Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
The Riddler will be taking next week off to celebrate the anniversary of the birth of the United States as well as of its editor, who will be officially in his mid-30s and none too happy about it. So enjoy the puzzles and the fireworks and we’ll see you back here in two weeks.
From Ralph Henrickson, a classic problem of probability manipulation:
You have two buckets and 100 ping-pong balls, 50 of which are red and 50 of which are blue. You get to arrange the balls into the two buckets however you’d like, but each bucket needs at least one ball. Your friend will blindly choose one of the two buckets and then select a ball at random from the chosen bucket.
How can you arrange the balls to maximize the probability that your friend chooses a red ball? What probability of success do you achieve?
Extra credit: What probability of success do you get with 25 balls of each color? 200 balls of each color?
From Benjamin Danard, the Superstring Scrabble Challenge:
The game of Scrabble has 100 tiles — 98 of these tiles contain a letter and a score, and two of them are wildcards worth zero points. At home on a lazy summer day with a bag of these tiles, you decide to play the Superstring Scrabble Challenge. Using only the 100 tiles, you lay them out into one long 100-letter string of your choosing. You look through the string. For each word you find, you earn points equal to its score. Once you find a word, you don’t get any points for finding it again. The same tile may be used in multiple, overlapping words. So ‘“theater” includes “the,” “heat,” “heater,” “eat,” “eater,” “ate,” etc.
The super challenge: What order of tiles gives you the biggest score? (The blank tiles are locked into the letter they represent once you’ve picked it.)
The winner, and inaugural Wordsmith Extraordinaire of Riddler Nation, will be the solver whose string generates the most points. You should use this word list to determine whether a word is valid.
For reference, this is the distribution of letter tiles in the bag, by their point value:
1: E×12 A×9 I×9 O×8 N×6 R×6 T×6 L×4 S×4 U×4
2: D×4 G×3
3: B×2 C×2 M×2 P×2
4: F×2 H×2 V×2 W×2 Y×2
8: J X
10: Q Z
Solution to last week’s Riddler Express
Congratulations to 👏 Sarah Mikkonen 👏 of Calumet, Michigan, winner of last week’s Riddler Express!
Last week you went on a very long walk from the eastern shore of Lake Tahoe, southeast into Utah, through the Four Corners Monument and into New Mexico, many miles east into Oklahoma, to Tulsa where you rested a bit, then east to Fayetteville, Arkansas, then south to Shreveport, Louisiana, where you ate a po’boy before turning back north, crossing back into Arkansas before stepping into Missouri near Branson, and then, in Iowa, passed a few miles east of Des Moines when, finally, too tired to go on, you ended your trek at the SPAM Museum in Austin, Minnesota. You walked through parts of nine states. What did these nine contiguous states have in common that none of the other 41 states share?
The nine states you visited are the only ones with multiple-word capital cities. Racking your brain back to third-grade social studies, those are: Carson City, Nevada; Salt Lake City, Utah; Santa Fe, New Mexico; Oklahoma City, Oklahoma; Little Rock, Arkansas; Baton Rouge, Louisiana; Jefferson City, Missouri; Des Moines, Iowa; and St. Paul, Minnesota.
In any case, at least you got your steps in!
Solution to last week’s Riddler Classic
Congratulations to 👏 Chris Gibson 👏 of San Francisco, winner of last week’s Riddler Classic!
Last week, there was a square table hidden from you behind a curtain with a quarter on each of its four corners. Your goal was to get all of the quarters to be heads up — if at any time that happened, you would immediately be told and would win. The only way you could affect the quarters was to tell the person behind the curtain to flip over as many quarters as you liked and in whatever corners you specified. (For example, you could’ve said, “Flip over the top left quarter and bottom right quarter,” or, “Flip over all of the quarters.”) Flipping over a quarter changed it from heads to tails or tails to heads. However, after each command, the table was spun randomly to a new orientation (that you didn’t know), and you must have given another instruction before it was spun again. Could you find a series of steps that guaranteed you would have all of the quarters heads up in a finite number of moves?
Indeed you could. In fact, you could guarantee it in as few as 15 moves.
For starters, there are 16 (\(2^4\)) different possible arrangements of quarters in the corners, one of which — all heads — is our winning position. However, because the table is rotated randomly to a position that we do not know on each move, some of these arrangements are indistinguishable from our point of view — they are just rotated versions of other arrangements. The random spinning actually simplifies things for us. There are really only six arrangements we need to consider: all heads, all tails, one head, one tail, two adjacent heads, and two diagonal heads.
Similarly, there are 15 different moves we could make (\(2^4\) minus the meaningless move in which we flip no quarters). But again, because of the unknown rotation of the table, there are really only five moves from our point of view: flip one, flip two adjacent, flip two opposite, flip three and flip all four.
Solver Laurent Lessard illustrated all the possible transitions of the quarters that result from these moves (the move “flip three” turns out not to be necessary):
Given these moves, we can now try to assemble a winning strategy. A couple of further observations help us here. For example, as solver Hector Pefo explains, before we do any non-flip-four move, we should flip four. That will allow us to test if the table is in the “all tails” state — if it is, we’ve secured a win, and if it’s not, we keep going.
Our other moves work similarly, allowing us to move between the possible states of the table — for example, “flip two adjacent” allows us to move from “two adjacent heads” to either “two opposite heads” or “all tails.” And “flip two opposite” allows us to move from “two opposite heads” to “all tails.” And so on. If we assemble these moves just so, we can travel through every state, eventually guaranteeing a win.
Here’s a recipe that does just that:
Flip four, flip two opposite, flip four, flip two adjacent, flip four, flip two opposite, flip four, flip one, flip four, flip two opposite, flip four, flip two adjacent, flip four, flip two opposite, flip four.
Finally, solver Zach Wissner-Gross animated a winning strategy in action. Notice, of course, that by the 15th move we’ve won every table.
Have a flippin’ great Fourth of July!
Want more riddles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
Want to submit a riddle?
Email me at firstname.lastname@example.org.