Nobel prize winners are chosen independently of past awards, so if Pauling’s chance of winning one prize was (say) 1 in 3,000,000,000, then he’d have the same chance of winning the second, and his chance of winning two would be 1 in 3,000,000,000². (Actually they’re considerably better than that — not everyone on Earth is equally deserving of the Nobel prize.)

If we knew that the Nobel committee would award the 1962 prize only to an existing laureate, then Pauling’s argument would hold.

(From The Ivan Morris Puzzle Book, 1972.)

The work train backs into the siding and uncouples its three rearmost cars. Then it moves forward a sufficient distance. The passenger train passes entirely through the station, backs into the siding, and couples these three cars. Then it moves forward again into the main railway, backs entirely through the station, and uncouples the three cars. The work train backs into the siding, which will now accommodate it, and the passenger train can move forward on its way. Now the work train can emerge from the siding and recouple the three waiting cars.

As we might have expected, it’s a trick question. It appears at first that the race will end in a tie, as the two animals are covering ground at the same rate. The trouble occurs at the far end of the course. The cat, which covers 2 feet at a bound, can turn around neatly at the halfway point. The dog, which covers 3 feet, cannot: It will arrive at the 99-foot mark and then must make an additional leap to get past the same point.

“In all, the dog must make 68 leaps to go the distance. But it jumps only two-thirds as quickly as the cat, so that while the cat is making 100 leaps the dog cannot make quite 67.”

The coordinates of each point can be categorized by their parity — for example, (1, 1) is odd-odd and (4, 17) is even-odd. The midpoint of a line segment between two points will occur at a lattice point only if the two endpoints fall into the same category; for example, the two points above won’t produce such a midpoint, but (6, 8) and (14, 2) will because both are even-even.

There are only four possible categories: even-even, odd-odd, even-odd, and odd-even. And because we are plotting five points, at least two points must fall into the same category. The midpoint between these two points will occur at a lattice point.

From L.C. Larsen, Problem Solving Through Problems, 1983.

After the surprising 1. Bd3!, Black will be mated by 2. Qd1, 2. Bf6, or 2. Qxe4.

I don’t have Wheeler’s solution, but I believe this is right:

As things stand, only Black’s a4 bishop can move. With 1. Rd7, White can limit that bishop to three squares, from each of which a queen sacrifice will force it to the mating diagonal b1-h7:

1. Rd7 Bb5 2. Qd3+ Bxd3#
1. Rd7 Bc6 2. Qe4+ Bxe4#
1. Rd7 Bxd7 2. Qf5+ Bxf5#

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.