This week the #puzzle is: Can You or “Cantor” You Find the Distance? #probabilities #infinity #coding #montecarlo #permutations #distance

I start with a number line from 0 to 1, and then I remove the middle third. Then I take each of my remaining pieces (the segment from 0 to 1/3 and the segment from 2/3 to 1) and remove their middle thirds. Now I have four segments (from 0 to 1/9, 2/9 to 1/3, 2/3 to 7/9, and 8/9 to 1) and remove their middle thirds. I do this over and over again, infinitely many times.

The points I’m left with are collectively known as the Cantor set. The Cantor set is not empty; in fact, it contains infinitely many points on the number line, such as 0, 1, 1/3, and even 1/4.

It’s possible to pick a random point in the Cantor set in the following way: Start with the entire number line from 0 to 1. Then, every time you remove a middle third, you give yourself a 50 percent chance of being on the left remaining third and a 50 percent chance of being on the right remaining third. Then, when you remove the middle third of that segment, you again give yourself a 50 percent chance of being on the left vs. the right, and so on.

Suppose I independently pick two random points in the Cantor set. On average, how far apart can I expect them to be?

And for extra credit:

Suppose I independently pick three random points in the Cantor set. Each point has some value between 0 and 1.

What is the probability that these three values can be the side lengths of a triangle?

Can You or “Cantor” You Find the Distance?

Solution, possibly incorrect:

Program

Method 1:

• Monte Carlo. I try different “sensitivities”. This corresponds with how many decimals (ternimals?), I have in my ternary number. Using 30 digits and 1000000 tries, I get an average of 0.40. Really? I am baffled.

Method 2:

• Brute force. For a couple of different numbers of digits (1-12), I try all the numbers with that many digits, and try these in all combinations. My average converges nicely to 0.40 again.

Method 3:

• Let’s try and look at this using probabilities. Let’s look at all the 2 digit ternary numbers.

Distance	003	023	203	223
003 = 0	0	2/9	6/9	8/9
023 = 2/9	2/9	0	4/9	6/9
203 = 6/9	6/9	4/9	0	2/9
223 = 8/9	8/9	6/9	2/9	0

• The average in this tiny example is 7/18 or about 0.3888. Very close to 0.40 already.
• In fact I should be looking at intervals. Not 00₃, but 00₃-01₃.

Distance	003-013	023-103	203-213	223-1003
003-013 or 0/9-1/9	0/9-1/9	1/9-3/9	5/9-7/9	7/9-9/9
023-103 or 2/9-3/9	1/9-3/9	0/9-1/9	3/9-5/9	5/9-7/9
203-213 or 6/9-7/9	5/9-7/9	3/9-5/9	0/9-1/9	1/9-3/9
223-1003 or 8/9-9/9	7/9-9/9	5/9-7/9	1/9-3/9	0/9-1/9

Distance interval	Distance max	Frequency	Acc. freq.
0/9-1/9	1/9	4/16	4/16
1/9-3/9	3/9	4/16	8/16
3/9-5/9	5/9	2/16	10/16
5/9-7/9	7/9	4/16	14/16
7/9-9/9	9/9	2/16	16/16

• Half of the distances are 3/9 or below.
• The other half of the distances aren’t distributed the same way (that would be half of them 6/9 or above). So the average should be dragged down a little from simply being 0.5.

I trust my results. The average is 0.40.

And for extra credit:

Same program, same logic etc. Some of the same bafflement.

The ratio is 0.40.

I actually got curious and conducted the same experiments without the cantor set limitation. Then my average would have been 1/3 and my ratio would have been 1/5. So it does make a difference.

This week the #puzzle is: Can You Shovel the Snow? #geometry #area #equidistant #probabilities #coding #montecarlo

Two teams of shovelers plan to remove all the snow from a parking lot that’s shaped like a regular hexagon. Team Vertex initially places each of its six shovelers at the six corners of the lot. Meanwhile, Team Centroid initially places all its shovelers at the very center of the lot.

Each team is responsible for shoveling the snow that is initially closer to someone on their own team than anyone on the other team.

