Numberplay: Denominations - NYTimes.com It's often said that the most efficient coin system would consist of seven coins in denominations 1, 2, 4, 8, 16, 32 and 64 cents. This would allow you to carry the fewest coins necessary to make change for a dollar. We ran into this basic idea (until shattered by Mike) in our recent Mixed-Up Wires. The problem with this system is that it involves numbers that aren't very easy to add in your head. That's troublesome if you're a cashier. The system also a lot of different coins, which is troublesome if you're a vending machine. So I wondered about something better. Fewer coins, for starters. What about two coins? Would certainly make sorting a lot easier. And the math? Probably depends on the denominations. Which brings us to our puzzle this week. If a currency system could only have two coins, what should the values of those coins be? Recap: Fat Coin Last week we flipped over Fat Coin. The puzzle: What is the shape of a coin that has probability 1/3 of landing on its edge? This puzzle had been suggested by mathematician John McKay, who added that the problem could be done in the head. The solution: the conversation last week coalesced around two main solutions, which are shown from the side next to a regular coin (the flat disk on the left) below. Which solution is right? It depends on how you flip your coin. One reasonable way to flip a coin would be to toss it end-over-end and have it land on one of its corners and then keel over. Like a coin flipped in slow motion onto flypaper. We're looking for a coin that will keel over onto its edge - as opposed to its head or tail - a third of the time. In other words we want a coin that, when balanced on a corner, will have twice the angle between its side and the floor as between a face and the floor. An easy way to get the dimensions of such a coin would be to draw it over a collection of packed circles. This coin would have a diagonal length of four circles and be two circles thick, and would therefore have a diameter of three times the thickness. But there are problems with this solution. First of all, it's too easy. Gary noted this in an early comment. OK - it's not exactly that easy, but John McKay saying a problem can be done in the head is a bit like Lance Armstrong saying a destination is reachable by bike. The second problem is the end-over-end assumption. Coins bounce all over when they land. A more accurate model might be to imagine the coin materializing in any possible configuration on the floor and then falling over. In this case you'd want a thinner coin. In particular, you'd want a coin with thickness one third of its diagonal dimension, which would mean a diameter eight times the thickness. This, astonishingly, is what you'd get if you had a giant coin with edges that coincided with the earth's Tropic of Cancer and Tropic of Capricorn (which divide the earth into three equal-area regions). Hans and Marco were among the first with this solution. Steve Kass offers this explanation: Imagine that the coin is inscribed symmetrically within a soap-bubble sphere, with the coin's two circular edges being latitude lines defining the sphere's "temperate zone." Now set the sphere down on a table and imagine how the coin would land if the soap bubble popped. The coin will land on edge if and only if the sphere-table point of contact is within the temperate zone. Simply stunning - a Tour de France based on the same idea as Archimedes" Hat-Box Theorem. Very different answers, with one coin being about 50 percent fatter than the other. Which is right? The official solution: a diameter eight times the thickness. The thinner of the fat coins. That's the solution John von Neumann came up with in his head, and the one John McKay intended as the result for this puzzle. My own sense, by the way, is that John von Neumann was way off. His coin would be far too thin. A coin of a diameter of three times the thickness is probably more accurate. And even that might not be fat enough. My argument is based on the nickel, which happens to be about a tenth as fat as it is wide. John von Neumann, using the Hat-Box idea, would give a nickel about a one in 10 chance of landing on its side. Actual odds? About one in six thousand. Essentially zero. A nickel tends to land either heads or tails for the same reason a tossed pencil won't ever land on its end: the moment generated by its rotation will tend to prevent precarious final resting positions. Ravi and others hit upon this point several times. The final answer wasn't worked out, but the door was opened. And, of course, an experiment or two can often be useful. Here's Gary from D.C. with some initial tests. Incidentally, Numberplay puzzles often roll out in waves, with conceptually similar puzzles emerging around the same time. I'd say Fat Coin is very different from this week's Two Coins, but its end-over-end interpretation is nearly identical to the core problem in Meeting Times, while the temperate zone interpretation is a lot like Far Side. (I had solved Meeting Times using a sine wave, so I tried the same approach with end-over-end Fat Coin and found it worked very well. Below are the coin shapes necessary to create any probability of edge-landing from 0 to 1.) Thanks again to John McKay for the Fat Coin puzzle. Dr. McKay is a professor of computer science at Concordia University. Transform your Numberplay experience with Gary Hewitt's Numberplay Commenting Fix. This browser add-in will number comments and turn TeX strings and image URLs into formulas and graphics. Have any puzzles that can be done in the head? Please e-mail to Numberplay@NYTimes.com.