#ThisWeeksFiddler, 20260904

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This week the #puzzle is: Can You Win at Asymmetric Bingo? #probabilities #statistics

Can You Win at Asymmetric Bingo?

Solution, possibly incorrect:

Program

Result of running program:

Bingos on small vs large cards: 0.66518 0.33482

Some fun histograms:

Distribution, how many numbers necessary for bingo.

Only looking at small card, igonoring numbers on large card.

+---------+---------+---------+---------+---------+---------+---------+
               X                                                 
              XXX                                                
              XXX                                                
              XXX                                                
             XXXX                                                
             XXXX                                                
             XXXXX                                               
             XXXXX                                               
            XXXXXX                                               
            XXXXXX                                               
            XXXXXX                                               
            XXXXXX                                               
           XXXXXXXX                                              
           XXXXXXXX                                              
           XXXXXXXX                                              
          XXXXXXXXX                                              
          XXXXXXXXX                                              
         XXXXXXXXXXX                                             
         XXXXXXXXXXX                                             
        XXXXXXXXXXXX                                             
+---------+---------+---------+---------+---------+---------+---------+
Average: 14.29625

Same, for large card.

+---------+---------+---------+---------+---------+---------+---------+
                     XXXX                                        
                     XXXXX                                       
                    XXXXXX                                       
                    XXXXXX                                       
                   XXXXXXXX                                      
                   XXXXXXXX                                      
                  XXXXXXXXXX                                     
                  XXXXXXXXXX                                     
                  XXXXXXXXXX                                     
                 XXXXXXXXXXXX                                    
                 XXXXXXXXXXXX                                    
                XXXXXXXXXXXXX                                    
                XXXXXXXXXXXXXX                                   
               XXXXXXXXXXXXXXX                                   
               XXXXXXXXXXXXXXX                                   
              XXXXXXXXXXXXXXXXX                                  
              XXXXXXXXXXXXXXXXX                                  
             XXXXXXXXXXXXXXXXXXX                                 
            XXXXXXXXXXXXXXXXXXXXX                                
          XXXXXXXXXXXXXXXXXXXXXXXX                               
+---------+---------+---------+---------+---------+---------+---------+
Average: 22.07363

Looking at small card bingos, including numbers on large.

+---------+---------+---------+---------+---------+---------+---------+
                                  XXXX                           
                                 XXXXXX                          
                                XXXXXXXX                         
                               XXXXXXXXXX                        
                               XXXXXXXXXX                        
                              XXXXXXXXXXXX                       
                             XXXXXXXXXXXXX                       
                             XXXXXXXXXXXXXX                      
                            XXXXXXXXXXXXXXX                      
                           XXXXXXXXXXXXXXXXX                     
                          XXXXXXXXXXXXXXXXXX                     
                          XXXXXXXXXXXXXXXXXXX                    
                         XXXXXXXXXXXXXXXXXXXX                    
                        XXXXXXXXXXXXXXXXXXXXXX                   
                       XXXXXXXXXXXXXXXXXXXXXXX                   
                      XXXXXXXXXXXXXXXXXXXXXXXXX                  
                     XXXXXXXXXXXXXXXXXXXXXXXXXXX                 
                    XXXXXXXXXXXXXXXXXXXXXXXXXXXXX                
                  XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX               
                XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX              
+---------+---------+---------+---------+---------+---------+---------+
Average: 34.29726

Same, for large card.