What fraction of the lot’s snow is Team Centroid responsible for shoveling?

And for extra credit:

Having completed the shoveling of the hexagonal lot, the two teams move on to a circular lot. This time, Team Vertex initially places all their shovelers at one random point inside the circle, while Team Centroid initially places all their shovelers at one other, independently chosen random point inside the circle. (To be clear, by “random point” in the circle, I mean that the probability of being in any given region of the circle is proportional to that region’s area.)

As before, each team is responsible for shoveling the snow that is initially closer to someone on their own team than anyone on the other team.

Inevitably, one team is responsible for shoveling a greater fraction of the lot than the other. On average, what would you expect this greater fraction to be?

Can You Shovel the Snow?

Solution, possibly incorrect:

Desmos

This image nicely captures my thoughts (click to enlarge):

• For each point within the hexagon, it will be closest to one particular point on the hexagon.
• The hexagon slices up into 6 parts, defined by being between 2 neighbor points on the hexagon and the center.
• On the image, a slice is shown with red borders, on the right part of the hexagon. The points on the hexagon are at (c,d) and (c,-d). c and d are calculated from the angle at the center.
• The slices are defined by lines passing through the center and a point on the hexagon.
• Within such a slice, there are further 2 halfs. In this case let’s concentrate on the upper half. In this half, points within the slice are either closer to the center or to (c,d). Find the midpoint between these 2 points and draw the perpendicular. (All the relevant perpendiculars, the parts we need at least, are drawn with blue lines.)
• On one side of the perpendicular, points are closest to the center.
• Calculate the area of the whole slice and of the purple triangle, divide, and we have our ratio.

The purple triangle is between these 3 points:

The area of the slice is:

The area of the triangle is:

The fraction of the points closest to the center:

The result is 1/3.

And for extra credit:

Program Desmos 1 Desmos 2

This could of course be solved using integrals. But I don’t know how… So I monte carlo.

In part my program is based on what this image shows:

Given 2 random points (red and blue — the red one was trying to leave the circle there, sigh), find the midpoint between them (green). Translate the midpoint so that it’s on a line going through the center. Then rotate that new midpoint around the center, so that it lands on the x axis. Then draw the perpendicular through that point (black). Calculate the area of the circular segment on the left and calculate the fraction. (I actually calculate the smaller fraction, so it’s then subtracted from 1.)

Do this a lot of times and calculate the average.

Some of the calculations. Find the midpoint between the 2 random points:

Translate that midpoint onto the line going through the center. First find the slant of the line through the 2 random points:

Then find the translated midpoint:

Distance from translated midpoint to center:

Find angle:

Find area:

The area of the whole circle:

The fraction, remember we need the largest one:

The result, from the program, about 0.64.

I plotted the distribution in Desmos too:

This week the #puzzle is: Can Every Day Be First? #calendars #patterns

In 2026, every day of the week is the first day of the month at least once:

– Monday is June 1. – Tuesday is September 1 and December 1. – Wednesday is April 1 and July 1. – Thursday is January 1 and October 1. – Friday is May 1. – Saturday is August 1. – Sunday is February 1, March 1, and November 1.

Is 2026 special in this regard? If so, when is the next year when one of the days of the week is not represented among the firsts of the month? Otherwise, if 2026 is not special in this regard, then why not?

And for extra credit:

As we just noted, in 2026, all seven days of the week appear as the first of the month at least once. But you know, I decided that I don’t like that at all. Instead, I want as few days of the week as possible to appear as the first of the month in a given year.

To accomplish this, I have been granted the authority to change the number of days in each of that year’s 12 months, provided that there are still 365 or 366 days in the year and each month has at least 28 days and at most 31 days.

What are the fewest days of the week that can appear as the first of the month in such a calendar year? (And for fun, rather than for credit: How many such calendars can you design with this property?)

Can Every Day Be First?

Intermission

Last week I claimed, that 0 wasn’t the only possible solution to the extra credit. This week it sparked a little discussion in the comment section (link just above).

While Zach of course knows what he intended, I still think the actual wording allows more than 1 solution. But what do you think?

