Futility Closet

You die and go to hell, where the devil proposes a game of chance. If you play on the first day you’ll have a one-half chance of winning, on the second day a two-thirds chance, and so on. If you win, you’ll go to heaven; if you lose you’ll stay in hell. When should you play?

Unfortunately, it seems you should wait forever. After a year of waiting your chance of winning would be 0.997268, but even then another day’s (presumably finite) torment would earn you 0.000007 greater chance of infinite joy.

Maybe the mere hope of heaven is worth something. But is it a valid hope if you never realize it?

When he wasn’t taming electromagnetism, James Clerk Maxwell wrote verse. Here’s “A Problem in Dynamics,” composed in 1854:

An inextensible heavy chain
Lies on a smooth horizontal plane,
An impulsive force is applied at A,
Required the initial motion of K.

Let ds be the infinitesimal link,
Of which for the present we’ve only to think;
Let T be the tension, and T + dT
The same for the end that is nearest to B.
Let a be put, by a common convention,
For the angle at M ‘twixt OX and the tension;
Let V_t and V_n be ds’s velocities,
Of which V_t along and V_n across it is;
Then V_n/V_t the tangent will equal,
Of the angle of starting worked out in the sequel.

In working the problem the first thing of course is
To equate the impressed and effectual forces.
K is tugged by two tensions, whose difference dT
Must equal the element’s mass into V_t.
V_n must be due to the force perpendicular
To ds’s direction, which shows the particular
Advantage of using da to serve at your
Pleasure to estimate ds’s curvature.
For V_n into mass of a unit of chain
Must equal the curvature into the strain.

Thus managing cause and effect to discriminate,
The student must fruitlessly try to eliminate,
And painfully learn, that in order to do it, he
Must find the Equation of Continuity.
The reason is this, that the tough little element,
Which the force of impulsion to beat to a jelly meant,
Was endowed with a property incomprehensible,
And was “given,” in the language of Shop, “inextensible.”
It therefore with such pertinacity odd defied
The force which the length of the chain should have modified,
That its stubborn example may possibly yet recall
These overgrown rhymes to their prosody metrical.
The condition is got by resolving again,
According to axes assumed in the plane.
If then you reduce to the tangent and normal,
You will find the equation more neat tho’ less formal.
The condition thus found after these preparations,
When duly combined with the former equations,
Will give you another, in which differentials
(When the chain forms a circle), become in essentials
No harder than those that we easily solve
In the time a T totum would take to revolve.

Now joyfully leaving ds to itself, a-
Ttend to the values of T and of a.
The chain undergoes a distorting convulsion,
Produced first at A by the force of impulsion.
In magnitude R, in direction tangential,
Equating this R to the form exponential,
Obtained for the tension when a is zero,
It will measure the tug, such a tug as the “hero
Plume-waving” experienced, tied to the chariot.
But when dragged by the heels his grim head could not carry aught,
So give a its due at the end of the chain,
And the tension ought there to be zero again.
From these two conditions we get three equations,
Which serve to determine the proper relations
Between the first impulse and each coefficient
In the form for the tension, and this is sufficient
To work out the problem, and then, if you choose,
You may turn it and twist it the Dons to amuse.

See this beaker? It contains 1 to 2 liters of water and 1 liter of wine. That means that the ratio of water to wine (call it r) is between 1 and 2. Thus there’s a 50 percent chance that r is between 1 and 3/2. Right?

But now consider the ratio of wine to water, or 1/r. That’s between 1/2 and 1, so there’s a 50 percent chance that 1/r is between 3/4 and 1.

Taking the reciprocal, that means there’s a 50 percent chance that r is between 1 and 4/3, which contradicts our earlier result. Where is the error?

In 1838, Samuel Rowbotham waded into a drainage canal in Norfolk and sighted along its length with a telescope. Six miles away, an assistant held a flag three feet above the water. If the earth were round, its curvature should hide the flag from him. But he decided he could see it clearly. “It follows,” he wrote, “that the surface of standing water is not convex, and therefore that the Earth IS NOT A GLOBE!”

Rowbotham’s triumphant result stood until 1870, when naturalist, surveyor, and obvious crackpot Alfred Russel Wallace attempted to disprove the result. His endeavor ended only in a heated argument — and eventually a libel suit against the “planists.” (Round-earthers are clearly desperate men.)

In fairness, we must note that not all observations have agreed with Rowbotham’s. In 1896 a newspaper editor conducted a similar experiment in Illinois and discovered that the earth is concave. Clearly more work is needed.

