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,^{1} 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.

## Riddler Express

There are 200 episodes in a season of *Riddler Jeopardy!*. The first episode of the season features three brand-new contestants. Each subsequent episode includes a returning champion (the winner of the previous episode) as well as two new challengers.

Throughout the season, it so happens that the returning champions are particularly strong, with each one winning five consecutive episodes before being dethroned on the sixth.

If you pick a contestant at random from the season, what is the probability that they are a *Riddler Jeopardy!* champion (meaning they won at least one episode)?

The solution to this Riddler Express can be found in the following column.

## Riddler Classic

On June 24, the following challenge was posted on Twitter:

In this challenge, the numbers from 1 to 11 are arranged in a circle in a particular order: 1, 4, 8, 7, 11, 2, 5, 9, 3, 6, 10. You then have to connect pairs of numbers with straight line segments that don’t intersect, and your score is the sum of the products of the joined numbers. For example, with the connections {1, 4}, {8, 10}, {3, 7}, {5, 9}, and {2, 11} (and the 6 left by itself), you get a score of 1·4 + 8·10 + 3·7 + 5·9 + 2·11, or 172.

The *best* score you can achieve with this ordering of 1 through 11 around the circle is 237, which you can get with the following connections: {6, 10}, {3, 4}, {7, 8}, {9, 11} and {2, 5} (and the 1 left by itself).

This got Friend-of-The-Riddler Tyler Barron and me thinking about possible extensions of this challenge. If you want the highest possible maximum score, then you can rearrange the numbers from 1 to 11 so that they are in numerical order around the circle. (With this arrangement, the maximum score is 250.)

But what if you want the *lowest* possible maximum score? That is, how can you order the numbers from 1 to 11 around the circle so that the maximum possible score is as low as possible? And what is the resulting score?

The solution to this Riddler Classic can be found in the following column.

## Solution to last week’s Riddler Express

Congratulations to 👏 Joshua Korn 👏 of Cambridge, Massachusetts, winner of last week’s Riddler Express.

Last week, you had a bag containing 100 marbles. Each marble was one of three different colors. If you drew three marbles at random, the probability that you got one of each color was *exactly* 20 percent.

How many marbles of each color were in the bag?

Suppose the colors are red, green and blue, and that *r*, *g* and *b* represented the number of marbles of each color. You immediately knew that their sum, *r*+*g*+*b*, equaled 100 — the total number of marbles in the bag.

Next, there were six different ways to draw three marbles of different colors:

- red, green, blue
- red, blue, green
- green, red, blue
- green, blue, red
- blue, red, green
- blue, green, red

Moreover, each of these six permutations was equally likely. For example, the probability of first drawing a red was *r*/100. The probability of *then* drawing a green was *g*/99, since there were now only 99 marbles remaining in the bag. Finally, the probability of drawing a blue was then *b*/98. That meant the probability of each permutation was *rgb*/(100·99·98). To find the probability of drawing three different colors, you simply had to add these six probabilities together, giving you 6·*rgb*/(100·99·98), or *rgb*/161,700.

Finally, you were told that this probability equaled 20 percent, which told you that *rgb*/161,700 = 1/5. Multiplying both sides by 161,700 gave you *rgb* = 32,340.

At this point, you were looking for three whole numbers whose sum was 100 and whose product was 32,340. For some solvers, It was helpful to write 32,340 as a product of primes: 2^{2}·3·5·7^{2}·11. From here, you could try out different ways of separating these primes into three larger factors. For example, suppose these factors were 2^{2}·5, 3·7 and 7·11. Their product was indeed 32,340 (no surprise), but their sum was 20+21+77, or 118. Because the sum was *not* 100, these factors were *not* the correct values of *r*, *g* and *b*.

Many solvers found a trio of factors that worked by trial and error. Pramod Mathai applied Maclaurin’s inequality to show that *r*, *g* and *b* were all at least 20 and at most 50, which significantly reduced the number of possibilities to check.

Matt Enlow, the puzzle’s author, restricted this range a little further. Suppose the largest of the three factors was 49. To maximize their product, the other two factors (which added up to 51) would have been 25 and 26. However, 25·26·49 = 31,850, a few hundred shy of the required product — 32,340. That meant all three factors were at most 48, and that the primes 7, 7 and 11 all had to occur in different factors. In the end, the unique trio of factors was **21, 35 and 44**.

By the way, another way to interpret the problem was to place each marble back in the bag after is was selected (known as picking *with replacement*). Again, you needed *r*+*g*+*b *to equal 100. However, this time, the product *rgb* had to equal approximately 33,333.33. Since there was no way to multiply integers and get a non-integer product, picking with replacement was not an option.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Chris Bedell 👏 of Ashtead, England, winner of last week’s Riddler Classic.

Last week, you were playing Riddler Solitaire, a game with 11 cards: an ace, a two, a three, a four, a five, a six, a seven, an eight, a nine, a 10 and a joker. Each card was worth its face value in points, while the ace counted for 1 point. To play a game, you shuffled the cards so they were randomly ordered, and then turned them over one by one. You started with 0 points, and as you flipped over each card your score increased by that card’s points — as long as the joker hadn’t shown up. The moment the joker appeared, the game was over and your score was 0. The key was that you could stop at any moment and walk away with a nonzero score.

What strategy maximized your expected number of points?

To solve this, Guy D. Moore and Ming Chan both asked what would happen with the very next card you picked. Suppose the cards you had already turned over had a sum of *S*. Since the sum of all 10 cards (excluding the joker) was 55, that meant there were 55−*S* points remaining in your hand. Furthermore, suppose you had *N* cards remaining in your hand. On average, how many points would the *next* card have been worth?

If the next card was the joker, it would have zeroed out your score. In this case, it would have technically been worth −*S* points. That meant the total number of points remaining in your hand was 55−*S* (from the non-joker cards) *plus* another −*S* (from the joker). In other words, there were 55−2*S* points among *N* cards, meaning the next card was worth an average of (55−2*S*)/*N* points. The expected value was positive as long as 2*S* was less than 55, or *S* was less than 27.5. This result translated into the following optimal strategy:

**Whenever your current score was 27 or less, flip over another card.****Whenever your current score was 28 or greater, stop.**

For extra credit, you had to determine *how many* points this optimal strategy would earn you on average. There were 11! — or 39,916,800 — equally likely orderings for the 11 cards in the deck, which meant a lot of processing power was required.

Or was it? Solver Goh Pi Han of Malaysia reduced this figure by nearly an order of magnitude, since you would always flip the first three cards (whose sum was at most 27). In the end, the average score was **10,709/693, or about 15.453 points**.

Several solvers, like Emma Knight, observed that if you played optimally, you flipped over the joker and scored 0 half the time. Allen Gu explained this phenomenon by pairing each ordering of cards with the reverse ordering. For example, in a certain ordering you might have 26 points *before* the joker and 29 points *after* it; with the reverse ordering, you would have 29 points *before* the joker and 26 points *after* it. Within each pair, the ordering with 28 or more points before the joker would net you those points! Meanwhile, the other ordering would always net you 0 points. In which case, the joker’s on *you*!

## 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 riddlercolumn@gmail.com.