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Can You Figure Out How To Beat Roger Federer At Wimbledon?

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! 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 link 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 Adam Kalinich ÑÑâÐ of Glen Ellyn, Illinois, 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, a centre court tennis puzzle that comes to us from Ted LeMoine, a professional poker player from Medford, Massachusetts.

Your wish has been granted, and you get to play tennis against Roger Federer in his prime in the Wimbledon final. You have only a 1 percent chance to win each point, but Roger, sporting gentleman that he is, offers to let you name any score and begin the match at that point. (So, if you’ve entertained a fantasy of storming back after being down three match points in the fifth set, now’s the time to live it.) What score can you name that gives you the best chance to win, and what is your chance of winning the title?

Extra credit: It’s been a while, so let’s offer up a ÑÑÐâ  Coolest Riddler Extension Award ÑÑÐâ . Improve your tennis skills, make more realistic assumptions (about serving, etc.), or something even 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.

And here’s the answer to last week’s Riddler, concerning a robot that slices your pizza, randomly and independently chooses two points on the pie’s circumference, and then slices along the chord connecting them. If you ask the robot to make three such cuts, you can expect your pizza to have five pieces, on average.

The key thing to notice for this problem is that the only thing that matters is the order of the points around the circle selected by the robot — the points’ exact positions within that order don’t matter at all. (Their positions will affect the slice sizes, but here we only care about the number of slices.)

Think about it this way: First, picture six random points around your circular pizza pie. Pick any one point to start with, then randomly choose one of the remaining five, and slice between them. Now pick another point to start with, and pair it randomly with one of the three remaining points, and slice between those. Finally, there are only two points left, so you slice between those. That’s 15 possibilities. Five of these yield four pieces, six yield five pieces, three yield six pieces and one yields seven pieces. Here is an illustration of the possible arrangements and their slice counts, from Laurent Lessard:

Thanks to the way the robot randomizes its cuts, each scenario above occurs with equal probability. Therefore, the average number of pieces is (5*4+6*5+3*6+1*7)/15 = 75/15 = 5.

(Note also that there are some weird possibilities we didn’t consider, like the robot choosing the exactly same chord twice, or all three chords intersecting at one point, like how you’d normally try to slice a pizza. However, these happen with probability zero, so don’t factor into our calculation of the expectation.)

For the more engineering-minded Riddler solvers, Gabe Pezanoski-Cohen put together a little pizza robot simulator that can make any number of cuts. There is also an elegant generalized solution: If the robot makes k random cuts, the expected number of slices is (k+2)(k+3)/6. As the puzzle’s contributor Zach Wissner-Gross explains, this can be shown by induction: If the expected number of pieces resulting from k cuts is E(k), then the expected number of pieces for k+1 cuts is E(k+1)=E(k)+1+k/3. Why? Slicing along a new random chord will always add one slice, and the new chord has a one-third chance of intersecting each of the already sliced chords. Combining that with the fact that one cut gives two pieces, or E(1)=2, we arrive at the generalized solution. Mmm, pizza.

And elsewhere in the puzzling world:

• The confusing triangle situation. [Wait But Why]
• The puzzle of the daring lottery scheme. [The New York Times]
• More tennis: Are you smarter than Andy Murray? [The Guardian]
• Some problems on reflection. [Expii]
• John Urschel, a Baltimore Ravens offensive lineman and MIT mathematics PhD student, relays his three favorite math puzzles. [The Players’ Tribune]

Have a splendid weekend!

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.