Dueling Pennies

A certain strange casino offers only one game. The casino posts a positive integer n on the wall, and the customer flips a fair coin repeatedly until it falls tails. If he has tossed n – 1 times, he pays the house 8n – 1 dollars; if he’s tossed n + 1 times, the house pays him 8n dollars; and in all other cases the payoff is zero.

The probability of tossing the coin exactly n times is 1/2n, so the customer’s expected winnings are 8n/2n + 1 – 8n – 1/2n – 1 = 4n – 1 for n > 1, and 2 for n = 1. So his expected gain is positive.

But suppose it turns out that the casino arrived at the number n by tossing the same fair coin and counting the tosses, up to and including the first tails. This presents a puzzle: “You and the house are behaving in a completely symmetric manner,” writes David Gale in Tracking the Automatic ANT (1998). “Each of you tosses the coin, and if the number of tosses happens to be the consecutive integers n and n + 1, then the n-tosser pays the (n + 1)-tosser 8n dollars. But we have just seen that the game is to your advantage as measured by expectation no matter what number the house announces. How can there be this asymmetry in a completely symmetric game?”

Visual Calculus

https://commons.wikimedia.org/wiki/File:Cycloid_(PSF).png

As a circle rolls along a line, a point on its circumference traces an arch called a cycloid. The arch encloses an area three times that of the circle, a result commonly proven using calculus. Now Armenian mathematician Mamikon Mnatsakanian has devised a “sweeping-tangent theorem” that accomplishes the same proof using intuition:

https://commons.wikimedia.org/wiki/File:Mamikon_Cycloid.svg
Image: Wikimedia Commons

Imagine a tangent to the rolling circle. As the circle rolls, the tangent sweeps out a series of vectors (approximated here using colors). If these vectors are then gathered to a common point while preserving their length and orientation, they form a sort of bouquet whose size and shape turn out to match exactly those of the original circle. Because the enclosing rectangle has four times the area of the rolling circle (2πr × 2r = 4πr2), this shows that the area under the arch has three times the circle’s area.

All this is proven rigorously in Mnatsakanian’s 2012 book New Horizons in Geometry, written with his Caltech colleague Tom Apostol. The two have now collaborated on some 30 papers showing that many surprising and useful results that heretofore had required integration can now be obtained using intuitive methods that can appeal even to a young student.

That’s a welcome outcome for Mnatsakanian, who found himself stranded in the United States when the Armenian government collapsed in 1990. Apostol writes, “When young Mamikon showed his method to Soviet mathematicians they dismissed it out of hand and said ‘It can’t be right. You can’t solve calculus problems that easily.'”

Hesiod’s Anvil

https://commons.wikimedia.org/wiki/File:Acme_anvil.gif

How far off is heaven? In the Theogony Hesiod gives us a clue:

For a brazen anvil falling down from heaven nine nights and days would reach the earth upon the tenth; and again, a brazen anvil falling from earth nine nights and days would reach Tartarus upon the tenth.

How far can an anvil fall in nine days? Galileo, who taught that “the distances measured by the falling body increase according to the squares of the time,” would have determined that the anvil starts 2.96 × 109 km from earth, a distance greater than that between the sun and Uranus.

But Galileo’s calculation assumes that gravitational force is independent of the object’s distance from the earth. If we assume instead that it varies inversely with the square of the distance between mass centers (and if we ignore all masses except those of the earth and the anvil, and assume that the anvil falls in a straight line), King College mathematician Andrew Simoson calculates that Galileo’s anvil wouldn’t reach us for

hesiod's anvil calculation

Instead, under this new assumption, to reach us in nine days an anvil would start 5.81 × 105 km away — about one and a half times the distance between the earth and the moon.

(Andrew J. Simoson, Hesiod’s Anvil, 2007.)

Rolling Average

http://www.freeimages.com/photo/549165

In a standard 10-frame game of bowling, the lowest possible score is 0 (all gutterballs) and the highest is 300 (all strikes). An average player falls somewhere between these extremes. In 1985, Central Missouri State University mathematicians Curtis Cooper and Robert Kennedy wondered what the game’s theoretical average score is — if you compiled the score sheets for every legally possible game of bowling, what would be the arithmetic mean of the scores?