+---------+---------+---------+---------+---------+---------+---------+
                                  XXXX                           
                                 XXXXXX                          
                                XXXXXXXX                         
                               XXXXXXXXXX                        
                              XXXXXXXXXXX                        
                              XXXXXXXXXXXX                       
                             XXXXXXXXXXXXX                       
                            XXXXXXXXXXXXXXX                      
                            XXXXXXXXXXXXXXX                      
                           XXXXXXXXXXXXXXXXX                     
                          XXXXXXXXXXXXXXXXXX                     
                          XXXXXXXXXXXXXXXXXXX                    
                         XXXXXXXXXXXXXXXXXXXX                    
                        XXXXXXXXXXXXXXXXXXXXX                    
                       XXXXXXXXXXXXXXXXXXXXXXX                   
                      XXXXXXXXXXXXXXXXXXXXXXXXX                  
                     XXXXXXXXXXXXXXXXXXXXXXXXXX                  
                    XXXXXXXXXXXXXXXXXXXXXXXXXXXX                 
                  XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX                
                XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX               
+---------+---------+---------+---------+---------+---------+---------+
Average: 34.00285

Fiddler: The small board (me) wins more than 50% of the time.

Extra credit: The small board (me) wins about 67% of the time.

#ThisWeeksFiddler, 20260828

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This week the #puzzle is: Can You Spell Like a Queen Bee? #probability #counting #period #pattern #repetition

… If you find all the words in a given day and therefore accrue the maximum number of points, you earn the “Queen Bee” ranking. Meanwhile, accruing smaller point totals earns you other rankings. In particular, the point cutoff for “Genius” is 70 percent of the maximum number of points, rounded to the nearest whole number. While the cutoff for “Genius” is clearly displayed in the puzzle, the “Queen Bee” point total is not readily shown.
Of course, this maximum total can be approximated by dividing the “Genius” cutoff by 0.7. Even so, the total may be ambiguous, since multiple “Queen Bee” values can result in the same “Genius” cutoff.
Suppose a given round of Spelling Bee has some very large, randomly chosen point total. What is the probability that this total can be precisely determined (i.e., without any ambiguity) from its point cutoff for “Genius”?

And for extra credit:

Let’s put the rank of “Genius” aside. Here are some other ranks you can attain in Spelling Bee:
– Amazing (if you get 50 percent of the maximum, rounded to the nearest whole number)
– Great (40 percent)
– Nice (25 percent)
– Solid (15 percent)
– Good (8 percent)
– Moving Up (5 percent)
– Good Start (2 percent)
Suppose a given round of Spelling Bee has some very large, randomly chosen point total. What is the probability that this total can be precisely determined from these cutoffs combined (i.e., from “Good Start” through “Amazing,” inclusive)?

Can You Spell Like a Queen Bee?

Solution, possibly incorrect:

Program

Method 1: Go through a lot of large queen bee values, and check whether a queen bee value either 1 higher or 1 lower would give the same genius cutoff. I do this with a program.

Result: 0.4.

A wonderfully exact number!

Method 2: Examine what’s actually happening.

Queen bee100001000110002100031000410005
QB * 0.77000.07000.77001.47002.17002.87003.5
Genius700070017001700270037004
Queen bee100061000710008100091001010011
QB * 0.77004.27004.97005.67006.370077007.7
Genius700470057006700670077008

  1. Every time the queen bee value is divisible by 10, the genius cutoff won’t be rounded at all.
  2. Every 10 queen bee values result in the same roundings.
  3. Adding 1 to the queen bee value = adding 0.7 to the unrounded genius cutoff.
  4. From 10000-10009 there are 10 queen bee values, but only 7 genius cutoffs, 7000-7006. In 3 cases rounding a number means hitting the same genius cutoff as the queen bee value below (7001.4, 7004.2, 7006.3). This creates 3 pairs of ambiguous cutoffs, or 6 ambiguous cutoffs.
  5. How often does all this result in an ambiguous genius cutoff? 6/10 times.
  6. So how often do we not have ambiguity? 4/10 times.

I wonder whether probability and statistics have something like programming’s one off error? But instead it’s “the probability is 1 – p, not p”.

And for extra credit:

Method 1: Basically the same as method 1.

Result: 0.7.

Method 2:

Because one of the multipliers is 0.02 and another 0.05, this pattern will probably repeat every 100 possible queen bee values. Let’s look at it.

