This week the question is: Can You Solve the Tricky Mathematical Treat?

It’s Halloween time! While trick-or-treating, you encounter a mysterious house in your neighborhood.

You ring the doorbell, and someone dressed as a mathematician answers. (What does a “mathematician” costume look like? Look in the mirror!) They present you with a giant bag from which to pick candy, and inform you that the bag contains exactly three peanut butter cups (your favorite!) [no, it isn’t!], while the rest are individual kernels of candy corn (not your favorite!).

You have absolutely no idea how much candy corn is in the bag—any whole number of kernels (including zero) seems equally possible in this monstrous bag.

You reach in and pull out a candy at random (that is, each piece of candy is equally likely to be picked, whether it’s a peanut butter cup or a kernel of candy corn). You remove your hand from the bag to find that you’ve picked a peanut butter cup. Huzzah! [Yuck!]

You reach in again and pull a second candy at random. It’s another peanut butter cup! You reach in one last time and pull a third candy at random. It’s the third peanut butter cup! [Triple yuck!]

At this point, whatever is left in the bag is just candy corn. How many candy corn kernels do you expect to be in the bag?

Highlight to reveal (possibly incorrect) solution:

Program. Wikipedia 1 and 2.

First let’s assume there is a finite number of kernels of candy corn (kcc), at most m. What is the probability I got 3 peanut butter cups (pbc) (and yuck)?

• If I have n kcc, that’s 3+n pieces of candy in all. How many different ways can they be sorted? This is a stars and bars question, and the answer is C(3+n n) = (3+n)!/n!3!
• Of these (3+n)!/n!3! permutations, only 1 is any good, because it has the 3 pbc at the top.
• P(3 pbc | n kcc) = 1 / (3+n)!/n!3! = n!3!/(3+n)!
• n has m+1 different possible values.
• P(n kcc) = 1 / (m+1)
• P(3 pbc) = sum of all P(3 pbc | n kcc) * P(n kcc)
• Use Bayes’ theorem.
• P(n kcc | 3 pbc) = P(3 pbc | n kcc) * P(n kcc) / P(3 pbc)
• The expected number (E) of kcc is the sum of all n * P(n kcc | 3 pbc).
• And I made a program to calculate that given m.

Varying m hopefully tells me something about the behavior of E. And yes, it seems to. When m grows, apparently E gets closer to 1. So my guess is the expected number of kernels of candy corn is 1.

Also, I’m still fascinated by the soup bowl last week. Here’s a video!

This week the question is: Can You Make the Biggest Bread Bowl?

I have a large, hemispherical piece of bread with a radius of 1 foot. I make a bread bowl by boring out a cylindrical hole with radius r, centered at the top of the hemisphere and extending all the way to the flat bottom crust.

What should the radius of my borehole be to maximize the volume of soup my bread bowl can hold?

And for extra credit:

Instead of a hemisphere, now suppose my bread is a sphere with a radius of 1 foot.

Again, I make a bowl by boring out a cylindrical shape with radius r, extending all the way to (but not through) the curved bottom crust of the bread. The central axis of the hole must pass through the center of the sphere.

What should the radius of my borehole be to maximize the volume of soup my bread bowl can hold?

Highlight to reveal (possibly incorrect) solution:

Calculations. WolframAlpha assistance.

In my calculations I create a formula for the volume of the cylinder, depending on r. I then find the maximum volume. At r = √(2/3) foot, approximately 0.817, the volume is 2π/3√3, approximately 1.209 foot³.

And for extra credit:

ETA: I interpret “a cylindrical shape” as being the same as a cylinder. Flat top, flat bottom. Others interpret it differently.

Everything is multiplied by 2. The radius ends up being the same.

Anmeldelse af October Daye-serien, af Seanan McGuire. Romanserie. 2009-23, foreløbig. Hugo-finalist, bedste serie.

Skitse: October Daye er ret langt nede. Hendes mor er fae, det var hendes far ikke, så hun kan magi, men ikke ret meget, altså ser fae i høj grad ned på hende, og overfor mennesker kan hun ikke fortælle alt om sin baggrund. Hun var så alligevel ved at få gang i noget, mand, barn, hus, job som privatdetektiv. Men så blev hun forvandlet til en fisk. I 14 år. Bagefter opgiver hun (delvis nødtvunget) sit tidligere liv og prøver at klare sig som ekspedient. Langt fra alt det der magi. Men så …

Er det science fiction? Fantasy fra morgen til aften.

