CyberChance

A puzzle by National Security Agency applied research mathematician Nicholas R., from the agency’s October 2016 Puzzle Periodical:

Eddie, Layne, Kurt, and Chris are locked in a heated game of CyberChance which only one of them will win. Their fans are on the edge of their seats: everyone knows that in CyberChance, things can change at any moment. One fan, who prefers Kurt or Chris, is feeling worried since there is only a 4-in-10 chance that one of them will win. An Eddie fan brags that Eddie is twice as likely as Layne to come out on top. Another Eddie fan concurs, and adds the following observation: Chris is Eddie’s main competition. He figures that, given that Chris doesn’t win, Eddie has a 4-in-7 chance of winning. A Kurt fan has been quietly standing in the corner, listening to all the other fans, and wonders: What chance does Kurt have of winning CyberChance?

	Select Click for Answer
Let E, L, K, and C be the chances that Eddie, Layne, Kurt, and Chris win the game, respectively. There is a 100% chance that one of them will win, so E + L + K + C = 100%. We know that K + C, the probability that Kurt or Chris will win, is 40%. This tells us that E + L = 100% – 40% = 60%. Now, we also know that the probability that Eddie will win is twice the probability that Layne will win. In numbers, this means that E = 2L, i.e. E – 2L = 0%. Knowing these two facts lets us find out what Eddie’s chance of winning is: add the equation E + L = 60% twice to the equation E – 2L = 0%. This tells us that 3E = 120%. Therefore E = 40%. We are almost there. We know that the conditional probability that Eddie wins, given that Chris loses, is 40%/70%. This probability is equal to the probability that Eddie wins and Chris loses, divided by the probability that Chris loses. But notice: CyberChance has only one winner, so the probability that Eddie wins and Chris loses is the same as the probability that Eddie wins. Also notice that the probability that Chris loses equals the probability that anybody else wins. In numbers, the probability that Chris loses is 100% – C. So we have the following equation: 40%/70% = E / (100% – C) = 40% / (100% – C) Solving for C tells us that C = 30%. Recall that K + C = 40%. Therefore K = 10%. So Kurt has a measly 10% chance of winning CyberChance. No wonder the Kurt fan was sulking in the corner!

Click for Answer

August 13, 2026August 12, 2026 | Puzzles

Tile Swap

You’re tiling a floor with 4×1 and 2×2 tiles when you accidentally break one. A tile of the other shape is available. Show that it’s not possible to cover the floor by rearranging the tiles.

	Select Click for Answer
Color the tiles as shown. Now every 4×1 tile must cover 0 or 2 black squares, and every 2×2 tile must cover one black square. So we can’t successfully exchange one for the other. From Arthur Engel’s excellent Problem-Solving Strategies, 1998.

August 12, 2026August 11, 2026 | Puzzles

Black and White

khanyan chess puzzle

Alexey Khanyan devised this “no-brainer” in 2008. Black is in check. His only legal response is to interpose one of his knights, which then checks White, who must capture the knight, and so on. The whole ensuing 22-ply bloodbath unfolds mechanically, always reaching the same conclusion.

Via Tim Krabbé’s chess diary. Krabbé writes, “As it is a no-brainer, I feel free not to give the moves.” See The Sure Thing.

August 11, 2026August 10, 2026 | Puzzles

Portrait

A problem from Joseph Madachy’s Mathematics on Vacation (1966): If these statements are all true, what can we deduce about Major Perkins?

Only bleary-eyed military men ever try to shave with a toothbrush.
No bleary-eyed military men are at all musical.
Men who never try to shave with a toothbrush never play golf.
Major Perkins plays the tuba on Wednesdays.
Only musical people who are golfers forget to change their socks.

	Select Click for Answer
The only thing we know for certain is that Major Perkins plays the tuba on Wednesdays. All the other seeming deductions (that Major Perkins is musical, not bleary-eyed, and not a golfer; that he does not try to shave with a toothbrush, and that he remembers to change his socks) require the assumption that Perkins is a man.