At a glance, the pattern repeats every 20. Ambiguous? 6/20. Unambiguous? 14/20. Result confirmed.

Science fiction predictions XXVII

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Foreword, Battlefield Earth, L. Ron Hubbard.

“Science fiction does not come after the face of a scientific discovery or development. It is the herald of possibility. It is the plea that someone should work on the future. Yet it is not prophecy. It is the dream that precedes the dawn when the inventor or scientist awakens and goes to his books or his lab saying, “I wonder whether I could make that dream come true in the world of real science.” “

#ThisWeeksFiddler, 20260821

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This week the #puzzle is: How Far Can You Roll? #arclength #estimate #coding #montecarlo #average

For her photography show, Frederica Fiddleria attaches a light to a point on the circumference of a circular wheel with a radius of 1 meter. She points a camera at the wheel and, during a single long exposure, rolls the wheel for one revolution along the ground.
When she develops the film, she is curious about the path the light took as the wheel rolled. What is the length of this path?

And for extra credit:

For her next show, Frederica wants to mix things up. Instead of placing the light on the circumference of the wheel, she will pick a random point inside the circle. (Before you ask, let me clarify what “random” means here: Any two regions with the same area are equally likely to contain the point.)
As before, she will roll the wheel for one revolution along the ground and capture the motion with a single long exposure on her camera. On average, what can she expect the length of the path to be?

How Far Can You Roll?

Intermission

Last week I didn’t really try to solve the fiddler. It involved learning rules to games, solving a chess puzzle, and picking up on subtle verbal cues. None of them are my strong suits.

Solution, possibly incorrect:

Cycloid

Okay, so this will be a little stream of consciousnessy. Apparently the curve, we’re looking for is a cycloid. And the length of the curve is an arc length. For a cycloid, the arc length is 8r8r (see source above). With r=1r=1, the arc length will be 8⋅1=88\cdot1=8.

Result: 8 meters. (Method 1.)

Oh. And I assume the distance between camera and light is large enough, that I don’t have to account for whatever it’s called, when the motive is distorted. Perspective? Parallax?

In general the length of a path (the arc length) is:

L=∫abx′(t)2+y′(t)2dtL=\int_a^b \sqrt{x'(t)^2+y'(t)²}dt

Just to test how this works, let’s look at a circle.

x(t)=r⋅cos⁡(t),x′(t)=−r⋅sin⁡(t)x(t)=r\cdot \cos(t), x'(t)=-r\cdot\sin(t)

y(t)=r⋅sin⁡(t),y′(t)=r⋅cos⁡(t)y(t)=r\cdot \sin(t), y'(t)=r\cdot\cos(t)

a=0,b=2πa=0, b=2\pi

And combining all this:

L=∫abx′(t)2+y′(t)2dtL=\int_a^b \sqrt{x'(t)^2+y'(t)²}dt

L=∫02π(−r⋅sin⁡(t))2+(r⋅cos⁡(t))2dtL=\int_0^{2\pi} \sqrt{(-r\cdot\sin(t))^2+(r\cdot\cos(t))²}dt

L=∫02πr2(sin⁡(t))2+r2(cos⁡(t))2dtL=\int_0^{2\pi} \sqrt{r^2(\sin(t))^2+r²(\cos(t))²}dt

L=∫02πr2((sin⁡(t))2+(cos⁡(t))2)dtL=\int_0^{2\pi} \sqrt{r^2((\sin(t))^2+(\cos(t))²)}dt

L=∫02πr(sin⁡(t))2+(cos⁡(t))2dtL=\int_0^{2\pi} r\sqrt{(\sin(t))^2+(\cos(t))²}dt

L=r∫02π1dtL=r\int_0^{2\pi} \sqrt{1}dt

L=r∫02π1dtL=r\int_0^{2\pi} 1dt

L=r(t|02π)L=r(t\big|_0^{2\pi})

L=r(2π−0)=r2πL=r(2\pi – 0)=r2\pi

This fits what we expect. So I guess I know how to use this formula.