Temaer: Henover de 18 bøger, det er blevet til (der kommer vist 2 til) bliver October (Toby) mere og mere veltilpasset. Hun får det bedre. I starten er hun en ret god privatdetektiv, men hun kaster sig ud i den ene dødsensfarlige situation efter den anden, bl.a. fordi hun er lidt ligeglad. Det går delvist over, i takt med at hun får en ny familie.

I øvrigt en noget utraditionel familie.

Hun kæmper gang på gang mod racisme.

Regelmæssigt viser det sig, at person A i virkeligheden er person B, eller at person C, en ven, faktisk er en fjende, eller omvendt. Og eftersom nogle af de her personer er store kanoner (konger og dronninger), så bliver det hele også mere og mere vigtigt. Som at opdage, at naboen i virkeligheden er Thor, og at kun jeg kan redde hans liv.

Er det godt? Ja. Jeg er nogle gange ved at få pip af stilen, der bruger aaaaalt for mange ord efter min smag. Men jeg har hængt på, mere og mere, fordi det er spændende. Hvem har gjort hvad? Hvordan får Toby løst det den her gang? 👽👽👽

Note: Udover den her serie, der altså var Hugo-finalist, har jeg i den her omgang kun læst en af de andre serier systematisk, som beskrevet i Et vidunderligt lys osv. Glemte i den forbindelse, at Toby også var finalist.

Jeg har løbende lavet mini-anmeldelser af bøgerne. Det har taget mig et års tid at læse dem.

• 1 Rosemary and Rue (2009) Mini (Facebook)
• 2 A Local Habitation (2010) Mini
• 3 An Artificial Night (2010) Mini
• 4 Late Eclipses (2011) Mini
• 5 One Salt Sea (2011) Mini (Mastodon)
• 6 Ashes of Honor (2012) Mini
• 7 Chimes at Midnight (2013) Mini
• 8 The Winter Long (2014) Mini
• 9 A Red-Rose Chain (2015) Mini
• 10 Once Broken Faith (2016) Mini
• 11 The Brightest Fell (2017) Mini
• 12 Night and Silence (2018) Mini
• 13 The Unkindest Tide (2019) Mini
• 14 A Killing Frost (2020) Mini
• 15 When Sorrows Come (2021) Mini
• 16 Be the Serpent (2022) Ikke nogen mini!
• 17 Sleep No More (2023) Mini
• 18 The Innocent Sleep (2023) Mini

Anmeldelse af A City on Mars: Can We Settle Space, Should We Settle Space, and Have We Really Thought This Through?, af Kelly & Zach Weinersmith. Fagbog. 2023. Hugo-finalist, best related.

Skitse: Hvis man kigger på emnet “at bygge en koloni på Mars” og andre nærliggende emner, så er der en masse underemner. Der er noget med teknologi, ja, og noget med medicin. Men der er fx også noget med lovgivning. Lad os tage et grundigt kig på det vigtigste!

Er det science fiction? Nej, det er en fagbog.

Temaer: Formen er sjov. Snurrige formuleringer og deciderede vittigheder. Uærbødige tegninger. Anekdoter. Og det er en god ting. Vi vil jo godt have folk til at læse den her bog. Det var i hvert fald en rigtig god ting for mig. Det er længe siden, jeg kværnede den her type fagbog.

Det hjælper også på, at hovedbudskabet er, nej, vi kan ikke det her endnu, og vi bør ikke prøve. Worst case får Enol Muks sendt en million mennesker til Mars, der prompte dør.

Det er pudsigt, at min sf-radar hele tiden slår ud. Der er rigtig mange historier, der tager fat i nogle af de her emner, mere eller mindre korrekt.

Er det godt? I sidste ende ikke godt nok til sådan en som mig. Der er masser af tankevækkende stof og sjove indslag, men det var stadig svært at komme igennem det hele. 👽👽💀

Note: Det her er den eneste finalist i den her kategori, jeg rigtig har gidet se på i sin helhed. (Omend jeg nok også vil skæve til en af de andre, en samling af anmeldelser.) Den vandt i øvrigt.

Jeg læste bogen som en pdf, der til tider blev konverteret til noget sært.

This week the question is: Will You Top the Leaderboard?

You’re doing a 30-minute workout on your stationary bike. There’s a live leaderboard that tracks your progress, along with the progress of everyone else who is currently riding, measured in units of energy called kilojoules. (For reference, one kilojoule is 1000 Watt-seconds.) Once someone completes their ride, they are removed from the leaderboard.

