Equal Opportunity

Can two dice be weighted so that the probability of each of the numbers 2, 3, …, 12 is the same?

	Select Click for Answer
I think this problem was first proposed by J.B. Kelly in the American Mathematical Monthly in 1950. The answer is no. Let a_j and b_j be the probabilities that the number j turns up on each of the two dice, and let P_j be the probability of getting the sum j on the pair of dice. So P₂ = a₁b₁ = a₆b₆ = P₁₂. Now, to maintain this balance, if the first die favors 1 over 6, then the second must favor 6 over 1, and vice versa, so that (a₁ – a₆)(b₁ – b₆) ≤ 0. But this means that P₂ + P₁₂ = a₁b₁ + a₆b₆ ≤ a₁b₆ + a₆b₁ ≤ P₇. The probabilities of rolling a 2, a 7, and a 12 cannot be equal.

August 27, 2014September 7, 2014 | Puzzles · Science & Math

Black and White

By Francis Healey. White to mate in two moves.

	Select Click for Answer
1. Be5! and the queen will mate. From A Collection of Two Hundred Chess Problems, 1866.

August 23, 2014August 31, 2014 | Puzzles

Half and Half

A bisecting arc is one that bisects the area of a given region. “What is the shortest bisecting arc of a circle?” Murray Klamkin asked D.J. Newman. Newman supposed that it was a diameter. “What is the shortest bisecting arc of a square?” Newman answered that it was an altitude through the center. Finally Klamkin asked, “And what is the shortest bisecting arc of an equilateral triangle?”

“By this time, Newman had suspected that I was setting him up (and I was) and almost was going to say the angle bisector,” Klamkin writes. “But he hesitated and said let me consider a chord parallel to the base and since this turns out to be shorter than an angle bisector, he gave this as his answer.”

Was he right?

	Select Click for Answer
“Unfortunately for him, the correct answer is a 60° arc of a circle.” Reflect the triangle about successive bases to make a regular hexagon. Now we need a figure that occupies half the hexagon’s area and has the shortest possible perimeter. That’s a circle: “Incidentally, it is a very rare occasion when I can catch Newman on problems such as these.” From Klamkin’s paper “Mathematical Creativity in Problem Solving and Problem Proposing II,” in Joby Milo Anthony, ed., In Eves’ Circles, 1994, citing George Pólya’s Mathematics and Plausible Reasoning, 1954.

Click for Answer

August 22, 2014August 31, 2014 | Puzzles · Science & Math

Change of Venue

A and B are playing a simple game. Between them are nine tiles numbered 1 to 9. They take tiles alternately from the pile, and the first to collect three tiles that sum to 15 wins the game. Does the first player have a winning strategy?

	Select Click for Answer
No. Arrange the numbers into a magic square: 8 1 6 3 5 7 4 9 2 Now A and B are playing tic-tac-toe! And, hence, best play always leads to a draw. From Elwyn R. Berlekamp, John H. Conway, Richard K. Guy, Winning Ways for Your Mathematical Plays, vol. 2, 2001.

August 19, 2014August 24, 2014 | Puzzles

The Wire Identification Problem

A conduit carries 50 identical wires under a river, but their ends have not been labeled — you don’t know which ends on the west bank correspond to which on the east bank. To identify them, you can tie together the wires in pairs on the west bank, then row across the river and test the wires on the east bank to discover which pairs close a circuit and are thus connected.

Testing wires is easy, but rowing is hard. How can you plan the work to minimize your trips across the river?

	Select Click for Answer
Suppose the wires are labeled W₁, W₂, etc., on the west bank and E₁, E₂, etc. on the east bank. Start by tying W₁ to W₂, W₃ to W₄, and so on, leaving W₄₉ and W₅₀ untied. Now row across the river and test the wires to see which are connected. We might find that E₁₆ is tied E₃₇, E₁₁ is tied E₂₆, E₇ is tied E₄₁, and E₁₉ and E₂₇ are unconnected. Row back to the west bank, untie the wires, and now connect W₂ to W₃, W₄ to W₅, and so on, leaving W₁ and W₅₀ unconnected. Now return to the east bank and test again. We might learn that E₄₁ is tied E₁₉, E₂₆ is tied E₇, E₃₇ is tied E₁₁, and E₁₆ and E₂₇ are unconnected. Happily, we can now deduce all the connections. There will be one wire that had been paired in our first test but not in our second — here that’s E₁₆. That wire corresponds to W₁. The wire that E₁₆ had been paired with in our first test, E₃₇, must then correspond to W₂. And E₃₇ is now connected to E₁₁, so E₁₁ must be W₃. And so on: We can work down the list from W₁, alternately consulting the current connections and our memory of the first connections to reveal each wire in succession. So only three rowings are necessary. From Roland Sprague, Recreation in Mathematics, 1963. Sort of related: There’s a light bulb in the attic, and you’re downstairs facing three on-off switches. How can you tell which switch controls the light with only a single trip to the attic? Turn on Switch 1, leave it on for a minute, then turn it off, turn on Switch 2, and run upstairs. If the bulb is on, then Switch 2 controls it. If the bulb is off and warm (ha!), then Switch 1 controls it. Otherwise it’s Switch 3.