For the cycloid, the formulas are:

x(t)=r(t−sin⁡(t)),x′(t)=r(1−cos⁡(t))x(t)=r(t-\sin(t)),x'(t)=r(1-\cos(t))

y(t)=r(1−cos⁡(t)),y′(t)=r(sin⁡(t))y(t)=r(1-\cos(t)), y'(t)=r(\sin(t))

a=0,b=2π,r=1a=0, b=2\pi,r=1

L=∫abx′(t)2+y′(t)2dtL=\int_a^b \sqrt{x'(t)^2+y'(t)²}dt

L=∫02π(1−cos⁡(t))2+(sin⁡(t))2dtL=\int_0^{2\pi} \sqrt{(1-\cos(t))^2+(\sin(t))²}dt

Or how WolframAlpha would put it:

integral square root((1-cos(t))^2+sin(t)^2) from 0 to 2pi

Everything seems to work okay. I guess that was method 2. Moving on.

And for extra credit:

Trochoid Arc length of a trochoid Program

This time the curve is a trochoid. A cycloid as a special case of that. And it’s my lucky day! The arc length of that one is already calculated Out There. A new parameter is b (again), the distance between the point on the circle and the center of the circle.

0≤t≤2π0\le t\le 2\pi

L=2⋅|r−b|⋅E(t2|02π−4rb(r−b)2)L=2\cdot|r-b|\cdot E\left(\frac{t}{2}\big|_0^{2\pi}-\frac{4rb}{(r-b)²}\right)

L=2⋅|r−b|⋅E((2π2−0)−4rb(r−b)2)L=2\cdot|r-b|\cdot E\left((\frac{2\pi}{2}-0)-\frac{4rb}{(r-b)²}\right)

L=2⋅|r−b|⋅E(π−4rb(r−b)2)L=2\cdot|r-b|\cdot E\left(\pi-\frac{4rb}{(r-b)²}\right)

Where E is an elliptical integral. Let me see… Ah yes.

E(t,k)=∫0t1−k2sin⁡(t)2dtE(t,k)=\int_0^t\sqrt{1-k²\sin(t)^2}dt

Hm. There’s something wrong. E can’t have both 1 and 2 parameters.

There’s another definition of L:

L=2⋅|r−b|⋅E(t2,2irb|r−b|)L=2\cdot |r-b| \cdot E \left( \frac{t}{2},\frac{2i \sqrt{rb}}{|r-b|} \right)

I think I’m supposed to read it like this actually:

L=2⋅|r−b|⋅E(t2|02pi,2irb|r−b|)L=2\cdot |r-b| \cdot E \left( \frac{t}{2}\big|_0^{2pi},\frac{2i \sqrt{rb}}{|r-b|} \right)

L=2⋅|r−b|⋅E(π,2irb|r−b|)L=2\cdot |r-b| \cdot E \left( \pi ,\frac{2i \sqrt{rb}}{|r-b|} \right)

And then combining:

E(t,k)=∫0π1−(2irb|r−b|)2sin⁡(t)2dtE(t,k)=\int_0^{\pi}\sqrt{1-(\frac{2i \sqrt{rb}}{|r-b|})²\sin(t)^2}dt

No. Scratch method 1. Let me go back to the definitions for the point.

x=r⋅t−bsin⁡(t)x=r\cdot t-b\sin(t)

y=r−bcos⁡(t)y=r-b\cos(t)

0≤t≤2π0\le t\le 2\pi

Then I code a monte carlo approach, where the arc length is estimated by assuming I can just look at some points along the curve and calculate at all the little straight distances between these. Just to make sure my approach works, I also code the fiddler and sort of get the same result. Then I run the program a few times.

Result 2: 7.109200
Result 2: 7.112045
Result 2: 7.111690
Result 2: 7.111431

I am confident I can say the rounded answer is 7.11. Method 2 worked out. And I guess also gave a method 3 for the fiddler.