Suppose many riders are doing the 30-minute workout right now, and that they all begin at random times, with many starting before you and many starting after. Further suppose that they are burning kilojoules at different constant rates (i.e., everyone is riding at constant power) that are uniformly distributed between 0 and 200 Watts.

Halfway through (i.e., 15 minutes into) your workout, you notice that you’re exactly halfway up the leaderboard. How far up the leaderboard can you expect to be as you’re finishing your workout?

As an added bonus problem (though not quite Extra Credit), what’s the highest up the leaderboard you could expect to be 15 minutes into your workout?

And for extra credit:

Again, suppose there are many riders starting their 30-minute workouts at random times, and that their powers are uniformly distributed between 0 and 200 Watts. Now, suppose you decide that you too will be pedaling with a random (but constant) power between 0 and 200 Watts.

If you look down at the leaderboard at a random time during this random workout, how far up the leaderboard can you expect to be, on average?

Highlight to reveal (possibly incorrect) solution:

Plot 1 and 2. Calculations. Program.

How high up you are on the leaderboard is determined by how many kJ you spent, which is in turn determined by how much time you spent and how hard you’re pedaling (power). Time: somewhere between 0 and 30 minutes. Power: somewhere between 0 and 200 W. Furthermore you look at the leaderboard at your personal time of 15 minutes.

Let’s simplify to Energy = Time * Power. E = T * P. Both T and P are evenly distributed. E can be somewhere between 0 kJ and 30 minutes * 200 W = 30 * 60 seconds * 200 W = 360000 Ws = 360 kWs = 360 kJ.

Trying to plot some sections of values for E, we see that a lot of space is taken up by the less than 100 kJ points. (The sections have borders corresponding to 90, 180 and 270 kJ.) While T and P are evenly distributed, E isn’t.

So, what we’re looking for here is some curve, f(x,y) = T * P = C, a constant, so that the area below the curve is half of the available rectangle. All those points below (or to the left of) this curve are the many, many people below you on the leaderboard. Because of the time or power they’ve spent, or maybe both, they’re behind you. The guy with 1 Watt is probably behind you the whole time. The gal beginning 1 minute ago is probably behind you too. They’re both in the lower left corner of the plot.

Some calculations reveal, that C = 67.2 kJ. A new plot sort of confirms this. A program also hits in this neighborhood. ETA: This uses the equation for the area C * (1 + log(360k/C) = E. In this case, C * (1 + log(360k/C) = 180 kJ, half the available area.

In other words, your energy use at 15 minutes is 67.2 kJ. This corresponds to 74.7 W. At the end of the 30 minutes, you will hit double this energy use, 134.4 kJ. Now we do the calculation the other way, T * P = 134.4 kJ. What is the area below this curve? ETA: 134.4k * (1 + log(360k/134.4k) = E = 266.8 kJ. Out of an area of 360 kJ in all, that’s 74.1%. So you can expect to rise above 74% of your competitors, just before you drop out of the leaderboard.

So, I’m not doing any more of this fiddler. I’m not in love with probabilities.

Bonus question:

The highest you can be on the leaderboard implies, that you are going full power, 200 W, ending up with 360 kJ. At 15 minutes you would be at 180 kJ. Again using the equation 180k * (1 + log(360k/180k) = E = 304.8 kJ. Out of the available area of 360 kJ, that means you are above 84.7% of the board.

And for extra credit:

We want the average value of E, because that’s the E you have at a random time. This requires calculating the “volume” of the figure in the coordinate system with axis T, P and E, and then dividing by the area, 360 kJ. A quick integral gives the volume as (1/2 * 360k)². This gives an average E of (1/2)² * 360k. = 90k. What is the area below E = 90 kJ? 90k * (1 + log(360k/90k) = 214.8 kJ. Out of an area of 360 kJ, that’s 59.7%. You will be above 59.7% of the competition.

And I am not too sure about those extra bits. ETA: My program says something else. Sigh.

This week the question is: How Many Dice Can You Roll the Same?

… To get started, you roll all 10 dice—whichever number comes up most frequently becomes your target number. In the event multiple numbers come up most frequently, you can choose your target number from among them. At this point, you put aside all the dice that came up with your target number. …

Now, consider a simplified version of the game in which you begin with three total dice (call it “THREEZI”).

On average, how many dice will you put aside after first rolling all three?

And for extra credit:

Let’s return to the original game of TENZI, which has 10 dice.

