Can The Riddler Bros. Beat Joe DiMaggio’s Hitting Streak?

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 Scott O’Neil, unriddle this sequence:

What number comes next?

2
6
10
3
8
9
4
7
?

Submit your answer

Riddler Classic

From Steven Pratt, where have you gone, Joe DiMaggio? To Riddler Nation, perhaps:

Five brothers join the Riddler Baseball Independent Society, or RBIs. Each of them enjoys a lengthy career of 20 seasons, with 160 games per season and four plate appearances per game. (To make this simple, assume each plate appearance results in a hit or an out, so there are no sac flies or walks to complicate this math.)

Given that their batting averages are .200, .250, .300, .350 and .400, what are each brother’s chances of beating DiMaggio’s 56-game hitting streak at some point in his career? (Streaks can span across seasons.)

By the way, their cousin has a .500 average, but he will get tossed from the league after his 10th season when he tests positive for performance enhancers. What are his chances of beating the streak?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 🎖 Alex Murray 🎖 of Philadelphia, winner of last week’s Riddler Express and the new Median Marquess of Riddler Nation!

Last week brought two interactive challenges. In the first, you submitted a whole number between 1 and 1 billion, and whoever submitted the number closest to 2/3 of the mean of all the submitted numbers won.

I received over 2,500 submissions. The mean of all the valid submissions was 163,918,246. Two-thirds of that is 109,278,831, or nearly 11 percent of the maximum submission. Our winner, Alex Murray, submitted 109,185,185 — off by less than 100,000. That particular number was chosen because, as Alex rather cryptically explained, it was “163,777,777 times two-thirds.” In any case, that worked! Well chosen, Alex.

Here is the distribution of all of the numbers submitted by the Nation, in bins of 10 million. Five of you said you just guessed your phone numbers.

I see you, people who submitted 1 billion. You had no chance of winning, but you did have an effect on everyone else. One such submitter, Zane from Pittsburgh, explained: “I seek chaos.” Another, Will Heath, said that he “just wants to watch the world burn.” Yet another, Emma Cooper, wrote, “I do not intend to be right but I intend to make other people wrong.”

As the benevolent leader of this column I am shocked and appalled. Shocked and appalled, I say!

This challenge is well known in game theory and it is called — also shockingly — guess 2/3 of the average. In this game, there is a unique (pure strategy) Nash equilibrium, meaning a set of actions for all the players from which no player would want to deviate. We … did not achieve it.

To achieve equilibrium, each player would submit the lowest possible number, and everyone would then share in the prize. Why? You can arrive at this result through a process known as the iterated elimination of weakly dominated strategies. For example, no player would want to submit a number above 666,666,666, because that’s 2/3 of the highest possible submission, which means 2/3 of the average submission cannot possibly be higher than that. Given that, no player would want to submit a number above 444,444,444 (2/3 of 666,666,666), and then no number above 296,296,296, and so on. Eventually, by this iterative logic, the only submission that makes any strategic sense is 1.

Therefore, I’m proud to share our new slogan. Riddler Nation: Out of Equilibrium Since 2015.

Solution to last week’s Riddler Classic

Congratulations to 👑 Vince Vatter 👑 of Gainesville, Florida, winner of last week’s Riddler Classic and the reigning king of Riddler Nation!

Last week also brought the third installment of the Riddler Nation Battle Royale. These were the rules: In a distant, war-torn land there were 10 castles. There were two warlords: you and your archenemy. Each castle had its own strategic value for a would-be conqueror. Specifically, the castles were worth 1, 2, 3, …, 9 and 10 victory points. You and your enemy each had 100 soldiers to distribute, in any way you like, to fight at any of the 10 castles. Whoever sent more soldiers to a given castle conquered that castle and won its victory points. (If you each sent the same number of troops, you split the points.) You didn’t know the distribution of the enemy forces until the battles began. Whoever won the most points won the war.

You submitted a plan distributing your 100 soldiers among the 10 castles. I took all these battle plans and adjudicated all the possible one-on-one matchups.

I received nearly 1,500 battle plans, nearly 1,200 of which were valid — i.e., they only used 100 soldiers and only whole numbers of soldiers and so forth. And without further ado, here were the top performers:

The top warlords relied heavily on the data from the first two such battles. And, amazingly, this is Vince’s second consecutive Riddler warlord victory! He explained that he “used a genetic algorithm (the same as in the last competition) to explore distributions that would be good against the second-round distributions and the first- and second-round distributions combined. Then I used the same algorithm to optimize against those and the first- and second-round distributions simultaneously.”

Our runner-up, Nate, explained: “I used k-medoid clustering to find median strategies that represent the most common strategies, then found an allocation of soldiers that beat the eight most common strategies. I then used that as an initial input to Robbie Ostrow’s simulated annealing code from Part 2, which spat out the above.” Well spat, Nate.

And our fourth-place finisher, Alex, explained simply: “A computer told me to.” Fair enough.

Most successful strategies, broadly speaking, devoted more troops to the high-value castles and ignored the low-value castles, sometimes entirely — none of which is too surprising. But that relationship is not perfect, and the win-loss record of any given strategy is highly sensitive to specific deployments. Our top performers tended to fight hard for Castles 3 and 6, for example. (Note also the tight similarity between the first- and second-place strategies above.)

Here is how the average soldier deployments broke down last week compared to the previous battles:

On the whole, this third edition of the battle royale saw a dramatic increase in soldiers sent to the highest-value castles — Castles 9 and 10 — with three and four more soldiers sent there on average. This additional firepower was, on average, siphoned mostly from the battles for Castles 3 and 7, where two fewer troops were sent.

This challenge is also well known in game theory, where it is called the Colonel Blotto game. However, unlike the 2/3-of-the-average game, there is no simple equilibrium — the analysis involves gnarly mixed strategies and often relies heavily on computation. It was analyzed as early as 1950 by researchers at the RAND Corporation, and it has applications in military strategy and political competition — including in yours truly’s doctoral thesis.

Though it is all quite easy, apparently, if your name is Vince Vatter.

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.