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

From Matt Enlow comes an original puzzle of his that previously appeared in Math Horizons. (At first it may seem like there’s information missing, but I assure you that is not the case.)

A bag contains 100 marbles, and each marble is one of three different colors. If you were to draw three marbles at random, the probability that you would get one of each color is *exactly* 20 percent.

How many marbles of each color are in the bag?

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

## Riddler Classic

This week’s Classic comes from Henk Tijms, author of *Basic Probability, What Every Math Student Should Know*.

Riddler solitaire is played 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 is worth its face value in points, while the ace counts for 1 point. To play a game, you shuffle the cards so they are randomly ordered, and then turn them over one by one. You start with 0 points, and as you flip over each card your score increases by that card’s points — as long as the joker hasn’t shown up. The moment the joker appears, the game is over and your score is 0. The key is that you can stop any moment and walk away with a nonzero score.

What strategy maximizes your expected number of points?

*Extra credit: *With an optimal strategy, how many points would you earn on average in a game of Riddler solitaire?

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

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐ Brian Gregorich ÑÑâÐ of Las Vegas, Nevada, winner of last week’s Riddler Express.

Last week, Magritte went bowling. Now, scoring in bowling can be a tricky matter. There are 10 frames, and in each frame you get two chances to knock down as many of the 10 pins as you can. Each pin knocked down is worth 1 point, and the pins are reset after each frame. Your score is then the sum of the scores across all 10 frames.

If only it were that simple. There are special rules for strikes (when you knock down all 10 pins with your first ball). Whenever you bowl a strike, that frame is scored as 10 *plus* the scores of your next two rolls. For example, if you bowled three strikes and missed every subsequent shot (i.e., they were gutter balls), your third frame would be worth 10 points, your second frame would be worth 20 and your first frame 30. Your final score would be 30 + 20 + 10, or 60 points.

Anyway, the bowling gods had decided that Magritte would bowl *exactly* three strikes in three randomly selected frames. All the other frames would consist of nothing but gutter balls. Magritte also lacked patience for bowling’s particular rules. If one of his three strikes came on the 10th and final frame and he was prompted to bowl further to complete the game, he would bowl gutter balls out of frustration.

What score could Magritte expect to achieve? (That is, with three randomly placed strikes, what was his average score?)

First off, there were 10 choose 3 — that is, (10·9·8)/(3·2·1), or 120 — cases to consider. Now if no two strikes were consecutive, Magritte would simply score 30 points. If *all three* occurred in a row, Magritte would score 60 points. And if exactly *two* were consecutive, then the second of them was worth 20 points, while the other two were worth 10 points — good for 40 points in total.

From here, all that was left was to figure out how many of the 120 cases were worth 30, 40 or 60 points.

There were only eight ways to score 60 points (i.e., three consecutive strikes), each corresponding to a first strike between the first and eighth frame. But counting up the 40-point and 60-point cases was trickier, and there were several ways to do this.

One clever approach for counting up the 40-point cases, suggested by solver Denward Chung of Louisville, Kentucky, was to group cases by where their consecutive strikes occurred. So if Magritte rolled strikes on the *first and second* frames, then to score 40 the third strike had to come between the fourth and 10th frames — seven cases in total. Had Magritte rolled strikes on the *second and third* frames, the third strike had to come between the fifth and 10th frame — six cases in total. As it turned out, there were seven ways to score 40 points when the strikes came on the first and second frames or the ninth and 10th frames; otherwise, there were six ways. Adding this up, there were 56 ways for Magritte to earn 40 points.

But what about three non-consecutive strikes? We already said among the 120 total cases, eight resulted in 60 points and 56 resulted in 40 points. That meant that the remaining number — 56 again! — resulted in 30 points. With an equal number of cases scoring 30 and 40 points, it was fair to say that the probabilities here were “split.”^{2}

Finally, to compute Magritte’s expected score, you had to average out all the cases, which gave you (8·60 + 56·40 + 56·30)/120, or **about 36.7 points**.

Disappointed with this rather low score, Magritte decided to give up on the sport of bowling and instead stick to bowler hats.

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐ David Ding ÑÑâÐ of Natick, Massachusetts, winner of last week’s Riddler Classic.

Last week, eight two-way roads all converged at a single intersection, as shown in the diagram below.

Two cars were heading toward the single intersection from different directions chosen at random. Upon reaching the intersection, they both turned in a random direction (where proceeding straight is a possible “turn”) — however, neither car pulled a U-turn (i.e., heading back the way it came).

In some cases, the paths of the cars could be drawn so that they did not cross. In this case, all was well.

However, in other cases, the paths *had to* cross. In this event, the cars crashed.

What was the probability the cars crashed? (If both cars headed off in the same direction, that also counted as a crash.)

As with last week’s Express, this was a matter of counting up cases. To simplify the problem, suppose the “red” car came in from the east. Then there were seven possible directions (i.e., *not* east) that it could have traveled upon leaving the intersection. Meanwhile, there were seven directions that the “blue” car could have come from (again, *not* east) and another seven (i.e., *not* the direction it came from) that it could have traveled upon leaving the intersection. In total, that was a whopping 7·7·7, or 343, cases to consider. Yikes!

Due to the ambiguity of the problem, there were some “edge cases” for which it wasn’t entirely clear if the cars would crash. For example, what if the cars exchanged directions? In theory, as long as each stayed in the right line, they would never crash.

Because of this ambiguity, I accepted multiple “correct” answers. David had a conservative definition of which cases constituted a crash, arriving at an answer of **16/49**. Meanwhile, Emma Knight assumed more edge cases resulted in a crash, giving an answer of **29/49**. And Rohan Lewis was in between the other two solvers, justifying an answer of **22/49**.

Suffice it to say that everyone was right last week.

Fortunately, all these edge cases disappeared in the grand scheme of things once you got to last week’s extra credit — the best part of the puzzle. As the number of two-way roads converging at the intersection approached infinity, what value did the probability of crashing approach?

Many solvers had already studied the general case of the riddle with *N* streets (rather than eight streets). This allowed them to explore the limit of their expressions as *N* approached infinity.

That was all well and good. But a few solvers, like Tim Curwick of Maple Grove, Minnesota, realized an elegant, equivalent way to restate this problem: Randomly pick four points on a circle. Then randomly pick two of them — these two will be the start and end locations for the red car. The remaining two points will be the start and end locations for the blue car. Now draw two chords that connect each pair of points. If the chords cross, that means the cars will crash!

For *any* four distinct points on a circle (whether chosen randomly or otherwise), there are exactly three ways to split them into two pairs of points. And the resulting chords will only intersect in one of these three ways. Therefore, the probability the chords will intersect — that is, the probability the cars will crash — is exactly **1/3**.another riddle from a few years back.

^{3}

If you’re still not convinced, Paulina Leperi simulated 100,000 drive-bys, finding that very nearly a third of them resulted in crashes.

Until next time, drive safely out there. And be sure to avoid intersections with infinitely many roads!

## 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.