Black and White

A chess problem from the Jamaica Gleaner. There’s no black king on this board. Find a place for him such that White can mate him in two moves.

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Put him on h5 and White can mate in two starting with 1. Qb3. From Arthur Ford Mackenzie, Chess: Its Poetry and Its Prose, 1887.

February 20, 2021February 19, 2021 | Puzzles

Route Cause

A problem submitted by France and shortlisted for the 17th International Mathematical Olympiad, Burgas-Sofia, Bulgaria, 1975:

A lake has six ports. Is it possible to arrange a series of routes that satisfy the following conditions?

Each route must include exactly three ports.
No two routes may contain the same three ports.
Any tourist who wants to visit two different arbitrary ports has a choice of exactly two routes.

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Yes. The thing to notice is that we can’t have three routes in the pattern (A,B,C), (A,B,D), (A,C,D), because in that case it’s impossible to fulfill the third condition: If the remaining ports are E and F, then we’ve left ourselves only one permissible route (not two) that visits A and E. That’s (A, E, F). Any other three-port route that includes A and E would “overload” another segment. This means that if two of the routes are (A,B,C) and (A,B,D), then the second one that covers A,C must be (without loss of generality) (A,C,E). With that in mind the other routes can be worked out uniquely: (A,D,F), (A,E,F), (B,D,E), (B,E,F), (B,C,F), (C,D,E), (C,D,F).

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February 11, 2021February 10, 2021 | Puzzles

Lucky Seven

A puzzle from the National Security Agency’s Puzzle Periodical, posed by NSA mathematician David G. in March 2017:

Each die of a pair of non-identical dice has six faces, but some numbers are missing, others are duplicated, and some faces may have more than six spots.

The dice can roll every number from 2 to 12.

What is the largest possible probability of rolling a 7?

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David’s solution: If one die has X different numbers and the other die has Y different numbers, there are at most XY possible totals. Since we need 11 totals from 2 to 12, XY must be at least 11. In particular, at least one die must have at least four different numbers. If the first die has six different numbers, then any number on the second die has at most a 1/6 probability of getting the right number on the first die to get a total of 7. Therefore, the probability of rolling a 7 is at most 1/6. If the first die has five different numbers, with one appearing twice, the second die must have at least three different numbers. There is only one possible number on the second die which totals a 7 with the duplicated number on the first die. The second die has at most four copies of the number which totals a 7 with the duplicated number and has two ways to roll a 7; all other numbers on the second die have only one way to roll a 7. This is a maximum total of 10 rolls for a probability of 10/36. If each die has four different numbers, they can be split 2,2,1,1 or 3,1,1,1. Each number on each die can pair with at most one number on the other die to make a 7. The largest possible sum of products, which gives the number of ways to roll a 7, is 3 × 3 + 1 × 1 + 1 × 1 + 1 × 1 = 12. The maximum probability is thus 12/36. If the first die has four different numbers and the second has three, there are only 12 different pairs of values, and thus there must be at most two different value pairs which total 7 in order to have 11 possible totals. Therefore, the largest possible sum of products is 3 × 4 + 1 × 1 = 13, not 3 × 4 + 1 × 1 + 1 × 1 = 14. And 13 is possible, with one die having 1,2,6,6,6,7 and the other having 1,1,1,1,3,5 (or 1,2,2,2,6,7 and 1,3,5,5,5,5); this gives a probability of 13/36.

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February 10, 2021February 10, 2021 | Puzzles

Black and White

white chess problem

“The Pyramid,” by A.C. White, from the British Chess Magazine. White to mate in two moves.

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1. d6! and White will mate next move.

February 2, 2021February 1, 2021 | Puzzles

Calendar Boy

Gary Foshee presented this puzzle at the 2010 Gathering for Gardner:

I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?

The first thing you think is ‘What has Tuesday got to do with it?’ Well, it has everything to do with it.

He proposed the answer 13/27, with this reasoning:

There are 14 equally likely possibilities for a single birth — (boy, Tuesday), (girl, Sunday), and so on.

If all we knew were that Foshee had two children, then it would seem that there are 14² = 196 equally likely possibilities as to their births.

