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Can You Win A Lifetime Supply Of Cranberry Sauce?

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 Keith Wynrow comes a clever little puzzle that stumped a few folks on social media:

On the Food Network’s latest game show, Cranberries or Bust, you have a choice between two doors: A and B. One door has a lifetime supply of cranberry sauce behind it, while the other door has absolutely nothing behind it. And boy, do you love cranberry sauce.

Of course, there’s a twist. The host presents you with a coin with two sides, marked A and B, which correspond to each door. The host tells you that the coin is weighted in favor of the cranberry door — without telling you which door that is — and that door’s letter will turn up 60 percent of the time. For example, if the sauce is behind door A, then the coin will turn up A 60 percent of the time and B the remaining 40 percent of the time.

You can flip the coin twice, after which you must make your selection. Assuming you optimize your strategy, what are your chances of choosing the door with the cranberry sauce?

Extra credit: Instead of two flips, what if you are allowed three or four flips? Now what are your chances of choosing the door with the cranberry sauce?

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

Riddler Classic

Trig the turkey cannot fly. And so, instead of flying, he decides to jump as far as he can with a running start up one of the famed Sinusoidal Hills, all of which are precisely the same size.

Trig knows that he cannot jump a horizontal distance of more than two hills. Also, he prefers a smooth takeoff and landing. That is, when he takes off from the ground and lands on it again, the slope of his parabolic trajectory through the air must perfectly match the instantaneous slope of the ground beneath him.

The animation below shows a jump where the takeoff and landing are precisely 1.2 hills apart.

A point representing Trig the turkey rolls over some sine waves. It then launches up -- smoothly -- into a parabolic trajectory, before smoothly returning down to the sine waves again.

What is the greatest horizontal distance Trig can jump, such that his takeoff and landing are smooth? Again, keep in mind that Trig cannot possibly jump a horizontal distance greater than two hills.

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

Solution to last week’s Riddler Express

Congratulations to 👏 Masa Miyazaki 👏 of Tokyo, Japan, winner of last week’s Riddler Express.

Last week, you decided to make some new friends at your local gym, which was open daily from 5 p.m. to 8 p.m.

Some gym members attended very often. Others barely showed up at all. As a matter of fact, there was a uniform distribution for how often the different members were in the gym — from 0 percent of the time that the gym was open to 100 percent of the time.

As a new member, you planned to be in the gym 50 percent of the time that it was open. While working out, you decided to make friends with the first person you saw. (Aw!)

What was the probability that this person visited the gym more often than you?

If you were to pick a random person from the gym’s membership list, then there was a 50 percent chance that person visited the gym more often than you. However, you weren’t picking a person from any list — you were comparing yourself to the first person you saw at the gym. And that made a difference!

For example, consider members who were at the gym close to 0 percent of the time. You would have been very unlikely to see them there at all. Meanwhile, members who spent close to 100 percent of the time at the gym were far more likely to be seen, since they were simply around more often.

In fact, the relative likelihood of seeing each person was proportional to how often they were at the gym. As noted (and sketched below) by solver Sion Verschraege, that meant the probability distribution was triangular:

Triangular probability distribution from 0 percent to 100 percent. The left part from 50 percent -- one quarter of the triangle -- is labeled "In gym less often." The right part is labeled "In gym more often."

The smaller triangle on the left represented the collective probability of seeing someone who was in the gym less often than you. Its dimensions were scaled down by a factor of 2, which meant it had 25 percent of the area of the larger triangle. The remaining area — the probability the person you saw was in the gym more often than you — was therefore 75 percent.

So the next time you’re at the gym and you feel like everyone there is fitter than you, worry not. Those people are just overrepresented in the probability distribution! Solver David Ding found that if you want to feel like the “median” (or should I say, “golden”) gym member, you would actually need to be there about 61.8 percent of the time.

Before moving on, it’s worth noting that the puzzle never explicitly said how many members the gym had. If you assumed there were many members, then the above reasoning was perfectly fine, and the answer was 75 percent. But if there were only two people (one of whom was you), there was only a 50 percent chance you were at the gym more often than your counterpart. For three, four or five members, the math got trickier, but solver Emily Boyajian was up to the challenge, finding that the exact probabilities rapidly converged to 75 percent.

Solution to last week’s Riddler Classic

Congratulations to 👏 Joseph Wetherell 👏 of San Diego, California, winner of last week’s Riddler Classic.

Four people were trying to escape from a room. Guards placed a hat on each person’s head, and each hat was one of three colors: red, yellow or blue. The four people were arranged at the vertices of a square, with an obstacle in the middle. Each person could see the hats on the heads of those on adjacent vertices of the square, but they could not see the hat of the person diagonally across from them. They also did not know the color of the hat on their own head.

Each person had to guess the color of the hat on their own head. If at least one person guessed correctly, they could all escape the room together. No communication was allowed once the hats were placed on their heads, but they could coordinate on a strategy beforehand. They also knew how they would be arranged in the square.

How could they be guaranteed to escape the room?

Like last week’s Express, this puzzle was a little ambiguous. Several readers asked whether the four people could hear each other’s guesses. If they could, and if they guessed at different times, then there was a straightforward strategy for success: They would agree on who should guess first, that person would say the color of one of their neighbors in the square, and then everyone else would say that same color. The neighbor whose color was said first was guaranteed to be correct. Huzzah!

But was it possible for them to escape without this simplifying assumption, or any others for that matter? Indeed it was.

The puzzle’s submitter, Eric Dallal, offered one such strategy. First, Eric assigned numerical values to each hat color: 0 for red, 1 for yellow and 2 for blue. Next, he labeled the four players in order around the square: p1, p2, p3 and p4, such that p1’s neighbors were p2 and p4, p2’s neighbors were p3 and p1, etc. He also labeled the actual number values of the hats on their heads: c1, c2, c3 and c4.

With these variables in place, a winning strategy was:

  • p1 guesses c2 + c4, modulo 3
  • p2 guesses -(c1 + c3), modulo 3
  • p3 guesses c2 − c4, modulo 3
  • p4 guesses c3 − c1, modulo 3

To ensure this works, check out the animation below, which loops through all 81 possible hat arrangements and verifies that at least one person guesses correctly for each arrangement.

Animation cycling through all 81 hat permutations. For each permutation, at least one person guesses their hat correctly.

As Eric noted, it was interesting that either one person would guess correctly or all four would do so — but never two or three.

In any case, the room has officially been escaped from!

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.

Footnotes

  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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