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How Badly Can A Car Salesman Swindle You?

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 at the bottom! I’ll reveal the solution next week, and a correct submission (chosen at random) will earn a shoutout in this column. Important small print: If you want to be eligible for the shoutout, I need to receive your correct answer before Sunday midnight EST — you now have the whole weekend to solve The Riddler!

Before we get to the new puzzle, let’s dwell on last week’s! Congratulations to ÑÑâÐ Emma Violet Todd ÑÑâÐ of Chicago, our big winner. You can find a full solution to the previous Riddler at the bottom of this post.

Now, here’s this week’s Riddler, which comes to us from Zach Wissner-Gross, a physics Ph.D. and entrepreneur from Brookline, Massachusetts:

You go to buy a specific car, whose fair price we’ll call N. You have absolutely no idea what N is and the dealer, sadistic capitalist that he is, won’t tell you. The dealer enjoys a good chase, though, so without directly revealing the value of N, he takes five index cards and writes down a number on each of them: N, N+100, N+200, N+300 and N+400. Important: the guy’s sadistic but not a math major. The numbers on the cards are numbers, not algebra equations.

He presents the cards to you, one at a time, in random order. (For example, if the price on the second card is \$100 more than the first, you can’t be sure if those are the two smallest prices, the two largest, or somewhere in between.) Each time he shows you a card, you must either pay the price on that card, or ask to see the next card. You cannot go back to previous cards. If you make it to the fifth and final card, then that is what you must pay. If you play the dealer’s game optimally, how much should you expect to pay on average above the fair price N?

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

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And here is the solution to last week’s Riddler, about calendar math, courtesy of James McTeague.

And for amusing illustrative purposes, check out this timely “F Minus” comic, also tapping into the calendar zeitgeist, published last week:

1. There are 14 different calendars. Two things are changing year to year — the day of the week on which the year begins, and whether or not the year is a leap year. Multiply those possibilities together (i.e. 7 x 2) and voila.
2. The next time we’ll use the 2015 calendar is in 2026. 2015 started on a Thursday. The first days of the week then progress, year by year, by two (since 2016 is a leap year), one, one, one, two (making the total progression divisible by seven, but this is a leap year so the calendar’s different), one, one, one, two (since 2024 is a leap year), one, and one, making the total progression divisible by seven again, so we’re back to starting on Thursday 11 years later, in 2026.
3. The smallest number of years between non-leap-year calendars is six. If there weren’t leap years, it’d be seven — the first day of the year would progress each year from Monday to Tuesday to etc. and eventually back to Monday after seven years, allowing us to reuse that calendar. But an intervening leap year can accelerate that process by a year. So, for example, the next time we’ll use 2017’s calendar is six years hence, in 2023, because 2020 is a leap year.
4. The largest number of years between non-leap-year calendars is 12. For example, the next time we’d use the 2097 calendar is in 2109. N.B. the year 2100 is not a leap year. As soon as the 2097 calendar is about to recycle, it hits a leap year in 2104. It’d need another six years to recycle again, but is “helped out” by 2108, also a leap year, so the full cycle, and in fact the largest possible cycle, is 12 years.
5. The smallest number of years between leap-year calendars is 12. The 2096 calendar, for instance, will next be used in 2108. The 2096 calendar “benefits” from leap years in just the right way. The first day of its week — a Sunday — progresses by four in 2100 (again, not a leap year), five in 2104, and five in 2108 (both leap years), for a total of 14, which is divisible by seven (the number of days in a week) bringing the first day back to Sunday in 2108.
6. The largest number of years between leap-year calendars is 40. The 2072 calendar, for instance, will next be used in 2112. Its first day — a Friday — progresses by five, then five, then five, then five, then five, then five, but then four (because 2100 is not a leap year) just missing being divisible by seven. Then it’ll be another 12 years before the first day cycles back around to a Friday.

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.