For this post, I'd like to turn to a variation of a classic Henry Dudeney puzzle, from his book Amusements in Mathematics. It can also be found in Martin Gardner's October 1960 Scientific American column, or his book, New Mathematical Diversions, as the 5th puzzle ,“Bisecting Yin and Yang”, in chapter 12, “Nine Problems”.

As you've no doubt guessed, this puzzle involved the yin (dark) yang (light) symbol. For this puzzle, I've drawn it in a very mathematically precise way over at Desmos. The outside is a unit circle (so, the radius is 1 unit), the main semicircular divisions of the design have a radius of ¹⁄₂ unit, and the opposite-color dots have a radius of ¹⁄₆ unit. Here's the challenge: What's the equation of the line that divides the design so that each side of the line contains exactly equal amounts of the dark and light portions?



As with many puzzles, this one seems hard, until you break it down into simpler steps. Let's start with a much easier puzzle: If the top half of the puzzle were dark, and the bottom half light, as in this rendering, where would you draw the line? The answer is easy. It should be a vertical line, so the equation would be x=0.



Next, we change the design a bit, so as to get closer to the yin yang symbol. Starting with the previous design, we cut a half circle (¹⁄₂ unit radius, remember?) of the dark portion from the right side, and add that to the left side, giving us the design below. The vertical line obviously won't work anymore, and we'll need to rotate that line by some amount to compensate, but how much?



Again, the secret is to take steps slowly. If you remember your high school geometry, you remember that the formula for a circle is A=πr², and that our design as a whole, being a unit circle, has an area equal to π.

The formula for the semicircle we've moved, then, is A=¹⁄₂πr². Plugging in ¹⁄₂ for the radius, we get ^π⁄₈ units. So, to compensate for ^π⁄₈ units out of a full circle with an area of π-units, we simply rotate the formerly-vertical line counterclockwise by ¹⁄₈ of the circle, or 45°! The upper left quadrant completely dark, so that makes this adjustment simple. The blue line is the dividing line for this design:



This, in fact, is the answer to the original puzzle as posed by Dudeney and Gardner. This is NOT the answer to the problem I posed above. When I first ran across this puzzle, it annoyed me that it wasn't done with the full yin yang symbol. The dots are a symbol of how, in nature, nothing is purely one thing or the other, and are a very important part of the design.

It's time to go back to the full design. Compared to the previous step, we're going to be removing some of the dark area from the right side to the left side. This means that we'll end up rotating our dividing line some distance clockwise this time, and we need to figure out by how much. Yes, once again, we'll be using our area formulas to work out the adjustment. We even know that the result should be easy to interpret, since the result will π over something, and this comes out of a π-unit circle.



The dots, of course, are full circles, so we use the formula for the area of a full circle once again. The dots have a radius of ¹⁄₆ of a unit, and plugging that into the formula, we get ^π⁄₃₆. In other words, the line needs to be moved back clockwise ¹⁄₃₆ of a full circle, or 10°. That brings the line to being 35° off of the original vertical line.

A 35° line must be our answer, right? Wrong. Go back and look at the original question: Here's the challenge: What's the equation of the line that divides the design so that each side of the line contains exactly equal amounts of the dark and light portions? We need to work out enough details for the line equation y = mx + b, where m is the slope, b is where the line crosses the y-axis, and y and x remain as variables. The line, of course, crosses the y-axis at 0, so b = 0. That reduces the equation to just y = mx, so we need to figure out the slope.

First, angles are usually measured in relation to the positive x axis, so we're actually talking about a 125° (35° + 90°), or ^25π⁄₃₆ radians (Confused? Read Intuitive Guide to Angles, Degrees and Radians). In geometry, we'd say we were trying to calculate rise over run (rise ÷ run). In trigonometry, we're trying to calculate the opposite site of the angle by the adjacent side (Confused? Read How To Learn Trigonometry Intuitively), and that means we need to use the tangent!

So, the equation is y = tan(^25π⁄₃₆)x, or y = tan(125°)x, if you prefer. The actual slope is an irrational number which is roughly equal to -1.428148. There you have it, the equation to a line which divides the design into equal parts of light and dark, as shown below.

Blame Marilyn vos Savant. Back in 1990, Craig F. Whitaker of Columbia, Maryland wrote to her with a probability puzzle, and found he'd kicked up a hornet's nest! He asked, “Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, 'Do you want to pick door #2?' Is it to your advantage to switch your choice of doors?”

Marilyn replied that, if you switch, your odds of winning the car are ⅔, and if you don't switch, your odds are only ⅓. It was difficult for many to believe. Even subsequent discussions about these probabilities, such as the Scam School episode on the Monty Hall Dilemma, find that the belief in a 50-50 chance prevails.

Despite all the numerous ways there are to explain it, practical demonstration if often the most effective way to see that the ⅔ odds of winning is correct.

Oxford Mathematics Professor Marcus Du Sautoy shows the effectiveness of practical demonstration to English comedian Alan Davies when it comes to the Monty Hall Dilemma:



While the practical demonstration in this video is effective, it's a little surprising that the switch approach won 4 times as often as it lost. This is one of the classic problems with using a small sample size (such as playing this game only 20 times). Over at Epanechnikov's blog entry on the Monty Hall Dilemma, he features a graph of repeated simulations that shows the problem with just 20 runs:



The probabilities don't even really start settling down to the calculations until about 300 trials have been run! To help see the true odds, why not use a computer to run thousands of simulations very quickly? Inspired by the spreadsheet approach used by Presh Talwalkar to simulate trials for a different probability puzzle, I decided to do the same for the Monty Hall Dilemma.

