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 Jamie Wilkowski, who suggests that you think outside of the box:
What is the missing letter in the sequence below?
Riddler Classic
From Michael Branicky, your card has been declined:
Lucky you! You’ve won two gift cards, each loaded with 50 free drinks from your favorite coffee shop, Riddler Caffei-Nation. The cards look identical, and because you’re not one for record-keeping, you randomly pick one of the cards to pay with each time you get a drink. One day, the clerk tells you that he can’t accept the card you presented to him because it doesn’t have any drink credits left on it.
What is the probability that the other card still has free drinks on it? How many free drinks can you expect are still available?
Solution to last week’s Riddler Express
Congratulations to 👏 Andrea Ceres 👏 of Glen Rock, New Jersey, winner of last week’s Riddler Express!
Last week, you played your first ever game of “Ticket to Ride,” a contest in which players compete to lay down railroad tracks on a map of the country. At the start of the game, you were randomly dealt a set of three Destination Tickets out of a deck of 30 different tickets. (Each revealed two terminals you had to connect with a railroad to receive points.) During the game, you eventually picked up another set of three Destination Tickets, so you saw six of the 30 tickets in the game. Later, because you enjoyed the game so much, you and your friends played again. The ticket cards were all returned and reshuffled. Again, you were dealt a set of three tickets. Which was more likely, that you had seen at least one of the tickets before or that they were all new to you?
Although the chances of each were very similar, it was ever slightly more likely that you had seen one of the tickets before.
Let’s work out the chances that each of the three tickets you drew in the second game had previously been unseen by you. There were 24 tickets in the deck of 30 that you hadn’t seen. Therefore, there was a 24/30 chance that you would draw one of them as your first ticket, a 23/29 chance that you would draw one of them as your second ticket assuming that your first ticket was new, and a 22/28 chance that you would draw one of them as your third ticket assuming that your first and second tickets were new. That’s a (24/30)(23/29)(22/28) ≈ 49.85 percent that all three tickets were previously unseen. Therefore, the chance that you had seen one before, in your first game, was 50.15 percent.
And by the way, good luck building that route from Seattle to New York!
Solution to last week’s Riddler Classic
Congratulations to 👏 Nir Jacoby 👏 of New York City, winner of last week’s Riddler Classic!
Last week, you and nine others were competing in a spelling bee. You could each spell words perfectly from a certain portion of the dictionary but would misspell any word not in that portion. Specifically, you had 99 percent of the dictionary down cold, and your opponents had 98 percent, 97 percent, 96 percent and so on down to 90 percent memorized. The bee’s rules were simple: You took turns spelling in some fixed order, and at the end of a round, you started again with the first surviving speller. If you missed a word, you were out. And the last speller standing won. The bee words were chosen randomly from the dictionary.
If the contestants went in decreasing order of knowledge, so that you went first, what were your chances of winning the bee? If the contestants went in increasing order of knowledge, so that you went last, what were your chances of winning?
Not surprisingly given your prodigious word knowledge, you were an odds-on favorite in this field of 10. Somewhat surprisingly, however, the order in which contestants take their turns doesn’t matter much. If you had gone first, your chances of winning were about 52.0 percent. If you had gone last, your chances of winning were about 52.5 percent.
Essentially what we are trying to calculate is the probability that all of your competitors err before you do in the two scenarios. Solver Anthony Mulieri explained how to do this with probability theory:
The probability of your getting N-1 words correct and then getting one wrong is modeled by the geometric distribution (\(.01 \cdot .99^{N-1}\)). Therefore, to win before that happens, we want to calculate the probability that none of the other contestants would have correctly spelled every word they were given by then. This can be modeled with the binomial distribution, where the contestant with \(X\) of the dictionary memorized has a \(1 – X^N\) chance of getting all of those \(N\) words correct. These probabilities can then be multiplied together for \(X\) from 0.90 to 0.98 (all of your competitors) and multiplied again with \((.01 \cdot .99^{N-1})\). Summing this from \(N=0\) to \(N=\infty\) (all the numbers of words you might be posed), we arrive at our answer of about 52.5 percent. Additionally, since this counted on your failing after getting the same number right as the other players, this is the solution for when you come last in the spelling order. To get the solution when you are first in the order, just take the same terms but use \((.01 \cdot .99^N)\) instead of \((.01 \cdot .99^{N-1})\). That gives about 52.0 percent.
You could also tackle this problem with computer spelling simulations, as many solvers did. Bradley Gannon and Josh Nees were kind enough to describe their approaches and share their code. As Josh explained, the slight advantage to going last is related to the “last player standing” rule: “The winning player does not actually have to spell a word in the final round because the game ends when their last competitor misspells a word.”
And finally, Tyler Barron plotted the chances of winning for not just you but each of your fellow spellers under both the best-first and best-last ordering regimes:
The “worst” speller in the bee has less than a 0.5 percent chance of winning, it turns out. So you’re telling me I memorized 90 percent of the dictionary for nothing?!
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.