Here’s a tip for your next craps game. You can find the odds of rolling any number with two dice by subtracting the number from 7, ignoring the sign, and subtracting the result from 6. The remainder is the number of chances out of 36 that the number will appear.

For example, there are (6 – (7 – 5)) = 4 chances in 36, or 1 chance in 9, that you’ll roll a 5.

We’re told that, in any right triangle, a² + b² = c². But consider:

In the figure above, the total length of the red line is 2(a/2) + 2(b/2), or a + b. And again:

Here the red line’s length is 4(a/4) + 4(b/4), which is still a + b.

With each iteration, the red line more closely approximates c, but its length remains a + b. At the limit, then, it seems, a + b = c. Was Pythagoras mistaken?

Solution to Two by Two:

Essentially the technique converts the first factor into binary, multiplies each of its constituents by the second factor, and sums the results.

Imagine that each line is associated with a power of 2: the first line with 2⁰, the second with 2¹,and so on.

The business in the first column, halving the first factor successively and crossing out those lines with even numbers, effectively reduces the first factor to its binary constituents — here, the lines that remain are those associated with 2⁰, 2⁵, and 2⁶, and, sure enough, 2⁰ + 2⁵ + 2⁶ = 97.

Now we need to multiply each of those constituents by the second factor, 23. In other words, we want to find:

(23 × 2⁰) + (23 × 2⁵) + (23 × 2⁶)

That’s what’s accomplished in the second column. Doubling the second factor successively, as we’ve done there, is equivalent to multiplying it by 2⁰ in the first line, by 2¹ in the second, and so on. That is, 23 = 23 × 2⁰, 46 = 23 × 2¹, etc. And the lines that haven’t been crossed out are precisely the ones we want:

23 (= 23 × 2⁰)
736 (= 23 × 2⁵)
1472 (= 23 × 2⁶)

So if we add those values, we’ll get the product of the original two numbers, which is what we sought: 23 + 736 + 1472 = 2231.

Here’s essentially what we’ve done, from the top:

97 × 23
= (2⁰ + 2⁵ + 2⁶) × 23
= (23 × 2⁰) + (23 × 2⁵) + (23 × 2⁶)
= 23 + 736 + 1472
= 2231

It works with any pair of numbers.

Here’s a curious way to multiply two numbers. Suppose we want to multiply 97 by 23. Write each at the head of a column. Now halve the first number successively, discarding remainders, until you reach 1, and double the second number correspondingly in its own column:

Cross out each row that has an even number in the left column, and add the numbers that remain in the second column:

That gives the right answer (97 × 23 = 2231). Why does it work?

(Answer)

This is rather amazing. Arrange a deck of cards in this order, top to bottom:

A♣, 8♥, 5♠, 4♦, J♣, 2♥, 9♠, 3♦, 7♣, Q♥, K♠, 6♦, 10♣,
A♥, 8♠, 5♦, 4♣, J♥, 2♠, 9♦, 3♣, 7♥, Q♠, K♦, 6♣, 10♥,
A♠, 8♦, 5♣, 4♥, J♠, 2♦, 9♣, 3♥, 7♠, Q♦, K♣, 6♥, 10♠,
A♦, 8♣, 5♥, 4♠, J♦, 2♣, 9♥, 3♠, 7♦, Q♣, K♥, 6♠, 10♦

Now:

1. Cut the deck and complete the cut. Do this as many times as you like.
2. Deal cards face down one at a time, stopping whenever you have a substantial pile.
3. Riffle-shuffle the two packs back together again.

Despite all this, you’ll find that the resulting deck is made up of 13 successive quartets of four suits–and four consecutive straights, ace through king.

The reasons for this are fairly complex, so I’ll just call it magic. You’ll find a full analysis in Julian Havil’s Impossible? Surprising Solutions to Counterintuitive Conundrums (2008).

Three condemned prisoners share a cell. A guard arrives and tells them that one has been pardoned.

“Which is it?” they ask.

“I can’t tell you that,” says the guard. “I can’t tell a prisoner his own fate.”

Prisoner A takes the guard aside. “Look,” he says. “Of the three of us, only one has been pardoned. That means that one of my cellmates is still sure to die. Give me his name. That way you’re not telling me my own fate, and you’re not identifying the pardoned man.”

The guard thinks about this and says, “Prisoner B is sure to die.”

Prisoner A rejoices that his own chance of survival has improved from 1/3 to 1/2. But how is this possible? The guard has given him no new information. Has he?