On average, how many dice will you put aside after first rolling all 10? (What if, instead of 10 dice, you have N dice?)

Highlight to reveal (possibly incorrect) solution:

Program.

Let’s call the 3 dice rolls a, b and c.

• There are 3 possible cases:
  • 3 dice are put aside. This means a = b = c. This again means a is something with p = 1, and b and c are the same, each with p = 1/6. So p(3) = (1/6)² = 1/36.
  • 2 dice are put aside. This means e.g. a = b != c. This again means a is something with p = 1, b is the same with p = 1/6, and c is different with p = 5/6. Furthermore, I could have chosen the different one in 3 different ways. So p(2) = (1/6) * (5/6) * 3 = 15/36.
  • 1 die is put aside. This means a != b, a != c and b != c. This again means a is something with p = 1, b is something else with p = 5/6, and c is something else again with p = 4/6. So p(1) = (5/6) * (4/6) = 20/36.
• A quick check: 1 + 15 + 20 = 36. The sum of our probabilities is 36/36 = 1. Good.
• The average we’re looking for is 3 * p(3) + 2 * p(2) + 1 * p(1) = 3 * 1/36 + 2 * 15/36 + 1 * 20/36 = (3 + 30 + 20)/36 = 53/36 = 1 17/36, about 1.472.
• A quick check: my program says the same.

And for extra credit:

Program 1 and 2.

I figure out a way to extend the previous program to run with more dice. Mainly by programming the dice as a number in base 6. With 10, the average number of dice set aside is 3.445… In general, with N dice, on average I set aside about √N dice.

This week the question is: Can You Win at “Rock, Paper, Scissors, Lizard, Spock?”

In a game of “Rock, Paper, Scissors,” each element you can throw ties itself, beats one of the other elements, and loses to the remaining element. In particular, Rock beats Scissors beats Paper beats Rock.

“Rock, Paper, Scissors, Lizard, Spock” (popularized via The Big Bang Theory) is similar, but has five elements you can throw instead of the typical three. Each element ties itself, beats another two, and loses to the remaining two. More specifically, Scissors beats Paper beats Rock beats Lizard beats Spock beats Scissors beats Lizard beats Paper beats Spock beats Rock beats Scissors.

Three players are playing “Rock, Paper, Scissors, Lizard, Spock.” At the same time, they all put out their hands, revealing one of the five elements. If they each chose their element randomly and independently, what is the probability that one player is immediately victorious, having defeated the other two?

And for extra credit:

The rules for “Rock, Paper, Scissors” can concisely be written in one of the following three ways:

• Rock beats Scissors beats Paper beats Rock
• Scissors beats Paper beats Rock beats Scissors
• Paper beats Rock beats Scissors beats Paper

Each description of the rules includes four mentions of elements and three “beats.”

Meanwhile, as previously mentioned, a similarly concise version of the rules for “Rock, Paper, Scissors, Lizard, Spock” (and adapted from the original site) is:

• Scissors beats Paper beats Rock beats Lizard beats Spock beats Scissors beats Lizard beats Paper beats Spock beats Rock beats Scissors

In this case, there are 11 mentions of elements and 10 “beats.” Including the one above, how many such ways are there to concisely describe the rules for “Rock, Paper, Scissors, Lizard, Spock

For my own sanity:

Highlight to reveal (possibly incorrect) solution:

Program solution.

Using the numbers shown above, 1 beats 3 and 5. In general, n beats n+2 and n+4, mod 5. This will be useful for the program.

Given 3 players (A, B and C), we might have one absolute winner, one absolute loser without an absolute winner, a circle (everybody wins once and loses once) and a threeway tie.

WLOG, simply assume A chose 1. The table above could be duplicated a bit, because B choosing either 2 or 4 is equivalent, and the same for B choosing either 3 or 5. When B chooses 1, there’s a 2/5 chance for an absolute winner. When B chooses 2 or 4, there’s a 2/5 chance for an absolute winner. When B chooses 3 or 5, there’s a 3/5 chance for an absolute winner. Added up, this is a (2+2*2+2*3)/25 = 12/25 = 48% chance for an absolute winner. The program confirms this.

And for extra credit:

Program assistance.

I go through all the ways to traverse the graph shown above, using each edge once. Each way corresponds to a new way to write out the long sentence with a beats b etc. My program finds 22 different ways to begin at 1. This gives 110 ways in all.

This week the question is: Can You Make a Toilet Paper Roll?

