Destiny

https://commons.wikimedia.org/wiki/Category:OpenClipart_ornamental_playing_cards

Arrange cards with values ace through 9 in a row, in counting order, with the ace on the left.

Take up a card from one end of the row — left or right, your choice.

Do this twice more, each time taking up either the leftmost or the rightmost card in the remaining row.

When you have three cards, add their values, divide the total by six, and call the result n. Count the cards that remain on the table from left to right.

The card in the nth position will be the 4.

Foursquare

http://natgeofound.tumblr.com/post/44756975727/alexander-graham-bell-and-mabel-kissing-within-a

Alexander Graham Bell kisses his daughter Daisy inside a tetrahedral kite, October 1903.

Bang’s theorem holds that the faces of a tetrahedron all have the same perimeter only if they’re congruent triangles. Also, if they all have the same area, then they’re congruent triangles.

Buckminster Fuller proposed establishing a floating tetrahedron in San Francisco Bay called Triton City (below). It would have been assembled from modules, starting with a floating “neighborhood” of 5,000 residents, with an elementary school, a supermarket and a few specialty shops. Three to six neighborhoods would form a town, and three to seven towns would form a city. At each stage the corresponding infrastructure would be added: schools, civic facilities, government offices, and industry. A full-sized city might accommodate 100,000 people in a single building. He envisioned an even larger tetrahedron, with a million citizens, for Tokyo Bay.

The moral of Fuller’s 1975 book Synergetics was “Dare to be naïve.”

Fuller Triton City

Two-Timing

The square root of 2 is 1.41421356237 … Multiply this successively by 1, by 2, by 3, and so on, writing down each result without its fractional part:

two-timing 1

Beneath this, make a list of the numbers that are missing from the first sequence:

two-timing 2

The difference between the upper and lower numbers in these pairs is 2, 4, 6, 8 …

From Roland Sprague, Recreations in Mathematics, 1963.

Kirkman Schoolgirls Problem

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“Fifteen young ladies in a school walk out three abreast for seven days in succession: it is required to arrange them daily, so that no two shall walk twice abreast.”

The Rev. Thomas Penyngton Kirkman posed this query innocently in The Lady’s and Gentleman’s Diary in 1850. It’s trickier than it looks — though seven solutions are possible, they’re difficult to discover by trial and error. Here’s one:

kirkman schoolgirl solution

Each letter appears once in each row, but no two letters share a cell more than once. The problem’s simplicity made it popular among Victorian amateur mathematicians, and Kirkman later complained that it had eclipsed his more serious work — though he took pains to dispute James Joseph Sylvester’s claim to have invented the problem himself.

Do such problems generally have solutions? Surprisingly, the answer remained unknown until just last January, when Oxford mathematician Peter Keevash showed that the answer is yes if certain basic requirements are met. Keevash’s result was “a bit of an earthquake as far as design theory is concerned,” said Cambridge mathematician Timothy Gowers.

Now that the schoolgirl problem is under control, a related challenge has taken its place. If 20 golfers want to arrange themselves into different foursomes on five successive days, is it possible to plan the groups so that each golfer plays no more than once with any other golfer? Formally the “social golfer problem” remains unsolved — though tournament organizers work out individual solutions every day.

(Thanks, Martha.)

Color Scheme

http://www.sxc.hu/photo/434247

Amateur magician Oscar Weigle invented this surprising effect in 1949. Assemble a deck of 20 playing cards, 10 red and 10 black, in strictly alternating colors. Hold this deck under a table. Now turn over the top two cards as one, place them on top, and cut the deck. Repeat this procedure as many times as you like — turn two, cut, turn two, cut. When you’ve finished, the deck will contain an unknown number of reversed cards, distributed randomly.

Now, still holding the deck under the table, shift the top card to the bottom, then turn over the next card and place it on the table. Do this repeatedly — shift a card to the bottom, then reverse the next card and put it on the table — continuing until you’ve put 10 cards on the table. Surprisingly, these cards are sorted by color — the face-up cards are of one color, and the face-down ones are of the other.

You’re still holding 10 cards under the table. Divide these into two stacks and weave them together under the table randomly. Do this as many times as you like — divide the 10 cards into two groups and merge them together however you like, so long as no card is turned upside down. Turn over the packet and shuffle it in the same way a few more times. Give it a final cut if you like.

Now deal these cards out as before: Shift the top card to the bottom, reverse the next card and put it on the table. Like the first group, this one will sort itself by color, with one color face up and the other face down.

Surprise Appearance

Suppose we put eight white and two black balls into a bag and then draw forth the balls one at a time. If we repeat this experiment many times, which draw is most likely to produce the first black ball?

Most people answer 4, but in fact the first black ball is most likely to appear on the very first draw:

surprise appearance table

By symmetry, the second black ball is most likely to appear on the final draw.

(A.E. Lawrence, “Playing With Probability,” Mathematical Gazette, vol. 53 [December 1969], 347-354.)

The Friendship Theorem

https://en.wikipedia.org/wiki/File:Friendship_graphs.svg
Image: Wikimedia Commons

If every pair of people in a group have exactly one friend in common, then there’s always one person who is a friend to everyone.

This rather heartwarming fact was proven by Paul Erdös, Alfréd Rényi, and Vera T. Sós in 1966. It’s sometimes more cynically known as the paradox of the politician.

The Erdös proof uses combinatorics and linear algebra, but in 1972 Judith Longyear published a proof using elementary graph theory.

See The Elevator Problem.

Three Coins

Three coins are lying on a table: a quarter, a half dollar, and a silver dollar. You claim one coin, I’ll claim the other two, and then we’ll toss all three. A coin that lands tails counts zero, and a coin that lands heads wins its value (in cents, 25, 50, or 100) for its owner. Whichever of us has the larger score wins all three coins. If all three coins land tails then we toss again.

Which coin should you claim to make the game fair — that is, so that each of us has an expected win of zero?

Click for Answer