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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

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.