Two envelopes contain unequal sums of money (for simplicity, assume the two amounts are positive integers). The probability distributions are unknown. You choose an envelope at random, open it, and see that it contains x dollars. Now you must predict whether the total in the other envelope is more or less than x.

Since we know nothing about the other envelope, it would seem we have a 50 percent chance of guessing correctly. But, El Camino College mathematician Leonard Wapner writes, “Unexpectedly, there is something you can do, short of opening the other envelope, to give yourself a better than even chance of getting it right.”

Choose a random positive integer, d, by any means at all. (If d = x then choose again until this isn’t the case.) Now if d > x, guess more, and if d < x, guess less. You’ll guess correctly more than 50 percent of the time.

How is this possible? The random number is chosen independently of the envelopes. How can it point in the direction of the unknown y most of the time? “Think of it this way,” writes Wapner. “If d falls between x and y then your prediction (as indicated by d) is guaranteed to be correct. Assume this occurs with probability p. If d falls less than both x and y, then your prediction will be correct only in the event your chosen number x is the larger of the two. There is a 50 percent chance of this. Similarly, if d is greater than both numbers, your prediction will be correct only if your chosen number is the smaller of the two. This occurs with a 50 percent probability as well.”

So, on balance, your overall probability of being correct is

That’s greater than 0.5, so the odds are in favor of your making a correct prediction.

This example is based on a principle identified by Stanford statistician David Blackwell. “It’s unexpected and ironic that an unrelated random variable can be used to predict that which appears to be completely unpredictable.”

(Leonard M. Wapner, Unexpected Expectations: The Curiosities of a Mathematical Crystal Ball, 2012, following David Blackwell, “On the Translation Parameter Problem for Discrete Variables,” Annals of Mathematical Statistics 22:3 [1951], 393–399.)

Raymond Smullyan offered this proof at a conference on self-reference:

A: Santa Claus exists, if I am not mistaken.
B: Well, of course Santa Claus exists if you are not mistaken!
A: So I was right.
B: Yes.
A: So I was not mistaken.
B: Yes.
A: Hence, Santa Claus exists.

Proctologist W.C. Bornemeier explains why the sphincter ani must be preserved when performing hemorrhoid surgery:

They say man has succeeded where the animals fail because of the clever use of his hands, yet when compared to the hands, the sphincter ani is far superior. If you place into your cupped hands a mixture of fluid, solid and gas and then through an opening at the bottom, try to let only the gas escape, you will fail. Yet the sphincter ani can do it. The sphincter apparently can differentiate between solid, fluid and gas. It apparently can tell whether its owner is alone or with someone, whether standing up or sitting down, whether its owner has his pants on or off. No other muscle in the body is such a protector of the dignity of man, yet so ready to come to his relief. A muscle like this is worth protecting.

— W.C. Bornemeier, “Sphincter Protecting Hemorrhoidectomy,” American Journal of Proctology 11 (1960), 48-52

In 2003, Danish computer scientist Peter Bro Miltersen discussed a surprisingly effective technique by which a player might guess the colors of slips of paper hidden in boxes (PDF). As this circulated in the mathematical community it evolved into a puzzle in which a group of 100 prisoners must find their own names on slips of paper. I wrote about it in 2011.

When Eugene Curtin and Max Warshauer wrote about the prisoner puzzle in The Mathematical Intelligencer in December 2006, reader A.S. Landsberg offered a variant called “The Return of Monty Hall.” On a new game show for couples, there are three curtains, which hide a key, a car, and a goat. One member of the couple is the “car-master” — she must find the car. The other is the “key-master” — he must find the key. If both succeed in their tasks, they win the new car. If either fails, they win the goat.

The key-master is led out of the room, where he can’t observe the proceedings, and then the car-master has two tries to find the car (open one curtain, and if the car isn’t there, open another curtain). If she finds the car, then all the curtains are closed again and the key-master is brought on to find the key. No communication at all is permitted between the two at this point. As before, the key-master has two tries to find the key by opening curtains.

If the couple play optimally, their odds of winning the car are a surprising 2/3. They do this using Miltersen’s technique. The car-master is Player #1, the key-master is Player #2, the car is Prize #1, the key is Prize #2, and the goat is Prize #3. The strategy is simply for each player to start by opening the curtain corresponding to his or her own player number, and if unsuccessful to open the curtain number corresponding to the prize number that the first curtain reveals. So, for example, the car-master, who is Player #1, begins by opening Curtain #1. If she finds the car then she’s done; if she finds the key (Prize #2) then she opens Curtain #2, and if she finds the goat (Prize #3) then she opens Curtain #3. When the curtains are reclosed, the key-master begins his turn by opening Curtain #2 (since he’s Player #2) and following the same plan.