Possibly incorrect solution:

Let’s look at just 8 of months of the year. (This is actually 2027. Click to enlarge.)

Or in other words, March = Monday, April = Thursday, May = Saturday, June = Tuesday, August = Sunday, September = Wednesday, October = Friday. (July is also a Thursday.) We have them all. This part of the year doesn’t change structure, even if the length of February changes. The only thing that changes is that the very first day, the Monday in March, might be something else, maybe a Sunday (as in 2026), but then all the other days change in the same way, in this case every day moving a step back in the order of weekdays.

A year will always go through all 7 weekdays for the 1st day of a month, specifically this will happen in March-October. 2026 isn’t special.

And for extra credit:

Program

I could probably do this in my head or on paper, but it’s easier to build a program. Output:

Nice calendars with very few different 1st days:
30,31,30,30,31,30,30,31,30,30,31,31
31,30,30,31,30,30,31,30,30,31,30,31
31,31,29,31,31,29,31,31,29,31,31,30
31,31,29,31,31,29,31,31,29,31,31,31

       0 calendars with 1 different 1st days
       0 calendars with 2 different 1st days
       4 calendars with 3 different 1st days
     112 calendars with 4 different 1st days
    4143 calendars with 5 different 1st days
   14718 calendars with 6 different 1st days
   19919 calendars with 7 different 1st days
16738320 calendars with a wrong length
16777216 calendars in all
16777216 expected

A calendar can have 3 different 1st days, as the lowest number. There are 4 different calendars working like this. Here’s the beginning of each:

Is it possible to have a calendar with only 1 kind of 1st weekday? No, because months all of length 28 wouldn’t give a year of 365/366 days. (Though having 13 months would almost get us there.)

Is it possible to have a calendar with only 2 kinds of 1st weekday? This would be a pattern, where 2 months would cover 63 days, so, no.

Is it possible to have a calendar with only 3 kinds of 1 weekday? Yes. Create a pattern where 3 months cover 91 days, and repeat. The patterns with 30, 30 and 31 in some order works, but 31 can’t be last, because then December would need to have 32 days. The patterns with 31, 31 and 29 works, in 2 different ways.

This week the #puzzle is: Can the Archers Coordinate? #probabilities

Two logicians are trying to earn a fabulous prize as a team. There are three targets, and, to win the prize, each logician must fire a single arrow and hit the same target as the other. Two of the targets are closer but are otherwise indistinguishable; the logicians know they each have a 98 percent chance of hitting either of these targets. The third target is farther away; the logicians know they each have a 70 percent chance of hitting that target.

The logicians can’t cooperate or consult in advance, and they have no knowledge of which target their counterpart is aiming for or whether they are successful.

What is the probability they will win the prize?

And for extra credit:

As before, there are still three targets, but their respective probabilities of being struck have changed. That said, two of the targets remain indistinguishable from each other and have the same probability of being struck. Moreover, all three probabilities are rational.

After doing some mental arithmetic, the logicians realize that it doesn’t matter which target they aim for—their probability of winning the prize is the same no matter what.

What is their probability of winning the prize?

Can the Archers Coordinate?

Possibly incorrect solution:

If the logicians could’ve consulted beforehand, they could have chosen 1 of 2 strategies. The 1st one involves aiming for one of the easy targets, the 2nd involves aiming for the hard target. Let’s look at the probability a strategy succeeds.

The 1st strategy includes, that the 2nd logician won’t know, which easy target the 1st logician aimed at, so there’s a 50% chance of aiming for the same one.

Both logicians will realize, that the 2nd strategy is superior. They both employ this strategy. The probability they will win the prize is 0.49 = 49%.

And for extra credit:

This time the probabilities depend on unknown numbers for the underlying probabilities. Based on integer choices for all the unknown numbers (so that the resulting probabilities are rational), this must hold:

It doesn’t matter which target is chosen, the probabilities will be the same. This means.

As the square root isn’t an integer, this must mean the rest of the equation is 0.

Therefore the probability of hitting anything is 0.

My only problem is, this doesn’t seem to require any mental arithmetic?