But we know that at least one of his children is a (boy, Tuesday), and only 27 of the 196 outcomes meet this criterion. (There are 14 cases in which the (boy, Tuesday) is the firstborn child and 14 in which he’s born second, and we must remove the single case in which he’s counted twice.)

Of those 27 possibilities, 13 include two boys — 7 with (boy, Tuesday) as the first child and 7 with (boy, Tuesday) as the second child, and we subtract the one in which he’s counted twice. That, Foshee says, gives the answer 13/27.

This generated a lot of discussion when it appeared — unfortunately because the meaning of Foshee’s question is open to interpretation. See the end of this New Scientist article and the comments on Columbia statistician Andrew Gelman’s blog.

January 28, 2021January 27, 2021 | Puzzles

Quickie

If I roll three dice and multiply the three resulting numbers together, what is the probability that the product will be odd?

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The product can be odd only if all three factors are odd: $\displaystyle \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ (Thanks, Nick.) 02/01/2021 UPDATE: From reader Paul-Georg Becker: Another dice quickie: I have a set of n dice, but unfortunately only one of them is fair. If I roll all dice and add the n resulting numbers together, what is the probability that the product will be odd? Answer: For natural n let S_n be the sum of the resulting numbers x₁, …, x_n. We assume that x_n is the number shown by the fair die. Then we have P(S_n even) = P(x_n + S_n-1 even) = P(S_n-1 even) P(x_n even) + P(S_n-1 odd) P(x_n odd) = P(S_n-1 even) /2 + P(S_n-1 odd) / 2 = (P(S_n-1 even) + P(S_n-1 odd)) / 2 = 1/2 It does not even matter what the result for S_n-1 is. (Thanks, Paul-Georg.)

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January 27, 2021February 1, 2021 | Puzzles

Disappearing Act

https://commons.wikimedia.org/wiki/File:Anatomy_of_a_Sunset-2.jpg — Image: Wikimedia Commons

A puzzle by Joseph Horton, from MIT Technology Review, January-February 1999:

If the sun takes two minutes to set, what angle does it subtend from Earth?

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Half a degree. Earth rotates through 360 degrees every 24 hours, or 1 degree every 4 minutes. 01/27/2021 UPDATE: This is oversimplified — time to set depends on time of year (declination) and the observer’s latitude. The Review posted a correction in the May/June issue: “Peter Blicher notes that we are erroneously assuming the sun sets at a 90 degree angle to the horizontal.” Thanks, Chris and Martin.

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January 25, 2021January 27, 2021 | Puzzles

Black and White

dobrusky chess problem

By J. Dobrusky. White to mate in two moves.

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1. Qe8! and there’s no stopping a fatal rook check next move. From Benjamin Glover Laws, The Two-Move Chess Problem, 1890. 01/19/2021 UPDATE: There’s a pretty variation I’d overlooked: 1. Qe8 Kb5 2. Nxc3#. So in that case the knight delivers mate.

January 16, 2021January 19, 2021 | Puzzles

“Problem for Poultry Farmers”

Eureka, the esteemed journal of recreational mathematics published at Cambridge University, has placed its archives online under a Creative Commons license. Here’s a problem from May 1940:

The chicken was twice as old when when the day before yesterday was to-morrow to-day was as far from Sunday as to-day will be when the day after to-morrow is yesterday as it was when when to-morrow will be to-day when the day before yesterday is to-morrow yesterday will be as far from Thursday as yesterday was when to-morrow was to-day when the day after to-morrow was yesterday. On what day was the chicken hatched out?

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“It can now be revealed that the chicken referred to in the last Eureka was hatched out on a Friday. (Clearly this piece of information cannot now be of any interest to anyone.)” A fuller solution is here. (Thanks, Bruce.)

January 10, 2021February 24, 2021 | Puzzles

“Self-Explanatory”

This impressive puzzle appeared recently in the Magpie, a magazine for advanced UK-style cryptic puzzles. See the file for instructions.

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The composer explains the puzzle’s creation here. (Thanks, Gareth.)

January 9, 2021 | Puzzles