It turns out it’s pretty low. There are (669)(241) possible games, which is about 5.7 × 1018. If we divide that into the total number of points scored in these games, we get

mean bowling score

which is about 80 (79.7439 …).

This “might make you feel better about your average,” Cooper and Kennedy conclude. “The mean bowling score is indeed awful even if you are just an occasional bowler. Even though this information is interesting, there are more difficult questions about the game of bowling that could be asked. For example, you might wish to determine the standard deviation of the set of bowling scores and hence know more about the distribution of the set of all bowling scores. But the exact determination of the distibution of the set of scores is, in our opinion, a difficult problem. For example, given an integer k between 0 and 300, how many different bowling games have the score k? This, we leave as an open problem.”

(Curtis N. Cooper and Robert E. Kennedy, “Is the Mean Bowling Score Awful?”, Journal of Recreational Mathematics 18:3, 1985-86.)

Golomb Rulers

https://commons.wikimedia.org/wiki/File:Golomb_ruler_conference_room.svg
Image: Wikimedia Commons

This conference room is 11 units long and has folding partitions at positions 2, 7, and 8. This gives it a curious property: For a meeting of any given size, the room can be configured in exactly one way. A meeting of size 6 must use partitions 2 and 8 — no other setup will work exactly.

This is an example of a Golomb ruler, named for USC mathematician Solomon Golomb. It’s called a ruler because the simplest example is a measuring stick: If we’re given a 6-centimeter ruler, we find that we can add 4 marks (at integer positions) so that no two of them are the same distance apart: 0 1 4 6. No shorter ruler can accommodate 4 marks without duplication, so the 0-1-4-6 ruler is said to be “optimal.” It’s also “perfect” because it can measure any distance from 1 to 6.

The conference room is optimal because no shorter room can accommodate 5 walls without equal-sized partitions becoming available, but it’s not perfect, because it can’t accommodate an assembly of size 10. (It turns out that no perfect ruler with five marks is possible.)

Finding optimal Golomb rulers is hard — simply extending an existing ruler tends to produce a new ruler that’s either not Golomb or not optimal. The only way forward, it seems, is to compare every possible ruler with n marks and note the shortest one, an immensely laborious process. Distributed computing projects have found the longest optimal rulers to date — the most recent, with 27 marks, was found in February, five years after the previous record.

Far From Home

https://commons.wikimedia.org/wiki/File:Peacock_served_in_full_plumage_(detail_of_BRUEGHEL_Taste,_Hearing_and_Touch).jpg

This is a detail from the allegorical painting Taste, Hearing and Touch, completed in 1620 by the Flemish artist Jan Brueghel the Elder. If the bird on the right looks out of place, that’s because it’s a sulphur-crested cockatoo, which is native to Australia. The same bird appears in Hearing, painted three years earlier by Brueghel and Peter Paul Rubens.

How did an Australian bird find its way into a Flemish painting in 1617? Apparently it was captured during one of the first Dutch visits to pre-European Australia, perhaps by Willem Janszoon in 1606, who would have carried it to the Dutch East Indies (Indonesia) and then to Holland in 1611. That’s significant — previously it had been thought that the first European images of Australian fauna had been made during the voyages of William Dampier and William de Vlamingh, which occurred decades after Brueghel’s death in 1625.

Warwick Hirst, a former manuscript curator at the State Library of New South Wales, writes, “While we don’t know exactly how Brueghel’s cockatoo arrived in the Netherlands, it appears that Taste, Hearing and Touch, and its precursor Hearing, may well contain the earliest existing European images of a bird or animal native to Australia, predating the images from Dampier’s and de Vlamingh’s voyages by some 80 years.”

(Warwick Hirst, “Brueghel’s Cockatoo,” SL Magazine, Summer 2013.) (Thanks, Ross.)