#ThisWeeksFiddler, 20260807

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This week the #puzzle is: How Lucky Can a Baseball Team Get? #probabilities #counting #montecarlo #coding

The Fiddler Baseball League consists of exactly two teams of equal skill: the Algebraists and the Geometers. Over the course of a season, these two teams play each other 162 times. Each team has an equal chance of winning each game, and the results of games are independent of one another.
At the end of the season, on average, how many games would you expect the team with the better record to have won? (If the teams have the same record, then you should include one of them in your calculation.)

And for extra credit:

After some expansion, the Fiddler Baseball League now boasts 30 teams. Over the course of a season, each team plays each other team five times. (Thus, each team plays a total of 145 games.) As before, each team has an equal chance of winning each game, and the results of games are independent of one another.
At the end of the season, on average, how many games would you expect the team with the best record to have won? (If more than one team has the same best record, then you should include one of them in your calculation.)

How Lucky Can a Baseball Team Get?

Solution, possibly incorrect:

Program

Method 1: Monte Carlo. Play through a tournament of 162 games. Do this in a loop a lot of times. Run the program a few times to find a stable result.

Method 2: Calculate the average directly. E.g., with 82 wins, there’s a probability of (16282)2162\frac{{162 \choose 82} }{2^{162}} of this situation arising. Multiply by 82. Sum over all win options, 0-162. Note that 82 wins and 80 wins both produce “winner won 82 times” situations.

E=Σn=0162(162n)2162⋅max(n,162−n)E=\Sigma_{n=0}^{162}\frac{{162\choose n}}{2^{162}}\cdot max(n,162-n)

Result: 86.06988.

And for extra credit:

Method 1: Definitely Monte Carlo.

Result: 84.98±0.01\underline{84.98 \pm 0.01}.

Hugo-relateret, 2026

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Anmeldelse af “best related”. 2025. Hugo-finalist.

Nebula nominees . Hugo finalists .

Colourfields: Writing About Writing About Science Fiction by Paul Kincaid

Skitse: En samling af anmeldelser og andre essays, der har med sf at gøre.

Er det science fiction? Absolut.

Er det godt? Ja, det er det vel egentlig. Velskrevet og interessante holdninger. Så hvis jeg virkelig godt kunne lide anmeldelser … ##-

The Hugo Spreadsheet of Doom, maintained by Renay

Skitse: En liste over værker af forskellig art, der kan inspirere til nominering til hugoen.

Er det science fiction? Sff osv.

Er det godt? Sådan noget som det her et regneark, men det er vel også arbejdet med at putte indhold i det, administrere det, tage stilling til om det enkelte værk egentlig kan nomineres osv. Tja. Godt arbejde. Men det føles også en anelse meta at have det med på stemmesedlen. ##-

Inventing the Renaissance by Ada Palmer

Skitse: Hvordan og hvorfor opstod begrebet renæssance? Fyldig historie.

Er det science fiction? Nix. Historie. Men Palmer har skrevet sf ved andre lejligheder.

Er det godt? Det virker velskrevet. Det virker bare også meget underligt at have det på stemmesedlen. Jeg tror, jeg vil undlade at give det min stemme på nogen måde. Dvs., jeg vil ikke placere det under “no award”, blot ikke stemme på det. #–

Last War in Albion: “The Cuddled Little Vice (Sandman)” by Elizabeth Sandifer

Skitse: En gennemgang af ham der, der skrev Sandman-bøgerne. Hvem er han, hvordan opstod tegneserierne, og ikke mindst, er han et godt menneske?

Er det science fiction? Eftersom Sandman mere er fantasy (og horror?), så nej.

