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

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.

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.

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 bee	10000	10001	10002	10003	10004	10005
QB * 0.7	7000.0	7000.7	7001.4	7002.1	7002.8	7003.5
Genius	7000	7001	7001	7002	7003	7004

Queen bee	10006	10007	10008	10009	10010	10011
QB * 0.7	7004.2	7004.9	7005.6	7006.3	7007	7007.7
Genius	7004	7005	7006	7006	7007	7008

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.

Elversprog, klingonsk og ‘Pig Latin’: Hvad kan opfundne sprog lære os om den menneskelige hjerne?

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 $8r$ (see source above). With $r=1$, the arc length will be $8\cdot1=8$.

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:

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

And combining all this:

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

For the cycloid, the formulas are:

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.

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

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

There’s another definition of L:

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

And then combining:

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

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.

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 $\frac{{162 \choose 82} }{2^{162}}$ 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.

Result: 86.06988.

And for extra credit:

Method 1: Definitely Monte Carlo.

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