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-sized 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 if you have a favorite puzzle collecting dust in your attic, find me on Twitter.
Riddler Express
From Mike Strong, a climate change problem:
In each of the last three years — 2014, 2015 and 2016 — a new global temperature record has been set. Assuming that accurate temperature records exist since 1880, what is the probability of this happening at random?
Riddler Classic
From Mont Chris Hubbard, a prediction puzzle inspired by his habit of trying to identify the funniest name as early as possible in a movie’s scrolling end credits:
From a shuffled deck of 100 cards that are numbered 1 to 100, you are dealt 10 cards face down. You turn the cards over one by one. After each card, you must decide whether to end the game. If you end the game on the highest card in the hand you were dealt, you win; otherwise, you lose.
What is the strategy that optimizes your chances of winning? How does the strategy change as the sizes of the deck and the hand are changed?
Solution to last week’s Riddler Express
Congratulations to ÑÑâÐ Bob Rietz ÑÑâÐ of Asheville, North Carolina, winner of last week’s Express puzzle!
In standard American bingo, a bingo card is a five-by-five grid of squares. The columns are labeled B, I, N, G and O, in that order. The five squares in the B column can be filled with the numbers 1 through 15, those in the I column with the numbers 16 through 30, those in the N column 31 through 45, and so on. The square in the very center of the grid is a “free space” on every card. How many different possible bingo cards are there?
There are 552,446,474,061,128,648,601,600,000.
For the four columns B, I, G and O — those without the free space — there are 15 ways to choose the first number in the column, then 14 ways to choose the second, 13 to choose the third, 12 to choose the fourth and 11 to choose the fifth. (The number of options descend because once a number is selected for one square, it can’t appear in another.) For the N column — with the free space in the middle — we need to pick only four numbers: There are 15 ways to choose the first, 14 ways to choose the second, 13 to choose the third and 12 to choose the fourth. Now, we multiply all those possibilities together!
\((15\cdot 14\cdot 13\cdot 12\cdot 11)^4\cdot (15\cdot 14\cdot 13\cdot 12)\approx 5.52\cdot 10^{26}\)
Solution to last week’s Riddler Classic
Congratulations to ÑÑâÐ Andrew Zwicky ÑÑâÐ of Cedar Falls, Iowa, winner of last week’s Classic puzzle!
Imagine that it’s the beginning of time, and the Supreme Court’s nine seats are empty. Assume further that seats on the bench are filled only if the same party controls both the presidency and the Senate. Every election, each of the two parties has a 50 percent chance of gaining control of the executive or legislative branch. Outcomes of the elections are independent, and the length of time for which a justice serves is uniformly distributed between zero and 40 years. What is the expected number of vacancies on the bench in the long run?
There are between 0.6 and 0.7 expected vacancies in this model.
Here’s a quick way to arrive at a good approximation of the expected number of vacancies. For starters, consider just a single vacant Supreme Court seat — we can multiply by nine when we’re done. Half the time, the same party will control both the presidency and the Senate, and a justice can be appointed immediately. The average wait this half of the time, then, is zero years.
The other half of the time, control will be split, in which case we’ll have to wait until at least the next election to fill the vacancy. On average, we’ll have to wait one year until the next election, since congressional elections happen every two years. But that election isn’t guaranteed to lead to a unified government. If it doesn’t, we’ll have to wait at least another two years to try again. By the second election, on average, we should have a unified government, since the chances of it happening are 50 percent. Put that all together, and the average wait to fill a vacancy if you start with a split government is three years.
Therefore, the average wait time to fill any given vacancy is \(0.5\cdot 0 + 0.5\cdot 3 = 1.5\) years. Since the average tenure of a justice in our model is 20 years, the probability that a given seat is vacant at any given time is \(\frac{1.5}{20+1.5}=\frac{3}{43}\). We can simply multiply that by nine to give the expected number of vacancies: \(9\cdot \frac{3}{43}\approx 0.628\).
Laurent Lessard went a step further and calculated the long-run distribution of vacancies:
But, as Hector Pefo points out, the precise answer is a bit more complicated. There’s a minor oversimplification in the assumptions behind the answer above. Specifically, it’s not exactly true that a justice’s term is equally likely to end in a unified government or a divided government. A justice necessarily begins his or her tenure in a united government, and if the tenure ends quickly, before any other elections have taken place, the two branches will still be aligned. After walking through some hairy math, including natural logarithms and integration, to account for that, Hector found an expectation of about 0.606 vacancies.
This week’s winner, Andrew Zwicky — along with many other solvers, including Daniel Thompson and Tiffany Washburn –– approached the problem using computer simulation and was kind enough to share his code. Here is Andrew’s chart of five centuries of the evolution of the simulated Supreme Court:
Brian Corrigan extended this problem to consider smaller courts of just three, five or seven justices. While the number of expected vacancies increased as the size of the court increased, Brian found that the percentage of the court that was expected to be vacant stayed roughly constant. He also found that, as the average tenure of a justice increases, the expected number of vacancies decreases.
But the ÑÑÐâ Coolest Riddler Extension Award ÑÑÐâ goes to Conor Smith, who complicated the outcome of elections. In the original problem, the outcomes of presidential and senatorial elections were independent, with a correlation of zero. But in reality, there is some correlation in the outcomes of these elections — if a party wins the Senate, it is more likely to have won the presidency. As that correlation increases, he found, the expected number of vacancies on the Supreme Court decreases, eventually settling at around 0.45 vacancies when the outcomes of the elections are perfectly correlated.
Want to submit a riddle?
Email me at oliver.roeder@fivethirtyeight.com.