there’s x already visited states on “my” side (WLOG the left side) of the diagram and y on the other?

$s_{33}=0$

If both sides are already full, there are 0 steps left.

$s_{23}=6$

This is the same situation as the fiddler. Being in one of the lower states on the left side and aiming for (WLOG) the top state of the left.

$$s_{32}=5$$

This was the intermediate situation above, taking 1 step away from the coveted state.

Let’s just summarize.

	$s_{x3}$	$s_{x2}$	$s_{x1}$	$s_{x0}$
$s_{3y}$	0	5		n/a
$s_{2y}$	6			n/a
$s_{1y}$
$s_{0y}$	n/a	n/a	n/a	n/a

Some of these situations are impossible. Typically, I can’t have visited 2-3 states on one side of the diagram and 0 on the other. Also, x can’t be 0, because I am in a state, that is therefore visited.

Right!

$s_{31}=\frac{1}{3}\cdot (1+s_{13})+\frac{2}{3}\cdot (1+s_{23})$

$s_{22}=\frac{1}{3}\cdot(1+s_{32})+\frac{2}{3}\cdot(1+s_{22})$

$s_{13}=1+s_{31}$

Let’s solve this.

$s_{23}=6,s_{31}=\frac{1}{3}\cdot (1+s_{13})+\frac{2}{3}\cdot (1+s_{23}),s_{13}=1+s_{31}$

$s_{31}=\frac{1}{3}\cdot (1+1+s_{31})+\frac{2}{3}\cdot (1+6)$

$s_{31}=\frac{2}{3}+\frac{1}{3}\cdot s_{31}+\frac{14}{3}$

$\frac{2}{3}\cdot s_{31}=\frac{16}{3}$

$s_{31}=8$

And:

$s_{31}=8,s_{13}=1+s_{31}$

$s_{13}=1+8$

$s_{13}=9$

And:

$s_{22}=\frac{1}{3}\cdot(1+s_{32})+\frac{2}{3}\cdot(1+s_{22}),s_{32}=5$

$s_{22}=\frac{1}{3}\cdot(1+5)+\frac{2}{3}\cdot(1+s_{22})$

$s_{22}=2+\frac{2}{3}+\frac{2}{3}\cdot s_{22}$

$\frac{1}{3}\cdot s_{22}=\frac{8}{3}$

$s_{22}=8$

Let’s summarize again.

	$s_{x3}$	$s_{x2}$	$s_{x1}$	$s_{x0}$
$s_{3y}$	0	5	8	n/a
$s_{2y}$	6	8		n/a
$s_{1y}$	9
$s_{0y}$	n/a	n/a	n/a	n/a

$s_{21}=\frac{1}{3}\cdot (1+s_{12})+\frac{2}{3}\cdot(1+s_{22})$

$s_{12}=\frac{1}{3}\cdot (1+s_{31})+\frac{2}{3}\cdot(1+s_{21})$

Solving.

And:

And summarizing:

	$s_{x3}$	$s_{x2}$	$s_{x1}$	$s_{x0}$
$s_{3y}$	0	5	8	n/a
$s_{2y}$	6	8	$9\frac{5}{7}$	n/a
$s_{1y}$	9	$10\frac{1}{7}$
$s_{0y}$	n/a	n/a	n/a	n/a

Almost there!

$s_{11}=\frac{1}{3}\cdot (1+s_{11})+\frac{2}{3}\cdot(1+s_{21})$

Solving:

	$s_{x3}$	$s_{x2}$	$s_{x1}$	$s_{x0}$
$s_{3y}$	0	5	8	n/a
$s_{2y}$	6	8	$9\frac{5}{7}$	n/a
$s_{1y}$	9	$10\frac{1}{7}$	$11\frac{3}{14}$
$s_{0y}$	n/a	n/a	n/a	n/a

Whew! Only one more.

$s_{10}=1+s_{11}$

Solving:

This was the number, we were looking for. And it confirms the previous result.

Just to finish the table:

	$s_{x3}$	$s_{x2}$	$s_{x1}$	$s_{x0}$
$s_{3y}$	0	5	8	n/a
$s_{2y}$	6	8	$9\frac{5}{7}$	n/a
$s_{1y}$	9	$10\frac{1}{7}$	$11\frac{3}{14}$	$12\frac{3}{14}$
$s_{0y}$	n/a	n/a	n/a	n/a

Forleden var vi på Ragnarock, et rockmuseum i Roskilde.