Er det godt? Jeg læste den her, da den var ny, og gik meget op i det. Nu er det hele sådan lidt blah. Endnu en fyr, der virker ret skummel. Suk. ##-

Positive Obsession: The Life and Times of Octavia E. Butler by Susana M. Morris

Skitse: Biografi om Butler, inklusive hvad der i øvrigt foregik i verden på de relevante tidspunkter, og hvordan Morris selv passer ind i historien.

Er det science fiction? Jeps, eftersom Butler mest skrev sf.

“A war waged with world-destroying weapons would be the pinnacle of the Human Contradiction, an example of both our stunning intelligence and our self-destructive hierarchal thinking.”

Er det godt? Ja. Der er lidt gentagelser hist og her, som om den sidste revision ikke bed helt nok. Men det meste er bare velskrevet og interessant. ###

“Ragnarök vs the Long Night” by Ashaya and Aziz

Skitse: Et afsnit af en podcast, der tager udgangspunkt i Game of Thrones/A Song of Ice and Fire. I lige det her afsnit skal vi diskutere ragnarok, fordi Martin har ladet sig inspirere af dette fænomen.

Er det science fiction? Nix.

Er det godt? Hvis man er fan af serien, hvis man kan goutere forkert udtale af Yggdrasil og Tolkien, hvis man kan håndtere 10 x “vi kommer tilbage til” på stribe, hvis man kan håndtere entusiasme (der i hvert fald på mig virker falsk), så kan man nok godt lide den her podcast. ##-

Butler-bogen tager ubesværet førstepladsen.

#ThisWeeksFiddler, 20260731

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This week the #puzzle is: Can You Sneak Past the Rings? #probabilities #montecarlo #coding #plot

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

And for extra credit:

Instead of requiring a vertical line down the cylinder’s surface, now any helix down the surface is allowed. An example of such a helix passing through all three gaps is shown below.
What is the probability that there exists at least one such helix that can pass through each ring’s gap?

Can You Sneak Past the Rings?

Solution, possibly incorrect:

Program Desmos

Method 1: Monte Carlo. It’s a bit fiddly, because when exactly do the 3 gaps overlap? I settle on:

  • Put the 1st gap at 0 – 1/8, WLOG.
  • If possible, create the overlap between the 2 1st gaps.
  • Then look for an overlap between that overlap and the 3rd gap.

467910 sneaks out of 10000000, prob = 0.04679
468490 sneaks out of 10000000, prob = 0.04685
468187 sneaks out of 10000000, prob = 0.04682

Once I had all those conditions in place, I could actually do method 2, plot it in Desmos.

It was fun to figure out how to plot it. How the cases combine. I look at the cases for the double overlap in the columns and the cases for the triple overlap in the rows.

Gap 1: a-b, 0 – 1/8
Gap 2: c-d, c = x, 0≤x≤10\le x\le1
Gap 3: e-f, e = y, 0≤y≤10\le y\le1
In general, overlap goes i1−i2i_1 – i_2		Overlap: c-b, x – 1/8
0≤x≤180\le x\le\frac{1}{8}		Overlap: a-d, 0 – [x – 1 + 1/8]
78≤x≤1\frac{7}{8}\le x\le1
i1≤e≤i2⟺i1≤y≤i2i_1\le e\le i_2\iff i_1\le y \le i_2x≤y≤18x\le y\le\frac{1}{8}0≤y≤x−1+180\le y\le x-1+\frac{1}{8}
i1≤f≤i2⟺i1≤y−1+18≤i2i_1\le f\le i_2\iff i_1\le y-1+\frac{1}{8}\le i_2x≤y−1+18≤18x\le y-1+\frac{1}{8}\le \frac{1}{8}0≤y−1+18≤x−1+180\le y-1+\frac{1}{8}\le x-1+\frac{1}{8}
e≤fe\le f
e≤i1,i2≤f⟺y≤i1,i2≤y+18e\le i_1,i_2\le f\iff y\le i_1,i_2\le y+\frac{1}{8}		y≤xy\le x
18≤y+18\frac{1}{8}\le y+\frac{1}{8}		Impossible
f≤ef \le e
0≤i1,i2≤f⟺i2≤y+780\le i_1,i_2\le f \iff i_2\le y+\frac{7}{8}		Impossible78≤y\frac{7}{8}\le y
x−78≤y−78x-\frac{7}{8}\le y – \frac{7}{8}

