Puzzles

Counterfeit Coins

You are given 15 piles of coins. Each pile contains 20 coins. All 300 coins look identical, but one pile consists entirely of counterfeit coins while the other 14 piles consist entirely of genuine coins. The genuine coins weigh exactly 2 grams each, whereas the counterfeit coins weigh 2.1 grams each. Your task is to determine which pile contains counterfeit coins by using a single produce-type scale and taking only one measurement. How do you do it?

Solution

If you take one coin from the first pile, two from the second, . . . , and 15 from the fifteenth, then weigh all of those coins together, the number of the pile consisting of counterfeit coins will be [(Weight) – 240] x 10. The preceding was adapted from the Ask Marilyn column in the March 18th, 1990 issue of Parade Magazine. I was thinking about this puzzle (which I’m pleased to say I solved immediately) and the way in which the solution is a primitive form of encoding information on the items being measured. If this was a practical application, then presumably we would want to reclaim all of the authentic coins from the scale while discarding all of the counterfeits. What would be the best way to do this? I came up with a few solutions, but would be interested to hear other suggestions.

Favorite Solution:

Assuming the scale has a flat platform with enough area for six to twelve stacks of coins, I would stack the coins sequentially. I would alternate between a heads-up, tails-up for each set of coins from a single stack. For example, one coin from stack one, heads-up, followed by two coins from stack two both tails-up, followed by three coins from stack three all heads-up, etc. If, for example, the seventh pile was counterfeit, simply locate the seven sequentially stacked coins and discard them.

If the scale won’t accommodate discrete stacks:

The coins may be numbered with the number of the stack from which they came with a marker (if permissible) although this is labor intensive for both the numbering and the post-weigh sorting. If marking the coins is not permissible , then each stack’s allotment of coins may be placed in a container (such as a baggie) before being placed in the scale, although one has to account for the mass of the bags in the total weight.

Trainspotting

 

There is a man who enjoys watching trains.  He lives a short distance from a railroad track, and has taken the hobby of watching trains go by.  After a few weeks of this, he observes that passenger trains appear much more frequently than do freight trains.  He resolves to record the exact frequency of each type of train by going to the track every day at a different time and waiting until he spots a train before he leaves.  After a few months of collecting data, he calculates that the ratio of passenger trains to freight trains is roughly 5 to 1.

 

Later on, during a discussion with a local railroad officer, the man is very surprised to learn that in fact the ratio of freight trains to passenger trains is 1 to 1. Assuming the man's math is correct and his times of visiting the track were randomly distributed over the 24 hours of the day, how do you explain the discrepancy between his data and the actual ratio?

Solution

The railroad officer explains that, while passenger trains run every hour throughout the day in order to serve the commuters on board, freight trains are always scheduled to run exactly 10 minutes behind the passenger trains. So the observer, arriving at the track at random times and waiting until the first train appears then leaving immediately thereafter, will only spot a freight train if he happens to arrive during the 10 minute window between the passing of a passenger train and the arrival of a freight train. He is about five times more likely to arrive between a freight train and a passenger train than vice-versa.

Three Boxes of Fruit

 

There are three boxes on a table before you. One contains only apples, another contains only oranges, and the third contains both apples and oranges. Each box is labeled incorrectly. The labels read Apples, Oranges, and Apples and Oranges. Your task is to correctly label each box. You may withdraw a single piece of fruit from a single box, no peeking or feeling around or other monkey business. How do you complete the task?

 

Solution

Draw a piece of fruit from the box labeled Apples and Oranges. We know that this box is labeled incorrectly, so it must be either apples or oranges, not both. Suppose you withdraw an apple. This means that the box labeled Apples must contain oranges (otherwise the box labeled Oranges would contain oranges, and we specified that each box is mislabeled), and the box labeled Oranges must contain apples and oranges.

I first encountered this venerable piece of puzzling in the book “More Games for the Super Intelligent” by James F. Fixx, although I have written it here from memory, not verbatim. I’ve no idea whether I’m required to disclose this kind of stuff, but I live in a world where lawyers prowl like hungry sharks, so I’m just trying to get by without any problems.

The Ship Problem

A ship is twice as old as it's boiler was when the ship was as old as the boiler is.  The sum of their ages equals 49 years.  How old is the boiler and how old is the ship?

Solution

The ship is 28 years old and the boiler is 21 years old. I struggled for a long time to discover a single elegant mathematical formula that would represent this problem, but when I gave it up as being beyond my abilities, I then moved on to technique #2: Brute-force it by plugging in numbers and adjusting until correct. Took me maybe 3 minutes to solve it this way. Extra nerd-cred if you post a formula for this problem in the comments.

Monty Hall Problem

Suppose Monty Hall presents you with three doors.  Behind one is a car (or for our purposes any desirable object) and behind the other two are goats (or any undesirable object).  He instructs you to choose one of the doors, and the prize it conceals shall be yours.  You pick a door at random, but instead of opening that door he walks to one of the other two doors and opens it to reveal a goat!  Now, he asks you if, with this new information, you would like to stick with your original choice or if you would like to change your mind and open the other remaining door.  What should you do?

Solution

 

It seems intuitive that with two doors remaining, there is a 50/50 chance that the door which you originally chose conceals the prize. In fact, this is not the case. The door which you originally chose has a 1/3 probability of concealing the prize, the same as it had before Mr. Hall opened one of the other doors. This means, of course, that changing your door selection will yield a 2/3 probability of winning the prize! Simply Amazing! If you are skeptical, I don't blame you, I was as well. This is not only extremely well documented and mathematically proven, but it also inspired some fascinating controversy when it was young. It duped thousands of self-proclaimed Ph.D holders who were utterly convinced that their intuition was correct. I encourage you to look this up and read about the fascinating history. Anyway, rather than transcribe a lengthy mathematical proof here, I encourage you to explore this problem with a friend or loved one. My wife and I used to play this game at a diner on Sunday mornings using ketchup bottles, juice glasses, sugar packets, whatever. It was a magical experience for me to conduct this investigation with her, but I guess my deep affinity for probability puzzles wasn't enough to keep her around. Oh well.

Newton-Pepys Problem

For my inaugural puzzle, I have chosen a (relatively) famous piece of probability theory.  In order to C.M.A., I am pulling this verbatim from The Greatest Puzzles Ever Solved, published by SevenOaks out of London, ISBN 978-1-86200-738-3.  Please don't sue me, you limey bastards.


"In 1693, famous English public servant and diarist Samuel Pepys entered into correspondence with Isaac Newton, the father of physics. The topic under discussion related to a wager that Pepys was considering.  Pepys wanted to know which of three dice rolls had the greatest odds of success.  These were to roll six dice and get at least one six, to roll twelve dice and get at least two sixes, and to roll eighteen dice and get at least three sixes.

Which is more likely?"
Solution

"Pepys' intuitive suspicion was that the largest roll was the easiest. It's not the case, though. You are more likely to make the six-die roll. Newton pointed out that you could imagine the twelve-die roll as two sets of the six-die roll, and the 18-die roll as three sets. To make the six-die set, you only have to achieve success once. The other sets effectively require you to make the roll more than once. The relative probabilities aren't that simple- you can roll more than three sixes in the eighteen-die roll, at which point you can't quite keep the principle holding true. The 6-die roll has a 0.66 chance, the twelve-die roll a 0.62 chance, and the 18-die roll a 0.60 chance."

This isn't the clearest explanation, although it is correct.  I don't blame you if you don't believe it- I didn't either.  Fortunately the solution is well documented and you can google it if you wish.