Rock og rock. En af udstillingerne gik tilbage til 1920’erne.

Men i hvert fald. Musik! Gæt et årstal.

Et fascinerende indslag var, at musikfans er fans på en for os genkendelig måde. Fans af Metallica laver et fanzine. Birthe skriver fanfic om Beatles. Emil laver Jackson-cosplay.

Selvfølgelig er der en souvenirshop. Helt bemærkelsesværdigt, så er der ikke en lille havfrue nogen steder. Til gengæld er der en hel sektion om Beatles. Hvor Dylan lige får sneget sig ind.

According to Ada Palmer, yes. But slowly. And sometimes good things have bad consequences for a while.

I’d like to quote chapters 66 and 67 from Inventing the Renaissance in their entirety, but that’s probably not within fair use etc. Let me try to pick some good quotes.

---

So when I hear, Has social progress failed? my zoomed-in self, my scared self living in these terrifying years, does feel afraid, but my zoomed-out self, my historian self, answers easily. Progress has done wonders! But it has also done what all change does in a dynamic system: had complicated consequences. Petrarch, Bacon, Marie Curie, all added something to the fish tank with many results. The same celebration of classical education which popularized the Platonic idea that sometimes humble parents produced kids with souls of gold both resulted in the Enlightenment ideal of universal equality, and fed the no columns, no culture side of European supremacy which enabled the atrocities of colonialism.

But if the consequences of our actions are completely unpredictable, would it be better to say that change over time is real, but that anything like human-controlled progress is an illusion?

No.

Why?

…

Because my mother’s cataract surgery—after 2,000 years of practice—saved her sight.

Because some of our civil rights have come by blood and war, but more recently more and more have come by negotiation and agreement.

Because my students who burn down imaginary Europe every year understand better afterward how to avoid collateral damage, and avoid burning things down in their own world. A lot of them even volunteer to come help with the next simulation, advising their peers on how to see the patterns, giving advice much more prudent than when they braved 1492’s Scylla and Charybdis the first time.

---

We have achieved some of what Machiavelli and Bacon wanted: a new method to make the next generation’s experience on this Earth a little better, and to help us avoid those moments when we lose so much so quickly.

---

Plague wasn’t the Horseman of the Apocalypse Petrarch hoped to defeat, but the man whose Remedies Against Fortune offered consolation, even for those who lose their homelands to war, would weep such happy tears to hear the plague for which he had no consolation is defeatable.

---

There was a baby in the bathwater of Whig history. We can find metrics for comparing times and places which don’t privilege particular ideologies. Metrics like infant mortality. Metrics like malnutrition. Metrics like the frequency of massacres. As I said, one of my favorite questions to ask an era is: What portion of the population can be murdered by a different portion and have the murderer suffer no meaningful consequences? The answer, in my homeland of America today, is not zero percent, but it’s not ninety percent, the way it was in the patronage-dominated Renaissance, when Benvenuto Cellini could stab you in the back and, if he’s working for the pope, there won’t be justice. That is progress. That, and infant mortality, and the conquest of smallpox, and soon of polio, and broadening access to education, these are genuine improvements of the sort the Try Everything Age believed would come if they kept striving to reveal the secret motions of things.

---

Progress is not inevitable, but it is happening.
It is not safe, but it is beneficial.
It is not linear, but it is ongoing.
It is not simple, but it can be studied.
It is not controllable, but it is us. In fact, it is nothing but us.

---

I mentioned earlier a debate, in the History Lab, between my friends Chris Celenza and Rocco Rubini, about whether Machiavelli is a modern man. Chris had just published his biography of Machiavelli, and he, much like me, described Machiavelli as the product of a desperate time, shaped by the profoundly un-modern violences and chaos of his day. Rocco objected that there are more murders in Chicago now than in Florence back then, so does the violence really make Old Nick un-modern? It is an important question, especially since so much of our own era’s daily violence tends to be erased or pushed off-camera, so to speak, by the social apparati which perpetuate it (racism, systemic inequality), in contrast with the grim glitter of Renaissance violence which TV dramas love to magnify. But there are also differences. It would have been easy for Chris to cite population, to say modern Chicago has thirty times the population of 1500s Florence, that our bloodshed is proportionately smaller. But Chris didn’t. Instead, pausing, smiling warmly, thinking through, Chris answered that we today believe the murder rate in Chicago is not acceptable, or natural, or immovable, that we believe we can fix it if we study and try, that we aspire to change it for the better. That we feel a humane duty to do better. Machiavelli didn’t think that was possible.