August 6, 2026August 5, 2026 | Puzzles

Level-Headed

A puzzle by Russian mathematician Boris Kordemsky:

A house has 6 stories, each the same height. How many times as long is the ascent to the sixth floor as the ascent to the third?

	Select Click for Answer
2 1/2 times (5 ÷ 2, not 6 ÷ 3). From The Moscow Puzzles, 1972.

August 5, 2026August 4, 2026 | Puzzles

Poser

https://commons.wikimedia.org/wiki/File:Aplusclick_Missing_Symbol.jpg — Image: Wikimedia Commons

Unusually, this puzzle may be easier for older readers.

The answer is R, the reverse gear on a manual gearshift.

August 1, 2026 | Puzzles

The Europa Redux Manuscript

In 2020 Brooklyn’s Honey & Wax Bookseller came into possession of a strange manuscript consisting of 7,000 illustrations, each 3 centimeters square. The images fill 106 loose sheets that often mimic European tourist brochures, and most bear captions that resemble no known language. The manuscript appears to have been created in Switzerland in the 1940s, but no one knows who created it or why. Notre Dame historian of science Robert Goulding wrote, “The fact that [the words] don’t seem to conform to any language or phonemes in any language is suspicious … suggesting to me that they are either a cipher, or there is some method or algorithm used to generate the words.”

The images can be viewed here; more information is here.

July 31, 2026July 30, 2026 | Oddities · Puzzles

The Magic Hat Puzzler

An old puzzle from the Car Talk radio show on NPR:

You and I each draw a number from a magic hat. We don’t know each other’s numbers, but the hat ensures that they’re adjacent positive integers.

An interlocutor now begins to ask each of us in turn whether he knows the other’s number. Eventually, one of us will answer yes. Why?

	Select Click for Answer
Suppose I’m asked first and I answer no. This tells you that I’m not holding the number 1, because if I were I’d know for certain that you held 2. The interlocutor turns to you and asks whether you know my number. You say no. This tells me that you don’t hold 2, by the same principle — I know that you just inferred that I’m not holding 1, so if you held 2 you’d know that I must have 3. And so on. Together we’ll work our way up the number line until one of us says yes.

Click for Answer

July 30, 2026 | Puzzles

Long Distance

de moraes chess puzzle

In this classic puzzle by Florencio Mendes de Moraes, White must checkmate Black in seven moves. The catch is that all three of his queens are confined to the leftmost file.

Either of two first moves will get things started, but beyond that there’s only one sequence that will work. What is it?

	Select Click for Answer
1. Qa4 (or Qa3) Kg8 2. Qa1 Kf8 3. Q1a2 Ke8 4. Q2a3 Kd8 5. Q6a4 Kc8 6. Qa8+ Kc7 7. Q4a7#

July 23, 2026July 22, 2026 | Puzzles

The B List

A problem from the Eighth International Mathematical Olympiad, held in Sofia, Bulgaria, in July 1966 (contributed by the Soviet Union):

In a mathematical contest, three problems, A, B, C were posed. Among the participants there were 25 students who solved at least one problem each. Of all the contestants who did not solve problem A, the number who solved B was twice the number who solved C. The number of students who solved only problem A was one more than the number of students who solved A and at least one other problem. Of all students who solved just one problem, half did not solve problem A. How many students solved only problem B?

	Select Click for Answer
Let N_a, N_b, N_c, N_ab, N_ac, N_bc, and N_abc denote the number of students who have solved exactly the problems identified in the index of each variable. Then: N_a + N_b + N_c + N_ab + N_bc + N_ac + N_abc = 25, N_b + N_bc = 2(N_c + N_bc), N_a – 1 = N_ac + N_ab + N_abc, and N_a = N_b + N_c. Combining the first and third equations tells us that 2N_a + N_b + N_c + N_bc = 26, and from the second and fourth we get 4N_b + N_c = 26, so we know that N_b ≤ 6. At the same time, the second equation tells us that N_b = 2N_c + N_bc, and thence N_c ≤ N_b/2 26 ≤ 9N_b/2, and N_b ≥ 6. Hence N_b must be 6. Via The IMO Compendium.

Click for Answer

July 10, 2026July 9, 2026 | Puzzles