Another solution could be, that actually all 3 targets are indistinguishable. But again, this wouldn’t require any mental arithmetic. Still, in that case the probability would be, for some integers:

The chance they choose the same target, and the chance of hitting it.

Result: m^2/3n^2, m and n integers, at least 0, at most 1/3.

Some time ago I was busy crocheting squares for Granny Life .

Then I preordered Art But Make It Sports . And as a thank you, the author promised to react to a photo with a piece of art. And this resulted. (The thought is, that the art resembles the photo.)

This week the #puzzle is: Can You Get the Rover Home? #trigonometry #spherical #3d #animation #desmos3d

A rover is dropped down on a spherical planet with a radius of 1000 miles. The rover has been programmed with a very specific set of motions:

– First, it moves straight forward a fixed distance s, and stops. – Without moving forward, it turns left 60 degrees. (Importantly, the rover turns 60 degrees, not 120 degrees.) – Next, the rover moves straight forward in this new direction another distance s, and stops. – Without moving forward, it again turns left 60 degrees. – Finally, the rover moves straight forward in this new direction another distance s, and stops.

To be picked up, the rover must complete its journey in the same place it was dropped down. What is the minimum value of s, with s > 0, for which this works?

To help you out, here’s an illustration, courtesy of Pierre, of what such a path might look like:

And for extra credit:

There are other values of s for which the rover will end its journey where it was dropped down. How many such positive values of s (including the answer you just found in the Fiddler) are less than 100,000 miles?

Can You Get the Rover Home?

Possibly incorrect solution:

Method 1: Turns out there’s a Spherical law of cosines! Who knew! This law says:

In our case:

So we’re just looking for s.

Solutions:

In our case the most interesting solution is this one:

Method 2: I looked at the sphere is Desmos . Looks good.

Result: 1911 miles.

Animations

I have made animations of the 3 situations! I am insanely proud of them!

Desmos 6 , Desmos 7 , Desmos 8 .

And for extra credit:

Method 1: The solutions are clustered around:

The next 3 solutions are:

The 3rd solution of these is a replica of the fiddler solution, it just goes all the way around the planet before taking the 1911 miles.

The 2nd solution is going all the way around the planet to the starting point. The trivial one.

The 1st solution emulates the fiddler. Imagine a fiddler solution. Start at one of the corners. Then instead of going to the next corner, go in the opposite direction, away from the triangle, going around the planet and arriving at the next corner. Etc.

There will be a set of solutions for n = 15:

And a partial set for n = 16:

And then the values grow larger than 100000.

So there’s 1 + 15 * 3 + 1 solutions.

Method 2: I also looked at these solutions in Desmos.

Result: 47 solutions.

This week the #puzzle is: How Long Is the All-Star Streak? #math #probabilities #InfiniteSums #coding #MonteCarlo

The Fiddler Basketball Association’s All-Star Game consists of two teams: “East” and “West.” Every year these two teams play a game, each with a 50 percent chance of winning that’s independent of the outcomes of previous years.

Many, many years into the future, you look at the most recent results of the All-Star Game. On average, what is the longest current winning streak that one of the teams is on? (Here, having won only the most recent game still counts as a “streak” of one game.)

And for extra credit:

To spice up the All-Star Game, the commissioner of the FBA has decided that there will now be three teams competing in All-Star Games: “Stars,” “Stripes,” and “International.” Each year, two of the three teams play each other. If one year has Stars vs. Stripes, the next year has Stripes vs. International, the year after that has International vs. Stars, and then the cycle repeats with Stars vs. Stripes.

Many, many years after this new format has been adopted, you look at the most recent results of the All-Star game. On average, what is the longest current winning streak that one of the teams is on? (As before, having won one game counts as a “streak.” Also, note that the team with the longest winning streak might not be one of the two teams that played in the most recent All-Star Game.)

How Long Is the All-Star Streak?

Highlight to reveal (possibly incorrect) solution:

Math StackExchange Program

Method 1:

• 1 win in a row has probability 1/2
• This is actually the probability of 1 team first losing, then winning, and there are 2 teams, 2 * 1/4 = 1/2
• 2 wins has probability 1/4
• n wins has probability 1/2ⁿ
• This checks out, as the sum of all these probabilities is 1.
• The average is 1 * 1/2 + 2 * 1/4 + 3 * 1/8 + … + n * 1/2ⁿ + …
• According to Math StackExchange (and probably a lot of other sources), this is 2.