Anyway. The result is 6 triangles. Each triangle has an area of 18⋅18⋅12=1128\frac{1}{8}\cdot\frac{1}{8}\cdot\frac{1}{2}=\frac{1}{128}. The area of all the triangles is 6128=364=0.046875\frac{6}{128}=\frac{3}{64}=0.046875. This agrees with method 1.

Result: 0.046875.

And for extra credit:

Desmos

I recreated the situation in Desmos and played around a little. Among other things, I construct the (orange) interval e has to fall in, in order to make a helix possible.

Interestingly, this interval always has length 0.5, corresponding to a probability of 0.5.

  • My x axis shows how far along the circumference of the cylinder I am. As with the fiddler, the first gap is frozen 0 – 1/8.
  • The helix has to go through a and d, or b and c, or somewhere in between. This corresponds to different slopes of the helix.
  • I construct a point on the line going through b and c and crossing the 3rd ring. This point will have x=c+(c−b)=2c−bx=c+(c-b)=2c-b.
  • This leads to 2c−b≤f⟺2c−b−18≤e2c-b\le f\iff 2c-b-\frac{1}{8}\le e.
  • I also construct a point on the line going through a and d and crossing the 3rd ring. This point will have x=d+(d−a)=2d−ax=d+(d-a)=2d-a.
  • This leads to e≤2d−ae\le 2d-a.
  • The interval for e to fall within is therefore 2c−b−18≤e≤2d−a2c-b-\frac{1}{8}\le e\le 2d-a.
  • The length of this interval is
    • 2d−a−(2c−b−18)2d-a-(2c-b-\frac{1}{8})
    • 2d−a−2c+b+182d-a-2c+b+\frac{1}{8}
    • 2d−2c+b−a+182d-2c+b-a+\frac{1}{8}
    • 2(d−c)+(b−a)+182(d-c)+(b-a)+\frac{1}{8}
    • 2⋅18+18+182\cdot \frac{1}{8} +\frac{1}{8}+\frac{1}{8}
    • 12\frac{1}{2}

So the result is 0.5. I am surprised, that the answer is so simple.

Hugo-noveller, 2026

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Her er årets nominerede noveller, med samt link til mine anmeldelser:

“In My Country” by Thomas Ha##-Nebula/Hugo-noveller
“Laser Eyes Ain’t Everything” by Effie Seiberg##-Nebula/Hugo-noveller
“Missing Helen” by Tia Tashiro###Hugo-noveller
“Six People to Revise You” by J.R. Dawson###Nebula/Hugo-noveller
“10 Visions of the Future; or, Self-Care for the End of Days” by Samantha Mills###Hugo-noveller
“Wire Mother” by Isabel J. Kim###Hugo-noveller

Den var værre. 4 topscorere. Helen og revisionerne er i top 2. Hm. Det kan jeg ikke bestemme mig for nærmere.

Nebula-noveller, 2026

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Her er årets nominerede noveller, med samt link til mine anmeldelser:

“Through the Machine” by P.A. Cornell##-Nebula-noveller
“Six People to Revise You” by J.R. Dawson###Nebula/Hugo-noveller
“In My Country” by Thomas Ha##-Nebula/Hugo-noveller
“The Tawlish Island Songbook of the Dead” by E.M. Linden###Nebula-noveller
“Because I Held His Name Like a Key” by Aimee Ogden##-Nebula-noveller
“Laser Eyes Ain’t Everything” by Effie Seiberg##-Nebula/Hugo-noveller

Blandt 2 topscorere, så er det alligevel revisions-historien, der løber med guldet.