Imagine a staircase with 1001 stairs, numbered -500 to 500. You’re standing in the middle, on stair 0, and you want to reach the top. On each step you can play either of two coin-flipping games — if the result is heads then you move up a step; if it’s tails then you move down a step:

• In game 1 you flip coin A, which is slightly biased: It comes up heads 49.5 percent of the time and tails 50.5 percent.
• In game 2 you use two coins, B and C. Coin B produces heads 9.5 percent of the time and tails 90.5 percent. Coin C produces heads 74.5 percent of the time and tails 25.5 percent. In game 2 if the number of the stair you’re on is a multiple of 3 then you flip coin B; otherwise you flip coin C.

Both of these are losing games — if you played either game 1 or game 2 exclusively, you’d eventually find yourself at the bottom of the staircase. But in 1996 Spanish physicist Juan Parrondo found that if you play the two games in succession in random order, keeping your place on the staircase as you switch between them, you’ll rise to the top of the staircase. It’s not, properly speaking, a paradox, but it’s certainly counterintuitive.

This example is from David Darling’s Universal Book of Mathematics. (Thanks, Nick.)

We’ve mentioned before that you can estimate π by dropping needles on the floor. (Reader Steven Karp also directed me to this remarkable solution, from Daniel A. Klain and Gian-Carlo Rota’s Introduction to Geometric Probability [1997].)

Here’s a related curiosity. If a circle of diameter L is placed at random on a pattern of circles of unit diameter, which are arranged hexagonally with centers C apart, then the probability that the placed circle will fall entirely inside one of the fixed circles is

If we put k = C/(1 – L), we get

And a frequency estimate of P will give us an estimate of π.

Remarkably, in 1933 A.L. Clarke actually tried this. In Scripta Mathematica, N.T. Gridgeman writes:

His circle was a ball-bearing, and his scissel a steel plate. Contacts between the falling ball and the plate were electrically transformed into earphone clicks, which virtually eliminated doubtful hits. With student help, a thousand man-hours went into the accumulation of N = 250,000. The k was about 8/5, and the final ‘estimate’ of π was 3.143, to which was appended a physical error of ±0.005.

“This is more or less the zenith of accuracy and precision,” Gridgeman writes. “It could not be bettered by any reasonable increase in N — even if the physical error could be reduced, hundreds of millions of falls would be needed to establish a third decimal place with confidence.”

(N.T. Gridgeman, “Geometric Probability and the Number π,” Scripta Mathematica 25:3 [November 1960], 183-195.)

(From Raymond F. Lausmann’s Fun With Figures, 1965.)

This trick seems to have been invented independently by Martin Gardner and Karl Fulves. A blindfolded magician asks a spectator to lay three pennies on a table, in any arrangement of heads and tails. The magician’s goal is to put all three coins into the same state, all heads or all tails.

If the three coins already match, then the trick is done. If not, then the magician gives three instructions: Flip the left coin, flip the middle coin, flip the left coin. After each step he asks whether the three coins now match. By the third flip, they will.

“It’s no surprise that the magician can eventually equalize all the coins,” writes MIT computer scientist Erik Demaine, “but it’s impressive that it always takes at most three moves.” The technique exploits a principle used in Gray codes, which are used to reduce errors when using analog signals to represent digital data. Demaine relates a similar trick involving four coins in the November-December 2010 issue of American Scientist.

See Lincoln Seeks Equality.

Cornell mathematician Robert Connelly devised this intuition-defying card game. I shuffle a standard deck of 52 cards and deal them out in a row before you, one at a time. At some point before the last card is dealt, you must say the word “red.” If the next card I deal is red, you win $1; if it’s black you lose $1. If you play blind, your chance of winning is 1/2. Can you improve on this by devising a strategy that considers the dealt cards?

Surprisingly, the answer is no. Imagine a deck with two red cards and two black. Now there are six equally likely deals:

RRBB
RBBR
BBRR
RBRB
BRBR
BRRB

By counting, we can see that the chance of success remains 1/2 regardless of whether you call red before the first, second, third, or fourth card.

Trying to outsmart the cards doesn’t help. You might resolve to wait and see the first card: If it’s black you’ll call red immediately, and if it’s red you’ll wait until the fourth card. It’s true that this strategy gives you a 2/3 chance of winning if the first card is black — but if it’s red then it has a 2/3 chance of losing.

Similarly, it would seem that if the first two cards are black then you have a sure thing — the next card must be red. This is true, but it will happen only once in six deals; on the other five deals, calling red at the third card wins only 2/5 of the time — so this strategy has an overall success rate of (1/6 × 1) + (5/6 × 2/5) = 1/2, just like the others. The cards conspire to erase every seeming advantage.

The same principle holds for a 52-card deck, or indeed for any deck. In general, if a deck has r red cards and b black ones, then your chance of winning, by any strategy whatsoever, is r/(b + r). Seeing the cards that have already been dealt, surprisingly, is no advantage.

(Robert Connelly, “Say Red,” Pallbearers Review 9 [1974], 702.)