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Can You Solve This Elevator Button Puzzle?

Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here — math, logic and probability. These problems, puzzles and riddles come from lots of top-notch puzzle folks around the world, including you, the readers. You’ll find this week’s puzzle below.

Mull it over on your commute, dissect it on your lunch break, and argue about it with your friends and lovers. When you’re ready, submit your answer using the form below. I’ll reveal the solution next week, and a correct submission (chosen at random) will earn a shoutout in this column. Important small print: To be eligible, I need to receive your correct answer before 11:59 p.m. EDT on Sunday — have a great weekend!

Before we get to the new puzzle, let’s return to last week’s. Congratulations to ÑÑâÐ Marc Tanis ÑÑâÐ of Kitwe, Zambia, our big winner. You can find a solution to the previous Riddler at the bottom of this post.

Now here’s this week’s Riddler, an elevator puzzle that comes to us from Ken Harlow of Orange County, California. Hey, that’s funny, I just wrote about elevators this week!

In a building’s lobby, some number (N) of people get on an elevator that goes to some number (M) of floors. There may be more people than floors, or more floors than people. Each person is equally likely to choose any floor, independently of one another. When a floor button is pushed, it will light up.

What is the expected number of lit buttons when the elevator begins its ascent?

Extra credit: I’m feeling generous, and am clearly on an elevator jag, so let’s offer up a ÑÑÐâ  Coolest Riddler Extension Award ÑÑÐâ  this week. Add some elevators, change the building, make interesting assumptions about its riders, or something far, far more creative. The winner gets a shiny emoji trophy next week.

Need a hint? You can try asking me nicely. Want to submit a new puzzle or problem? Email me.

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And here’s the solution to last week’s Riddler, concerning particularly proud partygoers, which came to us from Dominic van der Zypen.

The largest share of “proud” people at the party — those who have more friends than their friends, on average — is (N-2)/N. To see why, let N>3 be the number of people at the party, and call two particular people in attendance X and Y. Suppose everyone is friends with everyone else, except that X and Y are not friends with each other. (In math terms, you have a complete graph with one edge removed.) In this case, it’s easy to see that everyone is proud except X and Y. So the largest share of proud people is (N-2)/N, which converges to 1 as the party gets bigger, that is, as N goes to infinity.

Reader Saurabh Rane is going to put me out of a job — he posted all the solution work on his blog, including some nifty charts!

Tim Duff also goes deep into the math here. And this illustration of a solution, from Max Shaw, was too endearing not to share. Just look at those revelers!

Have a great weekend, and party hearty — and safely! I’m proud to be friends with each and every one of you.

CORRECTION (May 6, 9:44 a.m.): An earlier version of this article misstated the solution to last week’s Riddler. It is (N-2)/N, not (N-2)/2.

Oliver Roeder was a senior writer for FiveThirtyEight. He holds a Ph.D. in economics from the University of Texas at Austin, where he studied game theory and political competition.

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