Forget a new £1 coin, we need a £1.23 coin

Not the new coin we want, but the new coin we need

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Forget a new £1 coin... the £1.23 coin is what we should be replacing it with. (Formed from composite of public domain images)

Tomorrow, the Royal Mint—producer of British coins—is introducing a new, thinner, dodecagonal £1 coin. But they’re missing a trick. To cut the amount of change in our pockets, we don’t need a lighter £1 coin: we should replace it with a £1.23 coin.

New one pound coin

Coming soon to your pocket… the new pound coin. (Royal Mint)

The question you need to ask is: When you pay for something in a shop with a banknote, how many coins do you expect to get back in change? How much heavier will they make your purse?

The answers are different for different countries, so we’ve worked it out for your country as well!

How many coins do you expect to receive in change?

The smallest banknote in the UK is the meaty see-through £5 note. So if you pay with a note, you would hope to receive somewhere between 1p and £4.99 in coin change back. Supposing an even distribution of prices and that cashiers always give you the most efficient change (next week’s blog: when self-service machines don’t give efficent change), on average, then, how many coins do you expect to receive?

We’ve done the calculation for a few countries, for change amounts up to the smallest banknote.
You can find the code we’ve used at the bottom of the page.
Average number of coins expected as change in each country
Some variants on the UK system are highlighted in pink. Switching the £1 coin for a £1.23 coin reduces the coin expectancy by about half a coin, from 4.61 to 4.07: the best-performing option.

Some other observations:

  • Some countries do really well (India) because banknotes are used almost exclusively.
  • Low-scoring countries have abolished their smallest denominations (pennies etc.)
  • The US only has 4 common coins, yet averages the same number of coins as the UK with 8.
  • The pre-decimal UK system (pounds, shillings and pence) performs almost as efficiently as the current UK system.
  • The Harry Potter system (29 knuts in a sickle, 17 sickles in a galleon) is ridiculous.

And what about the weight of these expected coins in our pockets?

Average weight of coins expected as change in each country
Observations:

  • The UK has, on average, the heaviest coins out of the top 20 circulating world currencies. Removing 1p and 2p would reduce the weight by 20%.
  • Despite expecting the same number of coins in the UK and US, the weight of US coins is about half that of the UK.
  • Australian coins are heavy because their sizes are the same as pre-decimal UK ones.

If we could add just one coin, what would it be?

Suppose we were to keep all our coins, but could add one more. Which denomination would reduce the average number of coins you receive in change the most?
Average number of coins expected as change  if you add a new coin of a certain value
Adding a £1.33 or £1.37 coin would reduce the average number of coins from 4.61 to 3.93.

If you could choose any coins, what would they be?

So suppose we start over. A whole new set of coins up to £5. Designed to be the most efficient in terms of change. For a set number of coins, what would they be?

Number Coins Number expected
2 1p, 22p or 1p, 23p 21.3
3 1p, 14p, 61p 9.98
4 1p, 7p, 57p, 80p 6.82
5 1p, 6p, 20p, 85p, £1.21 5.45

Programming this yourself

I do this all with a bit of Python code I found on StackExchange:

def get_min_coins(coins, target_amount):
    n = len(coins)
    min_coins = [0] + [sys.maxint] * target_amount
    for i in range(1, n + 1):
        for j in range(coins[i - 1], target_amount + 1):
            min_coins[j] = min(min_coins[j - coins[i - 1]] + 1, min_coins[j])
    return min_coins

This is nice code because it avoids the lazy approach (‘greedy algorithm’) of trying the highest coin first and then dealing with the remainder. Such a lazy approach is quick but fails if you have coins of 1p, 3p, 4p and want to make 6p. The lazy approach would give you 4p, 1p, 1p; but of course the best option is 3p, 3p.

Questions to investigate

Next week we’ll use this code to answer the age-old question of

  • Why do supermarket self-checkout machines give such terrible change?

Plus, we’ll ask are quarters are better than 20-cent pieces?

Have a play with the code yourself. Comments are open below when you find something interesting.

Bonus: old £1 coin v new £1 coin

Adam is a postdoctoral researcher at Imperial College London, where he investigates weird, non-Newtonian fluids. If he’s not talking about the maths of chocolate fountains he is probably thinking about fonts, helping Professor Dirichlet answer your personal problems, and/or listening to BBC Radio 2.

Twitter  @Pecnut   Website  adamtownsend.com    + More articles by Adam

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2 thoughts on “Forget a new £1 coin, we need a £1.23 coin

  1. It’s a fancy idea, but would seriously discriminate against anyone with numeracy problems who handles money – so it would leave some people finding it harder to get employment, and also some with a higher chance of being ripped off by the unscrupulous (as happened back when decimalisation kicked in).

  2. Of course, we don’t actually need coins at all. Equip every shop with a random number generator, generating uniformly distributed random numbers between 1 and 500, which is spun for each purchase. If you make a purchase of £6.79 for example, and the generator lands on 179 or lower, you pay the next multiple of £5 above (ie £10). If it lands on 180 or higher, you pay the multiple of £5 below (ie £5). The expected value of your payment is £6.79, so the system is fair to both you and the shopkeeper. Although, somehow, I suspect you might have a bit of difficulty convincing the non-mathematical general public of this. It would also be a bit hard on those people on a very tight budget, who might get into difficulties due to an unfavourable spin on one occasion, even though on average they could just about cope.

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