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.
Riddler Express
From Gerald Dorrer, I reckon this here’s a math puzzle:
In the video game “Red Dead Redemption 2,” there is a side quest where the main character is supposed to collect 12 sets of cigarette cards, each consisting of 12 unique cards.
Some cards can be found lying around in the open world, but the easiest way to collect the cards is to buy cigarettes at the store and collect the single random card included in each pack. Suppose Arthur is too lazy to look for cards in the open world and tries to complete the set only by buying packs at the store.
At $5 a pack, how much money do we expect Arthur to spend to complete all 12 sets?
Riddler Classic
From Chadwick Matlin, the deputy editor of some website called FiveThirtyEight, a puzzle perfect for the polar vortex:
You’re snowed in alone with nothing to do but play a solitaire game of Bananagrams. As you spread its 144 lettered tiles out on the table in front of you, you begin to wonder:
What grid of words can you create that uses all of these tiles in the fewest possible words?
What grid uses all of the tiles in the most possible words?
(To test if a word is allowed, use the ENABLE word list, a variant of which is used in games such as Words With Friends, for the purposes of this problem. The letters in Bananagrams are distributed as shown here.)
Extra credit: How many completed grids are there, period, that use all 144 tiles?
Solution to the last week’s Riddler Express
Congratulations to 👏 Erin Seligsohn 👏 of Atlanta, winner of last week’s Riddler Express!
Last week, we found ourselves on an island with a curious property: All of its young people told the truth and all of its old people lied. Specifically, there was an age limit L — a positive integer — and all islanders who were younger than age L only told the truth, while islanders who were at least L years old only told lies. We met five of them, and they had this to say.
A: “B is more than 20 years old.”
B: “C is more than 18 years old.”
C: “D is less than 22 years old.”
D: “E is not 17 years old.”
E: “A is more than 21 years old.”
A: “D is more than 16 years old.”
B: “E is less than 20 years old.”
C: “A is 19 years old.”
D: “B is 20 years old.”
E: “C is less than 18 years old.”
What was L, and what could we learn about the ages of the islanders?
L was 19 years old. Islander A was 19 years old, B was 20, C was 18, D was at most 16 and E was at least 20.
The key is to ferret out the liars. Notice first that there are a few direct contradictions in the islanders’ statements: A and D contradict each other over B’s age, B and E contradict each other over C’s age, and E and C contradict each other over A’s age. So at least one member of the three pairs (A, D), (B, E) and (E, C) must be a liar. We also know that D and B aren’t both liars — that would make E both exactly 17 years old and more than 20 years old, which is impossible.
Given those facts, we have a few possibilities we can test. Let’s guess that A, B and E are liars and that C and D are truth-tellers, then check the consistency of their statements. There are two statements about each islander, which we can translate into true facts given our guess about the truthfulness of each of their speakers.
The statements about A check out: A is less than or equal to 21 and A is 19. That means A is 19.
The statements about B also check out: B is less than or equal to 20 and B is 20. That means B is 20.
The same goes for C: C is less than or equal to 18 and C is greater than or equal to 18. That means C is 18.
And D: D is less than 22 and D is less than or equal to 16. That means D is at most 16 years old.
And finally E: E is not 17 and E is greater than or equal to 20. That means E is at least 20.
Because all these statements check out, that means our guess worked! Since A, B and E are liars and C and D tell the truth, that means both of our truth-tellers are at most 18 and all of our liars are 19 or older. Therefore, L must equal 19.
And we’re done! And I don’t know about you, but I’m ready to get off this island.
Solution to last week’s Riddler Classic
Congratulations to 👏 Onufry Wojtaszczyk 👏 of Warsaw, Poland, winner of last week’s Riddler Classic!
Once we got off that island, we headed straight for some rest and relaxation at the card table, where we sat down to play a game of bridge. The game begins with an auction. There are four players, each sitting across from his or her partner. Simply put, the auction goes like this: Beginning with the dealer and orbiting around the table, players can place a bid or pass. Players who’ve passed can re-enter the bidding later. A bid is comprised of a number (one through seven) and a suit. Every legal bid must be higher than the one that came before, meaning either that its number is higher or that its number is the same and its suit is higher (from high to low, the suits go no-trump, spades, hearts, diamonds, clubs). The auction ends if the other three players pass after a bid, or if all four players pass right away.
But the R&R came to an abrupt end with this question: How many different legal bridge auctions are there?
There are 1,574,122,160,956,548,404,565, or more than a thousand billion billion, or more than the estimated number of grains of sand on the planet.
To get there, let’s start with a much smaller number: 35. A bid is made up of one of seven numbers and one of five “suits” (the four suits plus no-trump); multiplying numbers and suits gives us 35 possible bids, each with a unique rank.
From there, let’s build up to our final calculation. We know now that in every auction, the players will make somewhere between one and 35 bids. We will proceed by counting all the auctions with one bid, with two bids, with three bids and so on, and add up all of their possibilities in a big sum.
If an auction comprises \(N\) bids, say, then there are “35 choose \(N\)” choices for what specific bids those could be. And between those bids will be some number of passes. Before the first bid there could be zero, one, two or three passes (four possibilities), and between all subsequent bids there can be zero, one or two passes (three possibilities chosen at \(N-1\) points in the auction). We can account for passes by multiplying by those numbers of possibilities.
That leads us to our final, big sum:
\begin{equation*}1+\sum_{N=1}^{35} {35 \choose N}\cdot 4\cdot 3^{N-1}\end{equation*}
An online calculator can turn this into our numerical final answer. That lone “1+” at the front of the equation is the possibility that the auction ends immediately with all four players passing.
For extra credit, I asked how many different auctions there would be if we incorporated the bridge tactics known as doubling and redoubling. The calculation process is very similar, and the new formula — taking into account that there are now 21 intra-bid combinations of passes, doubles and redoubles, plus seven ways the auction ends with passes, doubles and redoubles — looks like this:
\begin{equation*}1+\sum_{N=1}^{35} {35 \choose N}\cdot 4\cdot 21^{N-1}\cdot 7\end{equation*}
And the final answer is, of course, bigger. Much, much bigger:
128,745,650,347,030,683,120,231,926,111,609,371,363,122,697,557.
But in the grand scale of gnarly game math, the number of possible bridge auctions isn’t all that enormous: It is only about 0.25 percent of the number of possible chess positions.
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 oliver.roeder@fivethirtyeight.com.