Suppose you have the parallelogram of cardboard shown below, which has side lengths of 2 units and 6 units, and angles of 30 degrees and 150 degrees:

By swirling two edges together, it’s possible to neatly (without any overlap) generate the lateral surface of a right cylinder—in other words, a toilet paper roll! (If you’re not convinced, try gently tearing a toilet paper roll along its diagonal seam and then unwrapping it into a flat shape. You get a parallelogram!)

Determine the volume of a cylinder you can make from this particular piece of cardboard.

And for extra credit:

Suppose you have a parallelogram with an area of 1 square unit. Let V represent the average volume of all cylinders whose lateral surface you can neatly make by swirling two edges of the parallelogram together.

What is the minimum possible value of V?

Highlight to reveal solution:

Figure 1.

Figure 1 shows, that if I construct my cylinder the traditional way, it will have a base with circumference 2. Therefore the radius of the base will be 2/2π = 1/π. Meanwhile the height h of the cylinder depends on x. The pink triangle demonstrates, that the length of the long edge, 2x, gives a height of x. 2x = 6 <=> x = 3. This means the volume of the cylinder is πr²h = π(1/π)²*3 = 3/π, about 0.955.

But there’s another way to construct a cylinder. The base could have circumference 6. (It would end up tilted in space, where the first one rests nicely on its base.) This base has radius 6/2π = 3/π. The green triangle shows, that the height h of this cylinder is y, and 2y = 2 <=> y = 1. The volume of this cylinder is πr²h = π(3/π)²*1 = 9/π, about 2.865.

The way this riddler is worded, both solutions should be valid.

This seems too easy! Like there are other solutions. But for now, these should be 2 valid solutions.

And for extra credit:

Calculation sheet 1 and 2. Wolfram Alpha result.

Let’s do the whole thing in more general terms. The parallelogram has sides a and b, a >= b, and angle α. b corresponds to height h₁, a corresponds to height h₂. In a small triangle, I have sides a*, b and h*. Further sin(α)/h* = sin(90°-α)/b = sin(90°)/a*. Rearranging gets h* = b * sin(α)/sin(90°-α) and a* = b * 1/sin(90°-α). Also, h₁/h* = a/a* <=> h₁ = h* * a/a* = b * sin(α)/sin(90°-α) * a * sin(90°-α)/b = sin(α) * a.

For symmetry reasons, h₂ = sin(α) * b.

(Sanity check: a = 6, b = 2, α = 30°. h₁ = sin(30°) * 6 = 0.5 * 6 = 3. h₂ = sin(30°) * 2 = 0.5 * 2 = 1. Same result as before.)

The first cylinder has circumference b, radius b/2π, height h₁ = sin(α) * a, volume πr²h = π * (b/2π)² * sin(α) * a = b² * a / 4π * sin(α). The other has circumference a, radius a/2π, height h₂ = sin(α) * b, volume πr²h = π * (a/2π)² * sin(α) * b = a² * b / 4π * sin(α).

Let’s take the average of these two: (b² * a / 4π * sin(α) + a² * b / 4π * sin(α))/2 = (b² * a + a² * b) * sin(α)/8π. And let’s try to minimise it. 8π is a constant and can be thrown away. Minimise (b² * a + a² * b) * sin(α).

The area of the parallelogram is 1. b * h₁ = a * h₂ = 1. From what we know about the heights, b * sin(α) * a = a * sin(α) * b = 1 <=> ab = 1/sin(α).

We’re minimising (b² * a + a² * b) * sin(α) = ab * (b + a) * sin(α) = 1/sin(α) * (b + a) * sin(α) = b + a. Using the expression for ab one more time, we’re minimising a + 1/(sin(α)*a). A quick trip to Wolfram Alpha later, and we find a minimum at a = 1, α = 90°, hence also b = 1. So min(a + b) = 2, and this occurs when the parallelogram is actually a square. ETA: Therefore the volume is 2/8π, approximately 0.0796.

All of this still based on: Every parallelogram can be turned into at most 2 cylinders.

And previously:

I didn’t get around to doing the fiddler last week. But it turns out I had the right idea. Here’s a sketch.

Denote the state of the tiles like this: xxxxxx. This means all 6 tiles have been flipped, and I won.

If the state is 1xxxxx, and I roll a 1, I win, otherwise I lose. So in this state p(win) = 1/6.

Similarly for x2xxxx, xx3xxx etc.

