Can You Solve Lewis Carroll’s Tricky ‘Pillow Problem’?

Lewis Carroll is best known to most as the whimsical author of Alice in Wonderland, but did you know that he was also an avid puzzler and published mathematician? Among his many contributions was a book of math puzzles he called “Pillow Problems.” They are so named because Carroll developed them in bed to help distract himself from anxious thoughts while falling asleep. He wrote that as he shifted in bed he had two choices: “Either to submit to the fruitless self-torment of going over and over a worrisome subject, or dictate to myself a subject sufficiently intriguing to keep the concern Bay. A math problem Is, such an issue for me…” Personally, I’m referring to Carroll’s situation. I fall asleep pondering a riddle most nights of my life and have found it to be a powerful antidote to a restless head.

Missed the challenge last week? Listen Here, and find the solution at the end of today’s article. Be careful not to read too far ahead if you’re still working on this puzzle!

Mystery #4: Lewis Carroll’s pillow problem

You have an opaque bag containing a marble that has a 50/50 chance of being black or white, but you don’t know what color it is. You take a white marble from your pocket and put it in the bag. Then you shake up the two marbles in the bag, reach in and pull one out at random. It happens to be white. What is the probability that the other marble in the bag is also white?

Don’t let the easy setup fool you. This puzzle is famous for defying people’s intuitions. If you’re having trouble cracking it, think about it while you fall asleep tonight. It might at least quell your worries.

We will post the solution next Monday along with a new puzzle. Do you have an awesome riddle you think we should cover here? Send it to us: gizmodopuzzle@gmail.com

Solution to Puzzle #3: Calendar Cube

Last weeks puzzle asked her to design a working pair of calendar cubes. Remember that a die only has six sides. Every month has an 11th and a 22nd day, so the digits 1 and 2 must appear on both dice, otherwise those days could not be rendered. Note that both dice also require a 0. This is because the numbers 01, 02,… and 09 all need to be represented, and if only one die had a 0, there would not be enough faces on the other die to accommodate all nine of the other digits. This leaves three unoccupied areas on each cube, i.e. a total of six additional spaces. However, seven digits remain that need a home (3, 4, 5, 6, 7, 8, and 9). How can we squeeze seven digits onto six faces? The trick is that a 9 is an inverted 6! Beyond this realization, several tasks work. For example, place 3, 4, and 5 on one die and 6, 7, and 8 on the other. When the 9th rolls around, flip that 6 upside down and, by the skin of our teeth, we’ve got every date covered.

This solution has an economy that I find beautiful. Two dice lack space for the task, and yet we squeak past, exploiting a whimsical symmetry in our digits. Some might find this a gimmick, but that’s how store-bought calendar cubes work. If even one month of the year were extended to 33 days, the market for calendar cubes would collapse.

There are two natural extensions of the calendar cube puzzle to other date information. Amazingly, this hair width issue persists with them. What if we want to add a cube representing the day of the week? Tuesday and Thursday start with the same letter, so we have to allow two letters on a single die face to distinguish them: “Tu” and “Th”. Likewise with Saturday and Sunday, which we represent with ‘Sa’ and ‘So’. Monday, Wednesday and Friday have no conflicts, so ‘M’, ‘W’ and ‘F’ will do. We find ourselves in a well-known conundrum. We need to stuff seven symbols onto just six faces of a die. do you see the solution The god of symmetry honors us again by letting the “M” stand for Monday and vice versa for Wednesday.

We have months left, which I gave you as an extra challenge last week. Can we issue all three letter month abbreviations: ‘jan’, ‘feb’, ‘mar’, ‘apr’, ‘may’, ‘jun’, ‘jul’, ‘aug’, ‘sep’, ‘oct’, ‘nov’ and ‘dec’, with three more dice containing lower case letters? There are 19 letters involved in a month abbreviation: ‘j’, ‘a’, ‘n’, ‘f’, ‘e’, ​​’b’, ‘m’, ‘r’, ‘p’, ‘y’ , ‘u’, ‘l’, ‘g’, ‘s’, ‘o’, ‘c’, ‘t’, ‘v’, ‘d’, again exactly one too many for the 18 faces three dice . Would you believe me if I told you it existed? Only enough symmetry in our alphabet to tie each month into three cubes? The method requires that we recognize ‘u’ and ‘n’ as inverses of each other, as well as ‘d’ and ‘p’. One version is shown below:

dice 1 = [j, e, r, y, g, o]

dice 2 = [a, f, s, c, v, (n/u)]

dice 3 = [b, m, l, t, (d/p), (n/u)]

Somehow the few symmetries in our numbering and labeling systems perfectly allow the construction of calendar cubes for days, weeks and months without leaving any leeway.

You might be wondering: if there are 19 letters for 18 slots, why isn’t it enough just to combine the “u/n” pair or the “d/p” pair? It seems that one of the two would save the extra slot. The rest of the article answers that question and is a bit complicated. So only stay on board if you’re curious about the answer and don’t want to find out for yourself. The reason is that if ‘d’ and ‘p’ were split into two different faces and only ‘u’ and ‘n’ shared one face, we wouldn’t be able to form ‘jun’ which is ‘ u’ required. and ‘n’ to be representable on different cubes. On the other hand, suppose that only ‘d’ and ‘p’ share a face, while ‘u’ and ‘n’ do not. June’s abbreviation insists that ‘j’, ‘u’, and ‘n’ are on different cubes:

dice 1 = [j, …]

dice 2 = [u,…]

dice 3 = [n,…]

Also, ‘a’ must share a cube with ‘u’ to form ‘jan’:

dice 1 = [j, …]

dice 2 = [u, a, …]

dice 3 = [n,…]

But then how do we do ‘aug’? The letters “a” and “u” share a face. The only way out is to use the ‘u/n’ symmetry as well.

Let us know how you did in this week’s challenge in the comments.