Black and White

steinwig chess problem

A simple but pretty “lightweight” problem by R. Steinweg. White to mate in three moves.

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1. Kc6 Ka7 2. c8=R! Ka6 3. Ra8# From Arthur Ford Mackenzie, Chess Lyrics: A Collection of Chess Problems, 1905.

March 10, 2021March 9, 2021 | Puzzles

“Piccadilly Underground Station”

This unusual puzzle by G.A. Roberts appeared in the January 1941 issue of Eureka, the journal of recreational mathematics published at Cambridge University. It concerns the Piccadilly Circus station of the London Underground, which lies on the Piccadilly line between Green Park and Leicester Square and on the Bakerloo line between Charing Cross and Oxford Circus.

At a given time there are on the platform, escalators and subways, and in the trains, 128 people, all of whom travel by train, and none of whom return immediately by the way they have come.

Those who have come via Leicester Square are equal in number to those who are about to travel via Leicester Square.

The number of people who arrived by Bakerloo Line is equal to the number who intend to leave by the Piccadilly Line.

The number of people who are travelling from the street to stations on the Piccadilly Line is equal to six-thirteenths of the number who change from the Piccadilly Line to the Bakerloo.

The number who arrive from Green Park and then change to the Bakerloo is equal to the number who are about to travel via Green Park.

The number who are travelling from the street to the Bakerloo is equal to four times the number who arrive in Piccadilly trains but do not use the Bakerloo Line, and of these, twice as many come from Green Park as from Leicester Square.

By how many does the number of people who use the Bakerloo Line exceed that of those who do not?

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The solution appears in the May 1941 issue: “Of the 128 people at Piccadilly, 104 either had used or were about to use the Bakerloo Line. Hence the answer required is 80.” (The Eureka archive is published under a Creative Commons license.)

February 26, 2021February 26, 2021 | Puzzles

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