Can You Win at Non-Traditional Blackjack?

If you’re committed to drawing more cards, how likely are you to reach 21? What if you never want to risk exceeding 21?

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “extra credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

You’re playing a modified version of blackjack, where the deck consists of exactly 10 cards numbered 1 through 10. Unlike traditional blackjack, in which the ace can count as 1 or 11, the 1 here always has a value of 1.

You shuffle the deck so the order of the cards is completely random, after which you draw one card at a time. You keep drawing until the sum of your drawn cards is at least 21. If the sum is exactly 21, you win! But if the sum is greater than 21, you “bust,” or lose. 

What are your chances of winning, that is, of drawing a sum that is exactly 21?

This Week’s Extra Credit

Playing for 21 or bust is a risky strategy. From this moment on, you decide to be risk averse.

You’re playing the same modified version of blackjack again, but this time, whenever there’s even the slightest chance you could bust on the next card, you quit the round and start over. On average, how many rounds should you expect to start until you finally win?

Making the Rounds

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m once again sharing one from a sequence of great questions posed by James Tanton. I’ll paraphrase here to avoid any spoilers:

An infinitely long line is the set of points that are equidistant from a point A and a second geometric object B. Must object B also be a point, or could it be something else?

For clarity, the “distance” between a point X and an object Y is defined as the minimum among the distances between X and all the points within Y.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Grant Hollis 🎻 from Easley, South Carolina. I received 64 timely submissions, of which 58 were correct—good for a 91 percent solve rate.

Last week’s puzzle started with circles. A circle with radius r has an area of 𝜋r² and a circumference of 2𝜋r. Now, if you take the derivative of the area formula with respect to r, you get the circumference formula! In other words, d(𝜋r²)/dr = 2𝜋r. Amazing, right?

Inspired by this fact, we defined the “differential radius” of a shape: The differential radius r of a shape with area A and perimeter P (both functions of r) had the property that dA/dr = P. (Note that A always scaled with r² and P always scaled with r.)

For example, consider a square with side length s. Its differential radius was r = s/2. The square’s area was s², or 4r², and its perimeter is 4s, or 8r. Sure enough, dA/dr = d(4r²)/dr = 8r = P.

What was the differential radius of an equilateral triangle with side length s?

The area of an equilateral triangle with side length s is A = s²√(3)/4, while its perimeter is P = 3s. Right off the bat, you knew the radius couldn’t have been s, since dA/ds = s√(3)/2, which was a good deal less than P.

That said, you knew the radius had to scale with s, so suppose the radius was s = kr, where k was some constant of proportionality. That meant A = (kr)²√(3)/4 and P = 3kr. Now, dA/dr = k²r√(3)/2. Setting this equal to P gave you the equation k²r√(3)/2 = 3kr. So which value of k made this equation true? Canceling out some r’s and k’s, and then rearranging, gave you k = 2√3. That meant r = s/(2√3). (Of course, I accepted equivalent expressions like s√(3)/6 and s/√12.)

While this algebraic approach resulted in the correct solution, it didn’t offer much in the way of insight. What did the triangle’s radius look like? Was there a particular segment within the triangle that had a length of s/(2√3)?

Indeed there was! A line segment connecting the triangle’s center and the midpoint of any of its three sides—known as the triangle’s “apothem”—had a length of precisely s/(2√3), as shown below:

So what did the circle’s radius, the square’s radius, and the triangle’s radius all have in common? They all had one endpoint in their shape’s center, and they were all perpendicular to the perimeter (or circumference)!

Why did this perpendicularity matter? Because when the radius increased by a small amount (as illustrated below), say, by dr, the increase in area, dA, was approximately P·dr—an approximation that approached exactness in the limit as dr went to zero.

And so, if you were trained in physics like I was, you’d feel perfectly comfortable writing dA = P·dr, or dA/dr = P. Again, this relationship worked when the radius was perpendicular to the perimeter for symmetric shapes like circles and regular polygons.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Rohan Lewis 🎻 from Cary, North Carolina. I received 30 timely submissions, of which 24 were correct—good for an 80 percent solve rate.

For Extra Credit, you had to determine the differential radius of a rectangle with sides of length a and b. (And your answer had to be in terms of both a and b.)