How do you set it up? The first column states the door which holds the car, and this is generated as a random integer ranging from 1 to 3. The second column states the door chosen by the player, and this is also generated as a random integer ranging from 1 to 3.

The next column is a little trickier. It's going to hold the door which is shown by the host, but there are restrictions on which door can be shown by the host, so we can't just randomly generate a number. The host will only show a door that was NOT chosen by the player and that the host knows will contain a goat. How do we communicate these restrictions to a spreadsheet? There are 2 cases to consider here. First, what happens when the door which contains the car DOESN'T match the door chosen by the player, such as when the car is behind door #1 and the player initially chooses door #2? In this case, the host can only show door #3. In fact, since 1 + 2 + 3 = 6, we can simply subtract the number of the door with the car and the number of the door chosen by the player from 6 to get the number of the door shown by the host.

That only works when the door chosen by the player and the door holding the car are different. What do we do when those two doors are the same? If the player chooses door #1 and the car is behind door #1, we can have the computer choose randomly between door #2 and door #3. A similar approach is used for the other 2 doors, of course. The final spreadsheet entry reads this way:

=IF(A6=B6,IF(RANDBETWEEN(1,2)=1,CHOOSE(A6,3,1,2),CHOOSE(A6,2,3,1)),6-A6-B6)

Would you believe tha another of my contributions has made it on to Scam School again? It was 2 other recent Scam School submissions that spurred me to restart Grey Matters, so it's looking like that was the right move.

Even if you've seen this week's Scam School episode, you may want to take a look at this post, as I'm going to give a few tips that may make this routine easier to learn.

Let's get started right away with this week's Scam School episode with a trick I dubbed “Out of Control”!



Quick side note: On one hand, I love being promoted as "the genius". On the other hand, I can't help but think of “genius” in this context.

This trick is actually a combination of two idea from two men who have far more a right to be called genius than me. The dealing procedure comes straight from Jim Steinmeyer's routine Remote Control, as published in Invocation #43 and the May 1993 issue of MAGIC Magazine. If you check those sources out, you'll see that not much has changed, as the original involves spelling the word C-O-L-O-R, and using the 9th card.

I combined this trick with a technique from Simon Aronson's “Try The Impossible” called Simon's Flash Speller. It's this part that may help make it easier to work out what you need to do. First, you'll need to quickly work out how many letters are in the name of the turned-up card. Here's the starting point:

• For clubs, remember: 11 letters
• For hearts or spades, remember: 12 letters
• For diamonds, remember: 14 letters

A little over 4 years ago, I wrote a post inspired by Alexander Craig Aitken's approach to dividing by numbers ending in 9, called Leapfrog Division. It's a remarkably fun and simple technique, so if you haven't already checked it out for yourself, give it a read.

I wrote 3 more posts in that series. Leapfrog Division II dealt with dividing by numbers ending in 1. My last 2 posts before this year were Leapfrog Division III, which dealt with dividing by numbers ending in 8, and Leapfrog Division IV, which dealt with dividing by numbers ending in 2.

These were fine, but the latter 3 posts employed rules that became increasingly cumbersome, and were tricky to apply quickly and without error. Recently, however, I learned a far simpler approach that makes these later posts much simpler!

Let's start with credit where credit is due. The thanks should go to Saurabh G over at Hubpages. He had a post at Owlcation titled Divide Numbers Easily Using Vedic Mathematics: Fast and Easy Division Techniques. Towards the end of the post, there's a section with the heading, “How Do You Use Vedic Division When the Divisor Is More Than One Digit?”. The approach taught with numbers ending in 9 is demonstrated a little differently, but is mathematically identical to the original Leapfrog Division post.

The next section, “Multi-Digit Divisor Ending in 8 Example,” is what really struck me the most. He teaches an almost identical approach, and casually mentions that the quotient needs to be doubled before adding the remainder to the 10s place. Click on that link, read through the example, and then compare that approach to what I taught in Leapfrog Division III. I do mention a doubling idea, but the rules I taught are far more complex.

Using Saurabh G's vedic math approach, here's how his example (73 ÷ 138) would look written in the style of the Leapfrog Division posts. We start with the idea that 14 won't go into 7.3, so we start with a 0 and decimal point at the beginning of the answer, and work from there:

• 73 ÷ 14 = 5 (remainder 3)
  (Double the 5 to get 10, add the 3 in the 10s place to get 40)
• 40 ÷ 14 = 2 (remainder 12)
  (Double the 2 to get 4, add the 12 in the 10s place to get 124)
• 124 ÷ 14 = 8 (remainder 12)
  (Double the 8 to get 16, add the 12 in the 10s place to get 136)
• 136 ÷ 14 = 9 (remainder 10)
  (Double the 9 to get 18, add the 10 in the 10s place to get 118)
• 118 ÷ 14 = 8 (remainder 6)
  (Double the 8 to get 16, add the 6 in the 10s place to get 76, and continue on from here, if desired....)