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: To be eligible for the shoutout, 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 ÑÑâÐ Taylor Hoogendoorn ÑÑâÐ of New Haven, Conn., 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 twist on the so-called “impossible puzzle” (happy April Fool’s Day!) which comes to us from Max Wahlund of Linköping, Sweden.
Three very skilled logicians are sitting around a table — Barack, Pete and Susan. Barack says: “I’m thinking of two natural numbers between 1 and 9, inclusive. I’ve written the product of these two numbers on this paper that I’m giving to you, Pete. I’ve written the sum of the two numbers on this paper that I’m giving to you, Susan. Now Pete, looking at your paper, do you know which numbers I’m thinking of?”
Pete looks at his paper and says: “No, I don’t.”
Barack turns to Susan and asks: “Susan, do you know which numbers I’m thinking of?” Susan looks at her paper and says: “No.”
Barack turns back to Pete and asks: “How about now? Do you know?”
“No, I still don’t,” Pete says.
Barack keeps going back and forth, and when he asks Pete for the fifth time, Pete says: “Yes, now I know!”
First, what are the two numbers? Second, if Pete had said no the fifth time, would Susan have said yes or no at her fifth turn?
Need a hint? You can try asking me nicely. Want to submit a puzzle or problem? Email me.
[googleapps domain=”docs” dir=”forms/d/15JTPF5kCiRZe9n9UufCmkJOdLPN_O7nI4wOLgFFRB80/viewform” query=”embedded=true” width=”760″ height=”1050″ /]
And here’s the solution to last week’s Riddler, concerning whether or not you should pay to play a new casino game. Thirty-eight percent of submitters knew how to bust the casino! The solution is adapted from an excellent submission by Tim Black, a math grad student at the University of Chicago. (Go Maroons!)
Yes, you should pay the $250 to play the game where the casino draws random numbers between 0 and 1, and you get $100 for every number drawn until their sum exceeds 1 — your expected winnings are about $271.83, or $100*e, where e is Euler’s number. Why?
Here’s one way to think about it. Let’s say the casino starts drawing numbers, writing down the sum, but only the digits after the decimal place. So, if the casino drew the numbers 0.4, 0.2, 0.1, 0.5, it would calculate:
0.4 = 0.4
0.4 + 0.2 = 0.6
0.4 + 0.2 + 0.1 = 0.7
0.4 + 0.2 + 0.1 + 0.5 = 1.2
But it would only write down _.4, _.6, _.7, and _.2. Now, let’s make two important observations:
- Given any item in the list, you can’t predict the next; it’s equally likely to be anything between _.00 and _.99…
- From this list, you can tell when the sum exceeded 1: It’s the first place where the decimal part decreases (from _.7 to _.2, in this case) instead of increases.
Observation 1 means that the casino’s list of decimal-parts-of-sums is a list of random numbers chosen between 0 and 1, uniformly. Observation 2 means that the amount of money you get is $100 for each nonnegative integer n such that the first n numbers are in increasing order. The probability that a list of n numbers chosen from the same distribution is in increasing order is 1/n!. So, the total expected value of your earnings is $100 times
1/0! + 1/1! + 1/2! + 1/3! + …
This is a famous infinite sum, which converges to e. So your winnings are $100*e = $271.83.
There is also an integration-based solution, which you can see at Tim’s blog. And who says print is dead? After my plea last week for pencil-and-paper solutions, I was inundated with your lovingly crafted scratch papers. (So many that my editor said we couldn’t include them all!) Riddler Nation: Long may you reign. They came in over Twitter:
https://twitter.com/MacMisterAych/status/713416930766548993
And they came in over email. This one is from Nancy Hansen:
And this, from Francesco Antognini, is, per Francesco, “full of stuff, some of it totally wrong. But I like it because it shows how nonlinear the path to enlightenment is.” Well said, Francesco.
And finally, dear Riddlers, I leave you with this, from Jacob Steel. Not exactly pencil-and-paper, but beautiful and simple and vexing and haunting enough to ponder this weekend — a Riddler’s koan. It depicts simplexes in hypercubes into which your first n numbers must fall for you to get the $100 payoff. Please share with me the worldly insights it inspires. Happy Friday.