Method 2:

• Monte Carlo. Same result.

And for extra credit:

Spreadsheet

Method 1:

• This time the probability of n wins is 3 *1/2 * 1/2ⁿ
• The spreadsheet shows, that the probability of a streak of 1 win is 1/4, as a special case
• The average is 1 * 1/4 + 2 * 3/8 + 3 * 3/16 + … + n * 3/2ⁿ⁺¹ + …
• The sum of n/2ⁿ from n = 1 is 2, as seen above; from n = 2, it’s 2 – 1/2 = 3/2
• Most of the above sum (from n = 2) can be rewritten as the sum of 3 * 1/2 * n/2ⁿ = 3 * 1/2 * 3/2 = 2 1/4
• The first part of the sum is 1 * 1/4 = 1/4
• The whole sum, the average, is 1/4 + 2 1/4 = 2 1/2 = 2.5.

Method 2:

• Monte Carlo. Same result.

This week the #puzzle is: Can You Hop in a Spiral? #geometry #trigonometry #angle #arctan #minimize #markovchain #coding

Frankie the frog is hopping on a large, packed grid of lily pads, shown below. The pads are circular and each is a distance 1 from its nearest neighbors. (More concretely: Each pad has a diameter of 1 and they are arranged in a hexagonal lattice.) Frankie starts at (0, 0), the center of the pad labeled A. Then she hops due east to pad B at (1, 0), and from there she hops to pad C at (1.5, √(3)/2).

She wants to continue hopping in a counterclockwise, spiral-like pattern. Each of her jumps is to the center of a neighboring pad, a net distance of 1. But there are two rules her spiral must follow:

– Each next pad must be in a more counterclockwise direction (relative to spiral’s origin at pad A) than the previous pad. – Each pad must be farther from A than the previous pad.

After a number of hops spiraling around, Frankie realizes she is, once again, due east from A. What is the closest to A she could possibly be? That is, what is the minimum possible distance between the center of the pad she’s currently on from the center of pad A?

And for extra credit:

Frankie has stored all of her food on lily pad A. However, her food has a tendency to “fly” away. Every second, the food that’s on every lily pad splits up into six equal portions that instantaneously relocate to the six neighboring pads.

At zero seconds, all the food is on lily pad A. After one second, there’s no food on pad A, and 1/6 of the food is on each of the surrounding six pads. After two seconds, 1/6 of the food is again on pad A, while the rest of the food is elsewhere.

After how many seconds N (with N > 2) will pad A have less than 1 percent of its original amount?

Can You Hop in a Spiral?

Highlight to reveal (possibly incorrect) solution:

Desmos 1 Program Desmos 2

Method 1:

• I try to construct the path in Desmos. At every step, is it possible to turn 60° counter clockwise and still obey the rules? If not, continue straight ahead.
• There are just too many steps!

Method 2:

• I write a program to assist. I look at all 6 directions. But I also know which 2 are 60° counter clockwise and straight ahead. I look at the distance and the angle. And then I choose.
• I reuse Desmos to show the result.
• I end up at a distance 64.

Method 3:

• I can vaguely see a method, where I keep going straight ahead, until I reach x = 0, y = x/2k or y = – x/2k. (k = √3/2.) (These guide lines are also shown in Desmos.) Let’s try it out. Desmos 2 assists.
• I want my spiral to be as tight as possible. From any point, I test going back the same way, or 60° less, or 120° less, or 180° less, or 240° less, or 300° less, in that order. I am looking for small additions to the distance to A, hopefully getting large jumps in angle.
• |AX| denotes the distance between points A and X, where points A and X are the centers of pads A and X. α_X denotes the angle to point A from X.
• Looking for D.
  • From C = (1.5,k) I could go back to D0 = B. But |AD0| = |AB| < |AC|, as the rules state. Likewise α_D0 = α_B < α_C, as the rules state. Rejected.
  • I could go to D1. But |AD1| < |AC|. Rejected.
  • I could go to D2. |AC| = 1.73, |AD2| = 2. α_C = 30°, α_D2 = 60°.
  • I could go to D3. |AD3| = 2.65. α_D3 = 40.89°. D2 would be better, more angle, less distance.
  • I could go to D4. α_D4 = 19.11°. Rejected.
  • I could go to D5. α_D5 = 0°. Rejected.
• I choose D = D2 = (1,2k). |AD| = 2. α_D = 60°.
• The line through C and D crosses y = x/2k in C. It is y = -2kx + 4k.
• Looking for E.
  • E1 = D1. |AE1| = |AD1| < |AC| < |AD|. Rejected.
  • E2. |AE2| < |AD|. Rejected.
  • E3. |AE3| = 2.65. α_E3 = 79.11°.
  • E4. α_E4 = 60°. Not smaller than α_D. Rejected.
  • E5 = D3. α_E5 = 40.89°. Rejected.
• There’s only 1 choice, E3 = E = (0.5,3k). |AE| = 2.65. α_E = 79.11°.
• Looking for F. F1 fails both distance and angle. F2 has the same distance. F3 is okay. F4 and F5 both fail the angle rule. So I choose F3 = F = (0,4k). I am now on the line x = 0 and the line y = 4k.
• Going forward, I will choose G3, H3 etc. until I reach L3, on the line y = – x/2k. All these points are on the line y = 4k. Then I will turn left a little, choosing M2. And then go straight forward a bit.
• Let’s look at some of these distances a bit more detailed.
• Being in D = (1,2k), on the line y = -2kx + 4k, I am looking for E.
  • |AD|² = 1² + (2k)²
  • E1 = D + (-0.5,-k).
  • |AE1|² = (1-0.5)² + (2k-k)² < 1² + (2k)² = |AD|²
  • E2 = D + (-1,0).
  • |AE2|² = (1-1)² + (2k+0)²
    • = 1² – 2*1*1 + 1² + (2k)²
    • = 1² + (2k)² – 2*1*1 + 1²
    • = |AD|² – 2*1*1 + 1²
    • = |AD|² – 2 + 1
    • = |AD|² – 1
    • < |AD|²
  • E3 = D + (-0.5,k)
  • |AE3|² = (1-0.5)² + (2k+k)²
    • = 1² – 2*1*0.5 + 0.5² + (2k)² + 2*2k*k + k²
    • = 1² + (2k)² – 2*1*0.5 + 0.5² + 2*2k*k + k²
    • = |AD|² – 2*1*0.5 + 0.5² + 2*2k*k + k²
    • = |AD|² – 1 + 0.5² + 4k² + k²
    • = |AD|² – 0.75 + 5k²
    • = |AD|² + 3
  • Distance wise, E3 is the only candidate. (E4 and E5 are rejected because of the angle.)
• For a more general case, let’s look at the search for H, I etc.
• X = (x, 4k), x < 0
• Y1 = (x+0.5, 4k-k)
• Y2 = (x-0.5, 4k-k)
• Y3 = (x-1, 4k)
• Y4 = (x-0.5, 4k+k)
• Y5 = (x+0.5, 4k+k)
• |AX|² = x² + (4k)²
• |AY1|² = (x+0.5)² + (4k-k)²
  • = x² + 2*0.5*x + 0.5² + (4k)² – 2*4k*k + k²
  • = x² + (4k)² + 2*0.5*x + 0.5² – 2*4k*k + k²
  • = |AX|² + 2*0.5*x + 0.5² – 2*4k*k + k²
  • = |AX|² + x + 0.5² – 8k² + k²
  • = |AX|² + x + 0.5² – 7k²
• We want |AX|² < |AY1|²
  • <=> |AX|² < |AX|² + x + 0.5² – 7k²
  • <=> 0 < x + 0.5² – 7k²
  • <=> – 0.5² + 7k² < x
  • <=> – 0.25 + 7*(√3/2)² < x
  • <=> – 0.25 + 7 * 3/4 < x
  • <=> 5 < x
• But x < 0, so that won’t happen.
• |AY2|² = (x-0.5)² + (4k-k)²
  • = x² – 2*0.5*x + 0.5² + (4k)² – 2*4k*k + k²
  • = x² + (4k)² – 2*0.5*x + 0.5² – 2*4k*k + k²
  • = |AX|² – 2*0.5*x + 0.5² – 2*4k*k + k²
  • = |AX|² – x + 0.5² – 8k² + k²
  • = |AX|² – x + 0.5² – 7k²
• We want |AX|² < |AY2|²
  • <=> |AX|² < |AX|² – x + 0.5² – 7k²
  • <=> 0 < – x + 0.5² – 7k²
  • <=> x < 0.5² – 7k²
  • <=> x < 0.25 – 7*(√3/2)²
  • <=> x < 0.25 – 7 * 3/4
  • <=> x < -5
• This should certainly happen at some point, say x = -6. In that case we would have M2 = (-6.5, 3k).
• Let’s also look at Y3.
• |AY3|² = (x-1)² + (4k)²
  • = x² – 2x + 1 + (4k)²
  • = x² + (4k)² – 2x + 1
  • = |AX|² – 2x + 1
• We want |AX|² < |AY3|²
  • <=> |AX|² < |AX|² – 2x + 1
  • <=> 0 < – 2x + 1
  • <=> 2x < 1
  • <=> x < 0.5
• We already have that as x < 0.
• The turn to M2 happens at L3 = (-6, 4k).
  • x = -6
  • k = √3/2
  • 1/k = 2/√3
    • = 2√3/3
  • Does L3 lie on y = – x/2k?
  • y = – (-6) / (2k)
    • = 6/2k
    • = 6 * 2√3/3 / 2
    • = 6 * √3/3
    • = 2 * √3
    • = 4 * √3/2
    • = 4k
  • Yes, it does.
• I think I need some more details here. But they should be trivial.
• Sketch: Rotate the situation around A. L3 is now on x = 0. Do similar calculations. Etc.
• In conclusion: Within the guide lines, travelling on a line perpendicular to x = 0, y = x/2k or y = – x/2k, it is safe to turn left when a guide line is hit.

