Heads and Tails

Let’s play a coin-flipping game. At stake is half the money in my pocket. If the coin comes up heads, you pay me that amount; if it comes up tails, I pay you.

Initially this looks like a bad deal for me. If the coin is fair, then on average we should expect equal numbers of heads and tails, and I’ll lose money steadily. Suppose I start with $100. If we flip heads and then tails, my bankroll will rise to $150 but then drop to $75. If we flip tails and then heads, then it will drop to $50 and then rise to $75. Either way, I’ve lost a quarter of my money after the first two flips.

Strangely, though, the game is fair: In the long run my winnings will exactly offset my losses. How can this be?

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It’s true that if the number of wins equals the number of losses, as expected, I’ll lose money. But the sequence of flips won’t always match our expectations so precisely — it may favor heads or tails. And when it favors heads, I’ll make a lot of money. Suppose we play four games in a row, with two flips per game, starting with $100: Here I lose fully 75 percent of the games we play — but my winnings are so great in the fourth case that they offset the whole foregoing series of (relatively small) losses. Such cases are rare, but they’re so lucrative that in the long run my bankroll is unchanged.

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August 28, 2012February 6, 2013 | Puzzles · Science & Math

Black and White

“The Forlorn Hope,” by J. Paul Taylor, 1878. White to mate in two moves.

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The White King looked around, and noted how Our ranks were thinned and broken, so that now Our small successes were not worth the cost; And said, “We must resign, the day is lost!” “But not the Knight!” so spake a horseman brave: “My steed says neigh, the day we yet may save.” Then turned the King, and, frowning darkly, said: “Such jesting, good Sir Knight, may cost your head. He who would make a pun at such a time Is capable of any other crime; But one more chance I give you — Forward ride, And lead a hope forlorn. If good betide, Perhaps I may forgive you for the pun, And if you fall but little harm is done!” Forth spurred the Knight, regardless of his life, And rode at once where thickest raged the strife. To such a spot he dashed, that whilst he braved Five sorts of slaughter, still the day he saved; For should the foeman dare to lay him low, He leaves his King a prey to vassal’s blow; And should the King himself the victim slay, A priest steps in with stroke that none can stay; Whilst if they should decide to spare the Knight, And try to shield their Monarch from his might, They yet would fail, for still his falchion bright Descends with fell effect from left or right.

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August 24, 2012February 6, 2013 | Puzzles

“Logic”

Cries logical Bobby to Ned, will you dare
A bet, which has most legs, a mare, or no mare.
A mare, to be sure, replied Ned, with a grin,
And fifty I’ll lay, for I’m certain to win.
Quoth Bob, you have lost, sure as you are alive,
A mare has but four legs, and no mare has five.

— The Panorama of Wit, 1809

August 20, 2012August 20, 2012 | Puzzles

Ball Juggling

http://commons.wikimedia.org/wiki/File:Geometry_of_Solar_eclipses_and_Lunar_eclipses.svg — Image: Wikimedia Commons

In a solar eclipse, the moon casts its shadow on Earth. In a lunar eclipse, Earth casts its shadow on the moon.

Solar eclipses are more common than lunar eclipses, but we tend to have the opposite impression. Why?

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Because the moon’s shadow is smaller than Earth’s, a total solar eclipse is visible only from within a relatively narrow latitude on Earth’s surface, while a lunar eclipse is visible from anywhere on the night side of the planet. For the same reason, a total solar eclipse lasts for only a few minutes at any given location, while a lunar eclipse lasts for several hours.

August 20, 2012May 6, 2013 | Puzzles

Black and White

By Philip Hamilton Williams, Evening News and Post, 1895. White to mate in two moves.

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1. Qb1 and every Black move loses.

August 17, 2012February 6, 2013 | Puzzles

Up From Down

Mount Everest rises 29,029 feet above sea level, and Ecuador’s volcano Chimborazo rises only 20,702 feet. But because Earth bulges at the equator, Chimborazo is actually farther from the center of the planet. If we could connect the two peaks with a water pipe, in which direction would the water flow?

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Sea level gives us a convenient equilibrium water level. Since Everest rises higher, the water would flow toward Chimborazo.

August 12, 2012February 6, 2013 | Puzzles

Black and White

pierce chess problem

By W. Timbrell Pierce, 1873. White to mate in two moves.

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1. Qh5 Ke4 2. Rxc4#. (If the knight moves then the queen can mate immediately from e5.)

August 10, 2012February 6, 2013 | Puzzles

Spin City

Imagine two concentric roulette wheels, each divided into 100 sectors. Choose 50 sectors at random on each wheel, paint them black, and paint the rest white. Prove that we can now position the wheels so that at least 50 of the aligned sectors match.

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Follow a sector on the inner wheel through a complete revolution: That sector will find exactly 50 matches. The same is true for each of the 100 sectors on the wheel; altogether, as the wheel turns through 100 positions, there will be 100 × 50 = 5,000 matches. This means that the average number of matches per position is 50, so a position must exist with at least 50 matches.

August 9, 2012February 6, 2013 | Puzzles

Horse Play

horse play 1

A carnival worker is asked to paint the deck of a carousel. Because the center of the carousel is occupied by machinery, he can’t measure its diameter or even its radius. The best he can do is to take the measurement shown in green, which is 42 feet.

He’s explaining this apologetically when his supervisor stops him. “That’s all the information we need,” he says. “That’s enough to tell us how much paint to buy.”

How did they go about it?

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The area of the deck is the area of the large circle minus the small circle. That’s πA² – πB², or π(A² – B²). We don’t know A or B, but we do know the length of the green line. That’s 42. So the Pythagorean theorem can tell us that B² + 21² = A², and hence that 21² = A² – B². Happily, we can plug this directly into the equation above and get our answer: The area of the deck is 21²π, or 441π square feet — regardless of the size of the carousel! Remarkably, this is true — so long as the green chord stays locked in position between them, the two circles can assume any size and the difference in their areas will remain constant. A similar principle holds in three dimensions. In The Universal Book of Mathematics, David Darling writes, “An even more amazing fact is that if you slide a chord of fixed length around any convex shape C so that the chord’s midpoint traces out another figure D, the area between C and D doesn’t depend on what shape you started with.” (Thanks, Dave.)

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August 8, 2012February 6, 2013 | Puzzles

Leap Day

Alice gets a rocket-powered pogo stick for her birthday. She jumps 1 foot on the first hop, 2 feet on the second, then 4, 8, and so on. This gets alarming. By judicious hopping, can she arrange to return to her starting point?

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No. After n hops, Alice will have traveled at most 1 + 2 + 4 + … + 2^n-1 = 2ⁿ – 1 feet from the starting point. But her very next hop will carry her 2ⁿ feet. So she must always overshoot the goal.

August 7, 2012February 6, 2013 | Puzzles