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Can You Find Your Pills?

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 or send me an email.

Riddler Express

If you’ve ever thought about buying a home, you might be familiar with the so-called “Rule of 72.” For an exponentially increasing quantity, like interest accrued on a principal, you can use the Rule of 72 to approximate the doubling time based on the interest rate. For example, for an interest rate of 4 percent — meaning the exponential growth is 1.04t, with t measured in years — the doubling time is approximately 72 divided by 4, or 18 years.

It turns out that there is only one interest rate for which the Rule of 72 tells you the exact doubling time. What is that interest rate?

Submit your answer

Riddler Classic

From Dave Moran comes a pill-popping puzzle:

I’ve been prescribed to take 1.5 pills of a certain medication every day for 10 days, so I have a bottle with 15 pills. Each morning, I take two pills out of the bottle at random.

On the first morning, these are guaranteed to be two full pills. I consume one of them, split the other in half using a precision blade, consume half of that second pill, and place the remaining half back into the bottle.

On subsequent mornings when I take out two pills, there are three possibilities:

  • I get two full pills. As on the first morning, I split one and place the unused half back into the bottle.
  • I get one full pill and one half-pill, both of which I consume.
  • I get two half-pills. In this case, I take out another pill at random. If it’s a half-pill, then I consume all three halves. But if it’s a full pill, I split it and place the unused half back in the bottle.

Assume that each pill — whether it is a full pill or a half-pill — is equally likely to be taken out of the bottle.

On the 10th day, I again take out two pills and consume them. In a rush, I immediately throw the bottle in the trash before bothering to check whether I had just consumed full pills or half-pills. What’s the probability that I took the full dosage, meaning I don’t have to dig through the trash for a remaining half-pill?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to ÐВЃЯСÐВџ Michael Goldwasser ÐВЃЯСÐВџ of St. Louis, Missouri, winner of last week’s Riddler Express.

Last week, you looked at hexominoes, which are shapes made of six identical, nonoverlapping squares that are connected by edges. Some hexominoes, like the one shown below, could be decomposed into an array of three straight squares, an array of two squares and an array of one square.

On the left, a hexomino with three left-aligned rows. The top row has one square, the middle row has two squares, and the bottom row has three squares. On the right, the same hexomino is now colored by row. The top row is white, the middle row is red, and the bottom row is green. There is an arrow pointing from the left hexomino to the right hexomino.

How many distinct hexominoes could you find that could not be decomposed into arrays of three, two and one squares? (For the purposes of this riddle, two hexominoes were considered equivalent if they could be turned into one another by rotation and/or reflection.)

First of all, how many distinct hexominoes were there? Fortunately, the tallying of polyominoes is a well-studied problem, and the number of each type of polyomino is an OEIS sequence. In particular, there are 35 distinct hexominoes that can’t be turned into each other by rotation or reflection (so-called “free hexominoes”).

From there, most solvers took out their felt-tip pens (or digital paint buckets) and got to coloring the 35 hexominoes. Rohan Lewis was able to carve out three straight arrays at least one way for 32 of the 35 hexominoes, as shown below. That meant there were three hexominoes that couldn’t be decomposed in this way.

All 35 free hexominoes are shown. 32 of them have been colored to show an array of three, an array of two and an array of one. The remaining three hexominoes cannot be decomposed in this way.

One is shaped like a Y or wishbone. A second has four squares in a row, with a square above the second square in the row and a square below the third square in the row. The third is a zigzagging pattern.

Solver Tom Keith went a step further and looked at how many other triangular-numbered polyominoes couldn’t be decomposed into straight arrays of 1, 2, 3, …, N squares. Tom found that as the size of the polyominoes increased, the ones that weren’t decomposable started to outnumber the decomposable ones — and rapidly.

By the way, the numbers of decomposable and non-decomposable polyominoes look like fodder for one or two new OEIS sequences, if anyone in Riddler Nation feels so inclined to submit them …

And finally, a quick shoutout to Jenny Mitchell, who submitted colored square emojis in the Show Your Work section of the answer form!

Solution to last week’s Riddler Classic

Congratulations to ÐВЃЯСÐВџ Steve Curry ÐВЃЯСÐВџ of Albuquerque, New Mexico, winner of last week’s Riddler Classic.

Last week, you were stranded in a casino and needed to purchase a flight home. Flights cost $250, but you had only $100 on hand. However, as I just said, you were in a casino! Surely, you could gamble your way to $250.

The casino had a particular game called “Riddler’s Delight,” in which you could bet any amount of money in your possession for an even greater amount of money. You could even bet fractional, irrational or infinitesimal amounts if you so desired.

The catch was that the odds were not in your favor. In Riddler’s Delight, whenever you bet A dollars in an attempt to win B dollars (with B > A), your probability of winning was not A/B, which you would have expected from a fair game. Instead, your probability of winning was always 10 percent less, or 0.9(A/B).

Which betting strategy would have maximized your probability of getting home, and what was that probability?

The first approach many readers tried was to bet it all in one go. When the odds are stacked against you, this is often a more reasonable strategy than betting in smaller amounts, where the casino will generally take your money thanks to the Law of Large Numbers. If you put all $100 on the line in an effort to win back $250, your probability of success was 0.9(100/250), or 36 percent. Not bad! But could you do better?

Instead of betting it all at once, you could instead have bet half your money, saving the other half in case you failed. If you bet $50, you only needed to win $200 (rather than $250), since you had another $50 set aside. Your chances of winning this first bet were 0.9(50/200), or 22.5 percent. And for the remaining 77.5 percent of the time you lost that $50, you could try betting the $50 you had left, which netted you $250 with probability 0.9(50/250), or 18 percent. Putting this all together, your probability of winning was 0.225 + (0.775)(0.18), or 36.45 percent, which was slightly greater than the 36 percent chance you had by betting all $100.

From there, a few solvers realized that breaking up larger bets into smaller ones always improved your chances. That meant the optimal strategy was the limit of making a series of infinitesimal bets to win $250 minus however much you still had on hand.

Solver Pradeep Niroula of Kathmandu, Nepal, calculated your probability of winning if you made this series of infinitesimal bets, each worth ÐВЃЭЬЦ dollars. On your jth such infinitesimal bet, you had to wager 250−(100−jÐВЃЭЬЦ) to reach $250, which meant your probability of winning was 0.9ÐВЃЭЬЦ/(250−(100−jÐВЃЭЬЦ)). Your probability of losing was then 1 minus this quantity, and multiplying all these terms together (from when j was 1 up to 100/ÐВЃЭЬЦ) gave you the total probability of failure. And subtracting this from 1 again gave you your probability of winning. Finally, taking the limit as ÐВЃЭЬЦ went to zero revealed a numerical answer, which Pradeep computationally determined to be about 36.8554 percent.

Meanwhile, Emma Knight and Shivam Sarodia were able to analytically solve that limit using calculus. Your precise chances of securing that $250 flight were 1−(3/5)9/10. I thought it was rather satisfying to see how each number from the puzzle (the ratio of $100 to $250 and the 10 percent shavings) appeared in this expression.

Finally, kudos to Benjamin Dickman who realized the answer was approximately 0.368 and guessed it was 1/e, which indeed rounds to 0.368 and was less than 1 percent away from the correct answer! But for once, the answer around here did not involve e or ÐВЃЭЬÐВ›.

Want more puzzles?

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

CORRECTION (Aug. 19, 2022, 11:39 a.m.): A previous version of this week’s Riddler Express gave the incorrect formula for doubling time with an interest rate of 4 percent. It is 1.04t, not e1.04t.


  1. Important small print: In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.


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