That’s it. It’s not guaranteed to work, but it’s a simple strategy that requires minimal preparation and no communication at all once the game has begun. The universe of possibilities is so small that we can simply count them — here are the various arrangements of prizes and the resulting outcomes:

car-key-goat: win
car-goat-key: win
key-goat-car: lose
key-car-goat: win
goat-key-car: win
goat-car-key: lose

Landsberg’s letter brought a comment by reader Eric Grunwald, who pointed out that a third person can be introduced to the Monty Hall game without reducing the overall chance of success. Replace the goat with a GPS system and add a third contestant, the “GPS-master.” Following the same rules, and again forbidding any communication among the contestants, Miltersen’s strategy ensures a 2/3 probability that all three players find their prizes.

The decimal expansion of 1/7 is 0.142857142857 …, a repeating decimal. Arrange the six repeating digits into overlapping ordered pairs, like so:

(1, 4), (4, 2), (2, 8), (8, 5), (5, 7) (7, 1),

and, remarkably, all six lie on an ellipse:

19x² + 36yx + 41y² – 333x – 531y + 1638 = 0

Even more remarkably, if we take the digits two at a time:

(14, 28), (42, 85), (28, 57), (85, 71), (57, 14), (71, 42),

these points also lie on an ellipse:

-165104x² + 160804yx + 8385498x – 41651y² – 3836349y – 7999600 = 0

That’s from David Wells, The Penguin Dictionary of Curious and Interesting Numbers (1986). Victor Hugo wrote, “Mankind is not a circle with a single center but an ellipse with two focal points, of which facts are one and ideas the other.”

The Fibonacci numbers are the ones in this sequence:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

Each number is the sum of the two that precede it. But now, interestingly:

“And so on!” writes James Tanton in Mathematics Galore! (2012). “The first relation, for instance, states that a line of slope 1/2 stacked with a line of slope 1/3 gives a line of slope 1. (Can you prove the relations?)”

(Ko Hayashi, “Fibonacci Numbers and the Arctangent Function,” Mathematics Magazine 76:3 [June 2003], 215.)

I hand you an ordinary deck of 52 cards. You inspect and shuffle it, then choose five cards from the deck and hand them to my assistant. She looks at them and passes four of them to me. I name the fifth card.

At first this appears impossible. The hidden card is one of 48 possibilities, and by passing me four cards in some order my assistant can have sent me only 1 of 4! = 24 messages. How am I able to name the card?

Part of the secret is that my assistant gets to choose which card to withhold. The group of five cards that you’ve chosen must contain two cards of the same suit. My assistant chooses one of these to be the hidden card and passes me the other one. Now I know the suit of the hidden card, and there are 12 possibilities as to its rank. But my assistant can pass me only three more cards, with 3! = 6 possible messages, so the task still appears impossible.

The rest of the secret lies in my assistant’s choice as to which of the two same-suit cards to give me. Think of the 13 card ranks arranged in a circle (with A=1, J=11, Q=12, and K=13). Given two ranks, it’s always possible to get from one to the other in at most 6 steps by traveling “the short way” around the circle. So we agree on a convention beforehand: We’ll imagine that the ranks increase in value A-K, and the suits as in bridge (or alphabetical) order, clubs-diamonds-hearts-spades. This puts the whole deck into a specified order, and my assistant can pass me the three remaining cards in one of six ways:

{low, middle, high} = 1
{low, high, middle} = 2
{middle, low, high} = 3
{middle, high, low} = 4
{high, low, middle} = 5
{high, middle, low} = 6

So if my assistant knows that I’ll always travel clockwise around the imaginary circle, she can choose the first card to establish the suit of the hidden card and to specify one point on the circle, and then order the remaining three cards to tell me how many clockwise steps to take from that point to reach the hidden rank.

“If you haven’t seen this trick before, the effect really is remarkable; reading it in print does not do it justice,” writes mathematician Michael Kleber. “I am forever indebted to a graduate student in one audience who blurted out ‘No way!’ just before I named the hidden card.”

It first appeared in print in Wallace Lee’s 1950 book Math Miracles. Lee attributes it to William Fitch Cheney, a San Francisco magician and the holder of the first math Ph.D. ever awarded by MIT.

(Michael Kleber, “The Best Card Trick,” Mathematical Intelligencer 24:1 [December 2002], 9-11.)