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Can You Flip Your Way To Freedom?

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 next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

Dakota Jones is back in action. In her quest to locate the Temple of Diametra, she has found another highly symmetric crystal. However, nefarious agents have again gotten wind of her plans, and now Dakota and the crystal are nowhere to be found.

And so, you must once again recreate the crystal using the data from Dakota’s laser scanner. As a reminder, the scanner takes a 3D object and records 2D cross-sectional slices along the third dimension. Here’s the looping animation file the scanner produced for the crystal this time:

What sort of three-dimensional shape is the crystal? No pressure — Dakota Jones, nay, the entire world, is counting on you to locate the lost temple!

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

## Riddler Classic

From Bart Wright comes a rhetorical question from a famed soliloquy, “To flip, or not to flip?”:

You are locked in the dungeon of a faraway castle with three fellow prisoners (i.e., there are four prisoners in total), each in a separate cell with no means of communication. But it just so happens that all of you are logicians (of course).

To entertain themselves, the guards have decided to give you all a single chance for immediate release. Each prisoner will be given a fair coin, which can either be fairly flipped one time or returned to the guards without being flipped. If all flipped coins come up heads, you will all be set free! But if any of the flipped coins comes up tails, or if no one chooses to flip a coin, you will all be doomed to spend the rest of your lives in the castle’s dungeon.

The only tools you and your fellow prisoners have to aid you are random number generators, which will give each prisoner a random number, uniformly and independently chosen between zero and one.

What are your chances of being released?

Extra credit: Instead of four prisoners, suppose there are N prisoners. Now what are your chances of being released?

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

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐ Tony Jackson ÑÑâÐ of Rohnert Park, California, winner of last week’s Riddler Express.

Last week, you and a friend were grilling two small square burger patties whose sides were 5 centimeters long. However, you only had one slice of cheese remaining, which was also square and whose sides were 7 centimeters long. You wanted to cut the slice so that all of the cheese was evenly split between the two patties, and no cheese was spilling over either patty and onto the grill.

What was the smallest number of cuts you needed to make? (You could only make straight cuts, and you were asked to assume that the cheese was stationary during the cutting process.)

First off, several solvers thought to stack layers of cheese on top of each other. If stacking had been allowed, then with just two cuts you could have sliced the larger square into four smaller squares, each of whose sides were 3.5 centimeters long. (You even could have made a single cut down the middle and then have folded each half into quarters.) While stacking wasn’t explicitly prohibited in the original puzzle, finding a solution that resulted in a single layer of cheese on the patty was a more interesting challenge.

If you restricted your cuts so they were parallel to the sides of the square, then you needed at least four cuts. Solver Michael Smith found one such way to do this:

But with diagonal slicing, you only needed two cuts. The puzzle’s creator, Andrew Heairet, cut along the main diagonals of the cheese and then rearranged the resulting quarters into two squares that each had an area of 24.5 square centimeters — just enough to fit onto a single patty. Andrew was even kind enough to illustrate this solution:

Solver Jason White arrived at the same result, but went the extra mile of cutting a real slice of cheese!

But that wasn’t the only way to cut the cheese! Solver Ethan Rubin found a whole class of additional two-cut solutions by rotating the diagonal cuts:

According to Ethan, as long as the diagonal cuts were within about 8.13 degrees of the diagonal, the resulting quarters could still be rearranged to form two cheesy toppings that stayed on the patties.

Delicious!

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐ Matt Maron ÑÑâÐ of Philadelphia, Pennsylvania, winner of last week’s Riddler Classic.

Last week, you tried your hand at a variation of the Monty Hall problem. This time, Monty randomly picked a number of goats to put behind the doors: zero, one, two or three, each with a 25 percent chance. After the number of goats was chosen, they were assigned to the doors at random, and each door had at most one goat. Any doors that didn’t have a goat behind them had an identical prize behind them.

At this point, you chose a door. If Monty was able to open another door, revealing a goat, he would do so. But if no other doors had goats behind them, he would tell you that was the case.

It just so happened that when you played, Monty was able to open another door, revealing a goat behind it. Should you have stayed with your original selection or switched? And what were your chances of winning the prize?

It helped to break the problem down into four cases, one for each possible number of goats:

• If there were three goats behind the doors, it didn’t matter if you switched or stayed — you’d always lose.
• If there were two goats behind the doors, then this reverted to the original Monty Hall problem. You had a two-thirds chance of winning the prize if you switched, but just a one-third chance of winning if you stayed.
• If there was one goat behind a door, then Monty just did you a huge favor by showing you which door it was behind. It didn’t matter if you switched or stayed — you’d always win.
• If there were zero goats behind the doors, then you’d always win.

Now you might have thought that each of these cases was equally likely — but wait just a minute! The fact that Monty was even able to open a door and reveal a goat meant you couldn’t have been in the zero-goat scenario. There had to have been at least one goat present.

But that wasn’t all. The trickiest part of the problem was around the relative likelihood of the one-goat scenario. If there had been two or three goats, then no matter which door you picked, Monty could always open a different door to reveal a goat. But if there was only one goat, then the one-third of the time you happened to pick that goat’s door, Monty wouldn’t have been able to open another door to reveal a goat.

All of that meant you were just two-thirds as likely to be in the one-goat scenario as you were to be in the two-goat or three-goat scenarios. In other words, the probability there were three goats was 3/8, the probability there were two goats was also 3/8, and the probability there was one goat was just 2/8.

By combining the probabilities of the different scenarios with your probability of winning the prize within each scenario, you found that, overall, you had a 50 percent chance of winning if you switched, but just a 37.5 percent chance of winning if you stayed.

Solver Geoffrey Lovelace verified these results by running a few hundred thousand computer simulations. And David Zimmerman, meanwhile, extended the problem by looking at the general case where there were N doors (rather than just three), with anywhere from zero to N goats behind those doors. He found that switching always gave you a 50 percent chance of winning the prize, no matter how many doors there were. However, your probability of winning when you stayed with your original door was N/(2(N+1)) — a value that’s always less than 50 percent.

And so, as with the original Monty Hall problem, your best bet was to switch doors. That is, unless a goat happens to be your idea of a prize.

## 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: Please wait until Monday to publicly share your answers. 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.