Click for Answer

August 17, 2014February 22, 2021 | Puzzles

Sleeping Alone

Lori has an icky problem: Worms keep crawling onto her bed. She knows that worms can’t swim, so she puts each leg of the bed into a pail of water, but now the worms crawl up the walls of the room and drop onto her bed from the ceiling. She suspends a large canopy over the bed, but worms drop from the ceiling onto the canopy, creep over its edge to the underside, crawl over the bed, and drop.

Desperate, Lori installs a water-filled gutter around the perimeter of the canopy, but the worms drop from the ceiling onto the outer edge of the gutter, then crawl beneath. (The worms are very determined.) What can Lori do?

	Select Click for Answer
She’s on the right track. Lori can stop the worms by installing a gutter around the canopy’s perimeter, but one that extends inward, under its lip, rather than outward: Now a worm that drops from the ceiling onto the canopy and creeps beneath will pass under this gutter and find itself surrounded by a moat, of sorts, that prevents it from reaching the underside of the canopy. From MIT’s Balint Virag, via Dartmouth mathematician Peter Winkler’s excellent Mathematical Mind-Benders (2007). Winkler adds, “If the worms have no higher-up way to get in her bedroom, Lori can more easily accomplish this same task by encircling the room itself with a water-filled gutter.”

Click for Answer

August 14, 2014August 24, 2014 | Puzzles

Black and White

pierce chess problem

By W. Timbrell Pierce. White to mate in two moves.

	Select Click for Answer
1. e6! and the pawn or bishop will mate. From Chess Problems, 1873.

August 6, 2014August 17, 2014 | Puzzles

Black and White

kirtley chess problem

Tim Krabbé calls this “one of the funniest chess problems I ever saw.” Its composer, M. Kirtley, won first prize with it in a Problemist tourney in 1986.

It’s a selfmate in 8, which means that White must force Black to checkmate him 8 moves, despite Black’s best efforts to avoid doing so.

The solution is a single line — all of Black’s moves are forced:

1. Nb1+ Kb3 2. Qd1+ Rc2 3. Bc1 axb6 4. Ra1 b5 5. Rh1 bxc4 6. Ke1 c3 7. Ng1 f3 8. Bf1 f2#

kirtley chess problem - solution

All of White’s pieces have returned to their starting squares!

July 31, 2014July 31, 2014 | Puzzles

Gear Trouble

A problem from the U.S.S.R. mathematical olympiad:

You’re given 13 gears. Each weighs an integral number of grams. Any 12 of them can be placed on a pan balance, with 6 in each pan, so that the scale is in equilibrium. Prove that all the gears must be of equal weight.

	Select Click for Answer
Each gear weighs either an even or an odd number of grams. Any set of 12 gears can be divided into two groups of equal weight, so a set of 12 gears has a total weight that’s even. This weight remains even if one of the 12 gears is exchanged with the 13th gear, and this is true no matter which of the 12 is exchanged. So either all the gears are of even weight or all are of odd. Now find the weight of the lightest gear and subtract it from all 13 gears. This gives us a set of 13 gears that still fulfill the conditions of the problem (though one or more of them now has zero weight). But now we know that each gear has an even weight: If all had originally been even, then we’ve subtracted an even number to produce an even number. And if all were originally odd, then we’ve subtracted odd from odd to produce even. Now, since each gear has even weight, we can divide the weight of each gear by 2 to obtain another set of 13 gears that still fulfills the conditions of the problem. If all the original gears had the same weight, we can continue to divide by 2 repeatedly without a problem: With each division, we produce a new set that satisfies the conditions, and every gear’s weight will have the same parity, even or odd. But if the original group of gears contained gears of differing weights, then repeated division by 2 will eventually produce a situation in which some weights are even and some odd. No such set can satisfy the original conditions of the problem. So all the original gears must have had the same weight.

Click for Answer

July 31, 2014August 10, 2014 | Puzzles

Coming and Going

A puzzle by Pierre Berloquin:

In my house are a number of rooms. (A hall separated from the rest of the house by one or more doors counts as a room.) Each room has an even number of doors, including doors that lead outside. Is the total number of outside doors even or odd?

	Select Click for Answer
Call a door internal if it opens between rooms or external if it opens outside. And designate each room either O (if it has an odd number of internal doors) or E (if even). Each E room has an even number of external doors, since all rooms have an even number of doors and even – even = even. So the E rooms, taken as a group, have a total number of external doors that’s even. The total number of O rooms is either odd or even. If it’s odd, then the total number of doors in the O rooms is odd, since the sum of an odd number of odd numbers is odd. But this is impossible: Every internal door must open on two rooms, and this would leave one door that has no second room to open on. So the total number of O rooms is even. Since each O room has an odd number of internal doors and an even number of total doors, each must have an odd number of external doors. Because the total number of O rooms is even, this means that the O rooms taken as a group have an even number of external doors (since the sum of an even number of odd numbers is even). Since both groups, the E rooms and the O rooms, have an even number of external doors, so does the whole house. L.A. Graham provided a simpler solution in 1959.

Click for Answer

July 26, 2014August 3, 2014 | Puzzles