Dozenal Money

a personal view by David James

		I was looking at the dozenal money by Shaun Ferguson when I came across his suggestion that the 1d, 3d and 6d coins be resurrected. These are not in fact the best denominations to use in terms of minimising the number of coins needed to make change. In a decimal currency, or at least one that is set up as well as such a currency can be, the ideal combination is 1, 2 and 5 (in North America we use 1 and 5, then 10 and 25 and then after that do we get to $1, $2 (not in US) and $5 - which probably explains why everyone over here complains about how many pennies they have). In a dozenal currency there are more options. I have found that {1, 3 and 4}, {1, 4, 6}, and, somewhat surprisingly, {1, 3, 5}, and even {1, 3, 7} to be all equally efficient and all are better than {1, 3, 6}. My personal preference would be {1, 4, 6} and then (1, 3, 4}. I would tend to stay away from series like {1, 2, 3, 6}, {1, 2, 4, 6}, {1, 2, 3, 4} and {1, 2, 3, 4, 6} since these just add extra coins that really aren't needed.
Here are tables showing my calculations, starting with a familiar example from the decimal system. (Averages, where approximate, rounded to 2 decimal places)

Comments by Andrew Sieber

David James expresses a preference for (1,4,6) but doesn't give reasons. Here are some reasons in support of that preference.

1. Counting up by 1 from value 0 to eleven using the minimum coins for each value (1,4,6), (1,3,5) and (1,3,7) allow 7 transitions by just adding coins and only 4 require replacing coins. In contrast (1,3,4) allows 4 transitions by just adding coins and 7 require replacing coins. The same holds for removing versus replacing coins while counting down.
2. If the next higher order of magnitude has no dedicated unit coins or notes (or none happen to be at hand), then forming a unit of it requires just two coins (two 6s) with (1,4,6) but requires three coins (three 4s) with (1,3,4), four coins (four 3s, or a 5, two 3s and a 1) with (1,3,5) and four coins (four 3s, or a 7, a 3 and two 1s) with (1,3,7), and in general forming any round value in that magnitude requires more coins with (1,3,4) or (1,3,5) than with (1,4,6) and on average slightly more with (1,3,7) than with (1,4,6). (1,3,7) requires less coins than (1,3,4) or (1,3,5) for units greater than 1 in the higher magnitude.
3. If coin sizes are proportional to value then the middle and large coins for (1,4,6) are easier to distinguish than for (1,3,4), since the large coin is 3/2 the size of the middle coin instead of just 4/3. But (1,3,5) is even better, and (1,3,7) the best in this regard.
4. For (1,3,4) and (1,4,6) all coins are even divisions of the next higher order of magnitude. This isn't the case for (1,3,5) and (1,3,7).

So:

1. (1,4,6), (1,3,5) and (1,3,7) are better than (1,3,4)
2. (1,4,6) is better than (1,3,4), (1,3,5) and (1,3,7)
3. (1,3,7) is better than (1,3,5), which is better than (1,4,6), which is better than (1,3,4).
4. (1,3,4) and (1,4,6) are better than (1,3,5) and (1,3,7).

Scores (2 points for winning - or tying for a win - an item, 0 for losing, 1 for coming in the middle):

• (1,3,4) : 2
• (1,3,5) : 3
• (1,3,7) : 4
• (1,4,6) : 6

So overall (1,4,6) wins, Intuitively, winning D is required; in this case only (1,3,4) and (1,4,6) qualify, and since (1,4,6) beats (1,3,4) in all three of the other items, (1,4,6) is the definite overall winner.

Another difference between (1,3,4) and (1,4,6), which could be an advantage for either one:
1,4,6) has a unique combination of coins for every value, but (1,3,4) has two minimum combinations for 9 (two 4s and a 1, or three 3s). But even if this were an advantage for (1,3,4) it still wouldn't affect the overall winner.