---

The year 2020 was the first time this planet faced a pandemic knowing what it is, and how to act. Forget 1348, even in 1918 we didn’t understand how influenza moved, and hand washing was still controversial. We didn’t take perfect action with COVID, and we should be condemning the failures, but I’m not just a historian of the plague of 1348, I’m a historian of the plagues of 1435, 1485, 1494, 1503, 1596, 1630, and 1656, those many generations who lived without hope that it could ever stop. We do have hope. Within two years we learned how to treat and vaccinate, achievements any other age in history would hail as miracles.

---

We also know about mental health now. We’re studying the mental health costs of pandemic, poverty, racial injustice, fear. The concept of mental health alone empowers us, as do related concepts: social safety net, welfare, social services, tools for thinking about how society can put structures in place to relieve suffering. There were social services in 1348, hospitals, city grain supplies, orphanages, but Florence’s public health response to the Black Death was to double the fine for selling meat from male and female animals in the same butcher’s booth, thinking that the mix of generative juices birthed disease. In Florence’s orphanage there hangs a painting by Michelangelo’s teacher Ghirlandaio, an Adoration of the Magi with the Virgin and Child and city’s patron John the Baptist, all surrounded by the Innocenti, dead babies and toddlers from the Massacre of the Innocents, tiny kids still in their swaddling clothes standing among the city’s patrons in poses of power—they’re there as role models to teach the orphans how to imagine their dead friends still watching over them, and how to imagine their place in the world after their deaths, since most of these kids would not survive to leave the orphanage. Our child services are far from perfect, but we aim higher than preparing kids for death.

---

Some say of science that Bacon built the scaffolding and Newton raised the cathedral, but it wasn’t as quick as that. Petrarch’s generation started us gathering materials, precious libraries, making worldly and social questions central to human inquiry at last. A century later, Machiavelli laid out the foundations, proposing a new way of using those libraries, and challenged the myth that there was nothing to be learned from that long, ever-changing Middle Age that followed Rome. Century three, Thomas Hobbes was starting on the ground floor, mixing Machiavelli with Bacon’s empiricism and social analysis. A century later Enlightenment voices—Voltaire, Beccaria, Montesquieu, dear Diderot—raised the upper sections of the wall, convincing people and nations to try changing their world on a large scale. It took another 100 years to realize big social experiments often had collateral damage, and readjust those upper sections to be safer and better, so they didn’t make the rain runoff flood other parts of town. Hope of a golden age to live in is still underway: it’s our job now to do the hardest part and build the dome. It took a long time to raise this cathedral, to get from Italia Mia via The Prince and the New Organon to RNA vaccines, but it’s the same project. Will we do it perfectly? No. Will we finish in our lifetimes and enjoy our golden age? No. We will add our layer to the structure, but there’s more to come, the trim, the lantern and the ball on top. Florence’s cathedral, begun 1296, is still not done today. But it is a wonder, as is every hand that helped to build it. We have tools in our hands now—tools that Machiavelli, Petrarch, and Alessandra Scala would have given anything to have. We can aim for better than the Renaissance, which is what our Renaissance friends did too.

Let’s try.

This week the #puzzle is: Can You Cheat on the Quiz? #probabilities #statistics #montecarlo #coding

Can You Cheat on the Quiz?

Solution, possibly incorrect:

Program

Ehm. Yes. I think part of the solution is to look at a diagram of how the states change.

At any given point in the quiz, g_i is a guess and a_i is the correct answer to the question q_i. I can choose the dashed strategy, which is to guess the same again, or I can guess something else.

At q₁, Charlie knows a₁ and therefore must guess g₁ = a₁, that is the best strategy. So now Charlie’s in the green bubble on the left, and he knows it. Therefore the best strategy for q₂ is not the dashed “choose the same again” strategy. With p = 1/3 he happens to guess the right answer and ends up in a new green bubble. With p = 2/3 he doesn’t and ends up in a red bubble. He doesn’t know which one happened, so for the next step of the strategy, his choices are “guess the same again” or “don’t guess the same again”.

Hang on. Let’s just simplify that diagram.

Nice symmetry, if you don’t use the dashed lines, regardless of whether Charlie guessed right or wrong, there’s p = 1/3 of guessing correctly the next time.

Charlie’s strategy should begin with “g₁ = a₁, g₂ ≠ g₁“. But then what?