And for extra credit:

Spreadsheet

Method 1:

• I fiddle around with a spreadsheet, just to see what happens.
• There are just too many steps!

Method 2:

• I write a program. Markov chain like.
• I end up with N = 28.

This week the #puzzle is: Bingo! #statistics #probabilities #combinatorics #counting #coding #program #montecarlo

A game of bingo typically consists of a 5-by-5 grid with 25 total squares. Each square (except for the center square) contains a number. When a square’s number is called, you place a marker on that square. The goal is to get “bingo,” which is five squares in a row, either across, down, or along one of the two long diagonals. The center square, which doesn’t have a number, is labeled “Free,” and begins with a marker on it before any numbers are called. Here’s an example of a winning 5-by-5 grid in which 10 squares (other than the “Free” square) have been marked:

Consider a smaller version of the game with a 3-by-3 grid: a “Free” square surrounded by eight other squares with numbers. Suppose each of these eight squares is equally likely to be called, and without replacement (i.e., once a number is called, it doesn’t get called again).

On average, how many markers must you place until you get “bingo” in this 3-by-3 grid? (The “Free” square doesn’t count as one of the markers—it’s “free”.)

And for extra credit:

Instead of a 3-by-3 grid, let’s return to the original 5-by-5 grid.

On average, how many markers must you place until you get “bingo”? (As before, the “Free” square doesn’t count as one of the markers—it’s “free”.)

Bingo!

Highlight to reveal (possibly incorrect) solution:

Program By hand

Method 1: Monte carlo. See program.

Method 2: Go through all options. See program.

Method 3: Go through all options by hand.

Result: 3.471429 markers.

And for extra credit:

Method 1 and 2, from above.

Result: 13.608084 markers.

I had to optimize the program quite a lot to get a reasonable run time. Memoization. And taking into account, that each setup exists in 8 versions, rotated and mirrored. As always it’s a joy when a recursive function suddenly works.