Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and 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
From Josh Vandenham, precipitation permutations:
Louie walks to and from work every day. In his city, there is a 50 percent chance of rain each morning and an independent 40 percent chance each evening. His habit is to bring (and use) an umbrella if it’s raining when he leaves the house or office, but to leave them all behind if not. Louie owns three umbrellas.
On Sunday night, two are with him at home and one is at his office. Assuming it never starts raining during his walk to his home or office, what is the probability that he makes it through the work week without getting wet?
Riddler Classic
From Randi Goldman and Zach Wissner-Gross, little misanthropic houses on the prairie:
Antisocial settlers are building houses on a prairie that’s a perfect circle with a radius of 1 mile. Each settler wants to live as far apart from his or her nearest neighbor as possible. To accomplish that, the settlers will overcome their antisocial behavior and work together so that the average distance between each settler and his or her nearest neighbor is as large as possible.
At first, there were slated to be seven settlers. Arranging that was easy enough: One will build his house in the center of the circle, while the other six will form a regular hexagon along its circumference. Every settler will be exactly 1 mile from his nearest neighbor, so the average distance is 1 mile.
However, at the last minute, one settler cancels his move to the prairie altogether (he’s really antisocial). That leaves six settlers. Does that mean the settlers can live further away from each other than they would have if there were seven settlers? Where will the six settlers ultimately build their houses, and what’s the maximum average distance between nearest neighbors?
Solution to the previous Riddler Express
Congratulations to ÑÑâÐ Brenda Holloway ÑÑâÐ of Manchester, Connecticut, winner of last week’s Riddler Express!
Last week, Alice and Bob were facing off in a footrace. The route was a straight path out followed by the same straight path back. But Alice literally wanted to face off — she wanted to run faster than Bob such as to maximize the amount of time they spent facing each other after she made the U-turn but before the two passed each other going in opposite directions. Assuming the turnaround time is negligible and the two each commit to running the race at a constant speed, how much faster than Bob should Alice run?
If Bob runs at speed \(B\), Alice should run at speed \((1+\sqrt{2})B\), or about 2.4 times his speed.
This (obviously) is a maximization problem, meaning we’re trying to find the value of a variable that maximizes some larger expression. That means we need a mathematical expression to maximize. Let’s try to piece that together.
Suppose the distance to the U-turn — that is, the halfway point of the race — is \(D\). (It won’t matter exactly what that distance is.) Suppose Alice’s speed, which we’re trying to calculate, is \(A\). The time it takes Alice to get there is \(D/A\). In that amount of time, Bob will have run a distance of \(B\cdot (D/A)\). So the distance between the two runners at the moment Alice turns around is \(D-B\cdot (D/A)\).
Now the two are running toward each other, so we must combine their speeds. They will, combined, cover that distance and pass each other after a time of \((D-B\cdot (D/A))/(A+B)\). That’s the amount of time we want to maximize.
A handy mathematical way to maximize things is to take derivatives. Specifically, we want to take the derivative of the expression above with respect to \(A\), set it equal to zero, and solve that for \(A\). That will give us a set of extremes, be they maxima or minima. The derivative of the expression that we’ll set to zero looks like:
\begin{equation*}\frac{D(-A^2+2AB+B^2)}{A^2(A+B)^2}=0\end{equation*}
That equation equals zero for two values of \(A\): \((1+\sqrt{2})B\) and \((1-\sqrt{2})B\). The latter of these doesn’t make sense for the case of our footrace (it’s negative), so we know the first must be the value we’re after. If Bob runs at speed \(B\), Alice should run at speed \((1+\sqrt{2})B\), or about 2.4 times his speed.
Solution to the previous Riddler Classic
Congratulations to ÑÑâÐ Peter Ingraham ÑÑâÐ of Brooklyn, New York, winner of last week’s Riddler Classic!
Last week, after facing each other down for so long in that footrace, Alice and Bob fell in love, got married and wanted to have kids. Potentially lots of kids. They figured out that the amount of work involved in having a child is equal to 1 divided by the age of that child. So if a child is 0, the work is infinite. If the child is 1, the work is 1; if a child is 2, the work is ½; and so on. They decided that they only wanted to have another kid if and when the total work involved for all of their other kids was equal to 1 or less. Ignoring real-world technicalities such as death, twins and the inability to decide precisely when one has a child: Will Alice and Bob have an infinite number of children? Can we predict, with an equation, when they will have their Nth child?
First, yes, Alice and Bob would have infinity children if they could. But they would have them very, very slowly. After about 10,000 kids, they’ll have to wait more than 10 years to have the next one, and the gap between children keeps increasing forever. Solver Scott Wu plotted how long they will wait to have their next child, based on how many children they’ve had so far, for their first 50,000 children. That looks like this:
Our winner this week, Peter, provided the following proof-by-contradiction of why the number of children the couple will eventually have is infinite: Suppose for the sake of argument that there was some point where Alice and Bob stopped having more kids, and that at this point they have \(k\) kids and \(n\) years have passed. Let’s examine the amount of work required at time \(2n\). All the kids are now age \(n\) or more. As an upper bound, each kid is worth at most \(1/n\) work, the amount required for the youngest child. Therefore the total work is at most \(k\cdot (1/n) = k/n\). But the number \(k\) is always less than \(n\): The gaps start at 1 year and always get longer. So Alice and Bob should have had another kid by this time since their total work \(k/n\) is less than 1. This is a contradiction. Therefore they never stop having kids.
There does not appear to be an exact, elegant formula for when Alice and Bob have their Nth child. But the Nth child does come at approximately N log N, a fact that we could ascertain by generating lots of simulated child data and running a regression. So, for example, the 1,000th child would be born approximately 1,000 log 1,000 — or 6,908 — years after the first child. Solver Laurent Lessard plotted the birthdates by the ordinal childbirths:
“Meet my brother, Bob Smith CMXLVII. Yeah, he’s my big brother, can’t you tell? He was born about 5,000 years ago.”
Want more riddles?
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Want to submit a riddle?
Email me at oliver.roeder@fivethirtyeight.com.