Warnsdorff’s Rule

The knight’s tour is a familiar task in chess: On a bare board, find a path by which a knight visits each of the 64 squares exactly once. There are many solutions, but finding them by hand can be tricky — the knight tends to get stuck in a backwater, surrounding by squares that it’s already visited. In 1823 H.C. von Warnsdorff suggested a simple rule: Always move the knight to a square from which it will have the fewest available subsequent moves.

This turns out to be remarkably effective: It produces a successful tour more than 85% of the time on boards smaller than 50×50, and more than 50% of the time on boards smaller than 100×100. (Strangely, on a 7×7 board its success rate drops to 75%; see this paper.) The video above shows a successful tour on a standard chessboard; here’s another on a 14×14 board:

While we’re at it: British puzzle expert Henry Dudeney once set himself the task of devising a complete knight’s tour of a cube each of whose sides is a chessboard. He came up with this:

http://www.gutenberg.org/files/16713/16713-h/16713-h.htm#X_340_THE_CUBIC_KNIGHTS_TOURa

If you cut out the figure, fold it into a cube and fasten it using the tabs provided, you’ll have a map of the knight’s path. It can start anywhere and make its way around the whole cube, visiting each of the 364 squares once and returning to its starting point.

Dudeney also came up with this puzzle. The square below contains 36 letters. Exchange each letter once with a letter that’s connected with it by a knight’s move so that you produce a word square — a square whose first row and first column comprise the same six-letter word, as do the second row and second column, and so on.

dudeney knight's tour word square puzzle

So, for example, starting with the top row you might exchange T with E, O with R, A with M, and so on. “A little thought will greatly simplify the task,” Dudeney writes. “Thus, as there is only one O, one L, and one N, these must clearly be transferred to the diagonal from the top left-hand corner to the bottom right-hand corner. Then, as the letters in the first row must be the same as in the first file, in the second row as in the second file, and so on, you are generally limited in your choice of making a pair. The puzzle can therefore be solved in a very few minutes.”

Click for Answer

Different Strokes

http://www.freeimages.com/photo/1063759

G.H. Hardy had a famous distaste for applied mathematics, but he made an exception in 1945 with an observation about golf. Conventional wisdom holds that consistency produces better results in stroke play (where strokes are counted for a full round of 18 holes) than in match play (where each hole is a separate contest). So if two players complete a full round with the same total number of strokes, then the more erratic player should do better if they compete hole by hole.

Hardy argues that the opposite is true. Imagine a course on which every hole is par 4. Player A is so deadly reliable that he shoots par on every hole. Player B has some chance x of hitting a “supershot,” which saves a stroke, and the same chance of hitting a “subshot,” losing a stroke. Otherwise he shoots par. Both players will average par and will be equal over a series of full rounds of golf, but the conventional wisdom says that B’s erratic play should give him an advantage if they play each hole as a separate contest.

Hardy’s insight is that the presence of the hole limits a run of good luck, while there’s no such limit on a run of bad luck. “To do a three, B must produce a supershot at one of his first three strokes, while he will take a five if he makes a subshot at one of his first four. He will thus have a net expectation 4x – 3x of loss on the hole, and should lose the match, contrary to common expectation.”

In general he finds that B’s chance of winning a hole is 3x – 9x2 + 10x3, and his chance of losing is 4x – 18x2 + 40x3 – 35x4, so that there’s a balance f(x) = x – 9x2 + 30x3 – 35x4 against him. If x < 0.37 -- that is, in all realistic cases -- the erratic player should lose. "If experience points the other way -- and I cannot deny it, since I am no golfer -- what is the explanation? I asked Mr. Bernard Darwin, who should be as good a judge as one could find, and he put his finger at once on a likely flaw in the model. To play a 'subshot' is to give yourself an opportunity of a 'supershot' which a more mechanical player would miss: if you get into a bunker you have an opportunity of recovering without loss, and one which you are naturally keyed up to take. Thus the less mechanical player's chance of a supershot is to some extent automatically increased. How far this may resolve the paradox, if it is one, I cannot say, and changes in the model make it unpleasantly complex." (G.H. Hardy, "A Mathematical Theorem About Golf," Mathematical Gazette, December 1945.)