If the state is 12xxxx, and I roll a 1, move on to x2xxxx. Roll a 2 and move to 1xxxxx. Roll a 3 and win. Otherwise lose. Here p(win) = 1/6² + 1/6² + 1/6 = 8/36 = 2/9.

If the state is 123xxx, and I roll a 3, I have a choice. If I flip 1 and 2, I move to xx3xxx and p(win) = 1/6. If I flip 3, I move to 12xxxx and p(win) = 2/9. So the optimal strategy is to flip 3.

Build a big table with all the states and all the probabilities and all the rolls. Begin with 6 flipped tiles, then 5 flipped tiles etc. Every time a roll gives me more than 1 choice, consider p(win) for the states, I can go to, choose the best one and make a note of it. My optimal strategy is all these good choices.

Finally I can read in this table what p(win) is for the state 123456. And that’s the probability I’m looking for.

The extra credit version has 9 tiles and 2 dice. Otherwise it’s the same procedure.

Anmeldelse af A Power Unbound, af Freya Marske. Roman, del af serie. 2023. Hugo-finalist.

Før vi begynder. De 3 bøger i serien har visse stærke træk tilfælles. Trilogien som sådan er én lang mission. Der er 3 magiske dimser, og i hver bog går jagten på én af dem. Vi får undervejs samlet en gruppe unge mennesker. I hver bog finder et par sammen, den ene en (nuværende eller tidligere) magiker, den anden umiddelbart ikke. Edwin er ikke nogen stor magiker, Robin bliver ret tilfældigt rodet ind i missionen. Edwin ~ Robin. Edwin kender (Jack) Hawthorn, tidligere magiker. Robin har en søster, Maud. Jack kender Violet, magiker, og sammen render de på Maud, bl.a. fordi Violet også er blevet viklet ind i missionen. Trioen render i øvrigt også ind i Alan. Maud ~ Violet. Oveni er der Adelaide, der er Robins sekretær, men hun er lidt mere på sidelinjen.

Skitse: Alan er sådan set ikke del af holdet, men da han stikker hovedet ind, så kan de sagtens finde på noget at bruge ham til. Jagten på den tredje dims er nemlig koncentreret om det hus, Violet har arvet. De gode leder huset igennem fra kælder til kvist. De onde prøver at bryde ind, for at gøre det samme. Jack og Alan sender absolut ikke intense blikke til hinanden, nej nej.

Er det science fiction? Selvfølgelig ikke.

Temaer: Hvor Robin og Edwin er noget med tillid, som i hvert fald en af dem har meget svært ved, og Violet og Maud er noget med at være ærlig og legesyg, så havner Alan og Jack i små rollespil.

Jack har meget lidt kontakt med nogen, inklusive familien. Alan er derimod tæt på sin halvstore og voksende familie. Til gengæld har de en “hobby” tilfælles, og de opdager, at de på sin vis har kendt hinanden længe.

Den ønskeliste, jeg har nævnt før, bliver fortsat krydset af. I et enkelt tilfælde irriterer det mig lidt.

Er det godt? Tja. Jeg kan lidt bedre lide den end nr. 2. Der var både overraskelser og absolut forudsigelige hændelser. Der var nogle virkelig mærkelige sætninger, se nedenfor. Alligevel vil jeg sige: 👽👽👽

Note: Så mange citater! Efterordet snakker om, at den her bog blev lavet forrygende hurtigt. Det kan til tider mærkes.

“Jack slipped his much-thumbed copy of Bootblacks and Groundskeepers into an inner pocket of his jacket to read on the train. He also bought a newspaper to disguise it in, on the slim chance that any of the other denizens of his first-class compartment on a midweek train heading to the southeastern corner of Essex were also readers of the Roman. … He had the compartment to himself after changing to the older, pokier car of the local line.”

“The blue eyes he’d inherited regarded him thoughtfully, hungrily, from within his mother’s face.”

“I stole opportunity. If the world were different, I wouldn’t have had to, but it’s the way it is, and so I’ve been scared my whole life and angry for even longer.”

“Oliver made a noise of outraged pride, but all of Jack’s eyes were for the way Alan’s face came alive with challenge.”

“They’d set out today to be ahead by two pieces and had instead lost all of them in a handful of minutes.” (Start: 1-1. Goal: 2-1. End: 0-3.)

“He did what everyone else was doing: racing full-tilt towards the huge wooden doors out onto the street, which had been flung entirely wide.”

“Alan glared and did not open his mouth in return. He was too busy breathing through it.”