There are 4 choices:

1. g₃ ≠ g₂, g₄ ≠ g₃
2. g₃ ≠ g₂, g₄ = g₃
3. g₃ = g₂, g₄ ≠ g₃
4. g₃ = g₂, g₄ = g₃

And there are these possible outcomes:

1. True True True True
2. True True True False
3. True True False True
4. True True False False
5. True False True True
6. True False True False
7. True False False True
8. True False False False

Any strategy leads to this for 1st 2 questions:

1. True True, p = 1/3
2. True False, p = 2/3

Strategy 1 leads to this for the 1st 3 questions:

1. True True True, p = 1/3 * 1/3
2. True True False, p = 1/3 * 2/3
3. True False True, p = 2/3 * 2/3
4. True False False, p = 2/3 * 2/3

And:

1. True True True True, p = 1/3 * 1/3 * 1/3 = 1/27
2. True True True False, p = 1/3 * 1/3 * 2/3 = 2/27
3. True True False True, p = 1/3 * 2/3 * 1/3 = 2/27
4. True True False False, p = 1/3 * 2/3 * 2/3 = 4/27
5. True False True True, p = 2/3 * 1/3 * 1/3 = 2/27
6. True False True False, p = 2/3 * 1/3 * 2/3 = 4/27
7. True False False True, p = 2/3 * 2/3 * 1/3 = 4/27
8. True False False False, p = 2/3 * 2/3 * 2/3 = 8/27

For this the expected value is:

Let’s try strategy 2, first different, then same. So the 3rd step looks the same.

1. True True True, p = 1/3 * 1/3
2. True True False, p = 1/3 * 2/3
3. True False True, p = 2/3 * 2/3
4. True False False, p = 2/3 * 2/3

And then we get something new for the 4th step:

1. True True True True, p = 1/3 * 1/3 * 0 = 0/27
2. True True True False, p = 1/3 * 1/3 * 1 = 3/27
3. True True False True, p = 1/3 * 2/3 * 1/3 = 2/27
4. True True False False, p = 1/3 * 2/3 * 2/3 = 4/27
5. True False True True, p = 2/3 * 1/3 * 0 = 0/27
6. True False True False, p = 2/3 * 1/3 * 1 = 6/27
7. True False False True, p = 2/3 * 2/3 * 1/3 = 4/27
8. True False False False, p = 2/3 * 2/3 * 2/3 = 8/27

For this the expected value is:

So far strategy 1 beats strategy 2.

Doing something similar leads to these expected values for strategy 3:

1. True True True True, p = 1/3 * 0 * 1/3 = 0/27
2. True True True False, p = 1/3 * 0 * 2/3 = 0/27
3. True True False True, p = 1/3 * 1 * 1/3 = 3/27
4. True True False False, p = 1/3 * 1 * 2/3 = 6/27
5. True False True True, p = 2/3 * 1/3 * 1/3 = 2/27
6. True False True False, p = 2/3 * 1/3 * 2/3 = 4/27
7. True False False True, p = 2/3 * 2/3 * 1/3 = 4/27
8. True False False False, p = 2/3 * 2/3 * 2/3 = 8/27

For this the expected value is:

And strategy 4:

1. True True True True, p = 1/3 * 0 * 0 = 0/27
2. True True True False, p = 1/3 * 0 * 1 = 0/27
3. True True False True, p = 1/3 * 1 * 1/3 = 3/27
4. True True False False, p = 1/3 * 1 * 2/3 = 6/27
5. True False True True, p = 2/3 * 1/3 * 0 = 0/27
6. True False True False, p = 2/3 * 1/3 * 1 = 6/27
7. True False False True, p = 2/3 * 2/3 * 1/3 = 4/27
8. True False False False, p = 2/3 * 2/3 * 2/3 = 8/27

Turns out the best strategy is to never choose the same twice in a row. And with this strategy, Charlie can expect 2 correct answers. If I have done all my calculations correctly… And even modelled the problem correctly.

Okay, let me try method 2. Monte Carlo. And this one gives… Something like 1.84. I have no idea why. And I do trust my Monte Carlo more. So that will be my official answer.

Result: 1.84.

And for extra credit:

This is strictly a guess on my part. But I think the best strategy is to cheat on question 4 and then construct the rest of the guesses from that. This gives 1 correct answer in the middle and an expectation of 0.84 correct answer above and below. 2.68 correct answers in all.