Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
Duke Leto Atreides knows for a fact that there are not one, but two traitors within his royal household. The suspects are Lady Jessica, Dr. Wellington Yueh, Gurney Halleck and Duncan Idaho. Leto’s advisor, Thufir Hawat, will assist him in questioning the four suspects. Anyone who is a traitor will tell a lie, while anyone who is not a traitor will tell the truth.
Upon interrogation, Jessica says that she is not the traitor, while Wellington similarly says that he is not the traitor. Gurney says that either Jessica or Wellington is a traitor. Finally, Duncan says that Jessica or Gurney is a traitor. (Thufir, being the logician that he is, notes that when someone says thing A is true or thing B is true, both A and B can technically be true.)
After playing back the interrogations in his mind, Thufir is ready to report the name of one of the traitors to the duke. Whose name does he report?
From Allen Gu comes a geometric puzzle that he recalls from a university challenge:
Suppose you have an equilateral triangle. You pick three random points, one along each of its three edges, uniformly along the length of each edge — that is, each point along each edge has the same probability of being selected.
With those three randomly selected points, you can form a new triangle inside the original one. What is the probability that the center of the larger triangle also lies inside the smaller one?
Solution to last week’s Riddler Express
Congratulations to 👏 Tevis Tsai 👏 of Baltimore, Maryland, winner of last week’s Riddler Express.
Last week, the American League Championship Series of Riddler League Baseball determined one of the teams that will compete in the Riddler World Series. This year’s teams — the Tampa Bay Lines and the Minnesota Twin Primes — were evenly matched. In other words, both teams were equally likely to win each game of the best-of-seven series.
On average, how many games would the series have lasted? (Note that the series ended as soon as one team had won four games.)
One key here was to recognize that not every sequence of seven games was possible. If L and T represented victories for the Lines and the Primes, respectively, then you couldn’t have a seven-game series that was LLLLTTT. Why? Because after the Lines won the first four games, the series would be over.
Right off the bat (no pun intended), you knew that the only possible four-game series were LLLL and TTTT, each of which had a (1/2)4 — or 1/16 — chance of happening. Putting those two scenarios together, there was a 1/8 chance the series went exactly four games.
For a series to go five games, solver Paige Kester of Southlake, Texas, recognized that you needed four Ls and one T, or vice versa. It appeared that there were five ways to order four Ls and one T, but one of those ways was LLLLT, in which case the series would have been over after just four games. That meant there were four ways to order four Ls and one T, and then another four ways to order four Ts and one L. That gave you a total of eight five-game series, each with a (1/2)5 — or 1/32 — chance of happening. In other words, the probability of a five-game series was 8/32, or 1/4.
Instead of worrying about six games next, let’s jump ahead to seven. In a seven-game series, the first six games had to be evenly split between the two teams. There were 6 choose 3 (or 20) ways to sequence three Ls and three Ts. For the seventh and final game, there were two ways to choose a winner (L or T) for each of those 20 sequences, giving you a grand total of 40. Each seven-game series had a (1/2)7 — or 1/128 — chance of happening. Therefore, the probability of a seven-game series was 40/128, or 5/16.
Since every series had to be four, five, six or seven games, the remaining probability had to be accounted for by a six-game series. This probability was 1−1/8−1/4−5/16, or 5/16.
The average number of games was then 1/8·(4) + 1/4·(5) + 5/16·(6) + 5/16·(7), or 5.8125 games.
Solver David Ding observed that the frequency with which Major League Baseball’s World Series went to seven games was very close to the 5/16 calculated above. Furthermore, the historical average number of World Series games has been about 5.87. That’s a pretty good sign that teams tend to be evenly matched.
Finally, what if Riddler League Baseball had a different length of series, such as three games, five games or more than seven games? Solver Gary Yane noted that the answer could be found in sequence A033504 of the OEIS, by dividing each term in the sequence by a corresponding value of 4. A one-game series lasted one game on average (surprise, surprise), a three-game series lasted 10/41 (or 2.5) games, a five-game series lasted 66/42 (or 4.125) games and a nine-game series lasted 1,930/44 (or about 7.539) games.
There’s no doubt that a future riddle will look at what happens as the series gets longer and longer …
Solution to last week’s Riddler Classic
Congratulations to 👏 Andrew Vaughn 👏 of Moraga, California, winner of last week’s Riddler Classic.
Last week, over in the National League Championship Series, the Washington Rationals and the St. Louis Ordinals were also evenly matched. Again, both teams were equally likely to win each game of the best-of-seven series.
You entered a competition in which you had to predict the winner of each of the seven games before the series began. If any or all of the fifth, sixth or seventh game were not played, you were not credited with predicting a winner of that corresponding game.
You won the competition if you predicted at least two games correctly. If you optimized your strategy for picking winners, what was the probability you would win the competition?
At first glance, it might not have been clear how a strategy could help you here. Why would you bother picking one team over another in any given game? So if you cast strategy aside and just picked one team to win all seven games, then what was the probability you would have predicted at least two games correctly? The only ways a team wouldn’t win at least two games were if they got swept or if they lost in a five-game series. From last week’s Riddler Express, these probabilities were 1/16 and 1/8, respectively, for a total of 3/16. That meant if you picked the Rationals to win all seven games, your probability of winning the competition was 13/16, or 81.25 percent.
But as it turned out, this was the worst possible strategy.
Suppose instead that you had picked the Rationals to win the first four games and the Ordinals to win the last three, a sequence we can designate as RRRROOO. Had this been your prediction, you would have lost the competition if the actual series had been OOOO (with probability 1/16) or one of ROOORRR, OROORRR, OORORRR and OOORRRR (each with probability 1/128). Adding these up, the probability you lost was 1/16 + 4/128, or 3/32. That meant your probability of winning was 29/32, or 90.625 percent.
Solvers like Jake Gacuan and Emma Knight, who checked all 128 possible seven-game strategies, found that this was the best you could do. This strategy was optimally balanced when you lost: in the relatively unlikely events of a four-game sweep or a comeback from a 3-0 or 3-1 deficit.
For extra credit, you entered a second competition in which you picked the winner of the first game and then of each next game, knowing who had won in all the previous games. Again, if you optimized your strategy, what was the probability you would have predicted at least two games correctly?
This week’s winner, Andrew, solved this via dynamic programming. Andrew found that the best strategy here was to pick the team that had won more games (and to pick either team for Game 1 or whenever else the series was tied). The one exception to this was when one of the teams had won three games, was leading in the series, and up to that point you had predicted zero games correctly. In that case, if you correctly predicted that the leading team would win the next game, the series would end with just one game correctly predicted; in other words, you would have lost. So you might as well have picked the trailing team and hoped for a comeback. With this strategy, Andrew determined that your chances of winning were 15/16, or 93.75 percent.
Looking back on this puzzle and its extra credit, it was rather remarkable that being able to update your predictions after each game only increased your chances of winning by about 3 percentage points.
Oh, and one more thing. Last week, I offered some “super extra credit” on Twitter, asking readers to come up with their own best math-y baseball team names. Beyond the four from last week’s puzzles, I also considered including the Arizona Rhombusbacks. Because, of course.
But Riddler Nation did not disappoint with some fantastic suggestions, including:
That’s some incredible creativity right there. We almost have a full league’s worth of names for Riddler League Baseball!
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 Zach Wissner-Gross at [email protected].