Don’t use data roaming when on holiday in Japan:
That’s a phone bill that cost as much as my last holiday.
Inspired by this analysis of coin efficiency at the Freakonomics blog, I’ve written a short Python script to calculate the efficiency of the Euro coin set. Like in the blog post, efficiency here means the lowest average number of coins needed to sum to any sub-Euro total (where all totals are assumed to occur equally frequently). For example it requires five coins to sum to 48¢: two 20¢, a 5¢, a 2¢, and a 1¢.
It turns out the average number of coins you’ll need is about 3.43, given the current set of six coins (I exclude the Euro and two Euro coins, since they’re no help in reaching sub-Euro totals). The maximum number of coins you ever need to reach any total is six. Note that we get this efficiency, much higher than the US’s 4.70, by cheating. We have six coins to play with, where they have only four.
The interesting part is in figuring out which coin would have the least negative impact on efficiency if it were removed. You might like to take a moment to ponder which is your least favourite coin before I reveal the answer. If you’re anything like me you probably hate the one cent coin, but sadly that’s the only one this analysis won’t let us remove: it’s required to form the total of 1¢, obviously. Every other coin is expendable, but with differing effects on the efficiency of the coin set.
Or so you might think. In fact, the coins barely differ at all in their effect. The 2¢, 5¢, 20¢, and 50¢ coins would each reduce the efficiency of our coinage by 0.81 if they were to disappear overnight. Only 10¢ differs, having a smaller effect of reducing efficiency by only 0.40.
So if we wanted to reduce the number of different coins all of us have to deal with (the principals are the same for me in the UK with Sterling, having 1p, 2p, 5p, etc.) with the minimum impact on efficiency, we should get rid of the 10¢ and 10p coins.