Lots of online generators will gladly spit out hoards that are heavily inclined toward higher-denomination coins, gems, and other, highly-concentrated instruments of wealth; I want to go heavily into the mid-denomination coins for some treasure hoards, especially where mechanically and/or thematically appropriate, such as in the hoard of a Silver Dragon or in the purse of a tax collector: the former because the Silver wyrm likes silver, the latter because the average person being taxed would not have had high-denomination coins or gems with which to render payment.

Here's an example problem: let us say there exists a hoard of gold, silver, and copper coins which have a combined value of 873 gp; and, the value of 1 gp = 10 sp; and, the value of 1 sp = 20 cp; and, the percentage of coins in this hoard are 13% gold coins, 60% silver coins, and 27% copper coins. Given the above— 1. How many coins of each type are in this hoard? 2. What is the mathematical formula to determine these numbers?

Here's another example problem: let us say there exists a hoard of gold, silver, and copper coins which have a combined value of 212 gp; and, the value of 1 gp = 10 sp; and, the value of 1 sp = 20 cp; and, the percentage of coins in this hoard are 17% gold coins, 54% silver coins, and 29% copper coins. Given the above— 1. How many coins of each type are in this hoard? 2. What is the mathematical formula to determine these numbers?

N.B.—I am not looking for napkin math or juggling numbers to brute force the answers through guessing and elimination. I've already done that to arrive at the answers to part 1 for the above problems; what I don't know, and want, is a maths formula I can use to solve different problems by simply inputting the total coin value and the coin percentages.

  • 4
    \$\begingroup\$ Is there tolerance on those %s, or on the total value? In general there isn't going to be an integer answer to this \$\endgroup\$
    – Caleth
    Nov 26, 2021 at 17:05
  • 5
    \$\begingroup\$ I think any solution to this is going to be algebraic - I'm not sure 'statistics' is an appropriate tag. \$\endgroup\$
    – Kirt
    Nov 27, 2021 at 3:48
  • 3
    \$\begingroup\$ I think OP should highlight in bold whether they're stipulating desired ratios for the number of coins or the value of the coins. We currently have multiple answers going in either direction (with mutually contradictory results, obv.) \$\endgroup\$ Nov 27, 2021 at 15:53
  • \$\begingroup\$ @Kirt Is there another tag that you think would fit better? AFAICT there is no more general tag for mathematics. \$\endgroup\$ Nov 28, 2021 at 0:20
  • 1
    \$\begingroup\$ @HighDiceRoller Not really - I don't know the tags that well, otherwise I would have suggested something. My first thought was that we should have a tag called 'mathematics' or 'calculations' - but looking at other questions tagged 'statistics', at least all the first page are things that deal with probability functions in a way this question does not, such that 'statistics' is most likely an appropriate and well-used tag. I think at this point 'statistics' should just be dropped from this question, but I think it should be brought up on meta. \$\endgroup\$
    – Kirt
    Nov 28, 2021 at 3:58

6 Answers 6


You can come up with the formula by figuring out the average value of adding one coin to the hoard. You know the percentage and value of each type of coin, so for each type of coin, multiply that percentage by the value of the coin in gold pieces, and add them all up or $$AvgValue = \sum_{t=Types}{Percentage_t \times ValueInGP_t}$$ Then the total number of coins in the hoard is just the total value divided by the average value. You can figure out the number of each type by multiplying the total number by the percentages.

In the example you gave (which can't be solved using an integer number of coins), the average value would be

$$AvgValue = \underbrace{13\% \times 1}_{Gold} + \underbrace{60\% \times 0.1}_{Silver} + \underbrace{27\% \times 0.005}_{Copper} = 0.19135$$ And the total number of coins in the hoard would be $$Number\;of\;Coins=873/0.19135=4562.32$$ Resulting in

$$ \begin{align} Gold &= 4562.32 \times 13\% = 593.1 \\ Silver &= 4562.32 \times 60\% = 2737.4 \\ Copper &= 4562.32 \times 27\% = 1231.8 \end{align} $$

  • 2
    \$\begingroup\$ Welcome to RPG.SE! Take the tour if you haven't already and see the help center or ask us here in the comments (use @ to ping someone) if you need more guidance. Good Luck and Happy Gaming! \$\endgroup\$
    – Someone_Evil
    Nov 26, 2021 at 17:46
  • 4
    \$\begingroup\$ And here’s an implementation you can play around with. \$\endgroup\$ Nov 27, 2021 at 12:49

To slightly challenge the framing of your question, I'd like to suggest that it will be much easier if you start by deciding what percentage of the value of the hoard is in coins of each type.

For example, let's say that 75% of the value is in gold, 20% in silver and 5% in copper, and the total value of the hoard is 800 gp.

  • 75% of 800 gp is 600 gp, so your hoard contains 600 gold coins.
  • 20% of 800 gp is 160 gp, so your hoard contains 160 × 10 = 1600 silver coins.
  • 5% of 800 gp is 40 gp, so your hoard contains 40 × 200 = 8000 copper coins.

With reasonable round numbers like this, you can even do the math in your head. (I did, above.) And your players will probably appreciate the round numbers too.

If that still feels too unrealistic to you, just arbitrarily fudge the numbers a bit after you've done the math and say that the hoard contains, say, 589 gold coins, 1712 silver coins and 8475 copper coins.

(As long as you only fudge the numbers by less than 10%, and make sure not to fudge them all in the same direction, the total value of the hoard won't change too much. The randomly fudged numbers above, for example, sum to 589 + 171.2 + (84.75 / 2) = 802.575 gp, which is actually surprisingly close to 800 gp.)

By the way, note how the copper coins in the example above ended up being almost 80% of the total number of coins, despite having only 5% of the total value of the hoard.

That's a thing that naturally happens when you have a mixture of coins of vastly different values: a single gold coin is worth 200 coppers, so if more than about 0.5% of the total value is in copper, there will necessarily be more copper coins than gold coins in the hoard.

It's up to you to decide if that's what you want, and that decision is likely to depend on who set up the hoard and why.

In general, a useful rule of thumb is that high-value coins like gold are the best for storing lots of wealth in a small space, while low-value coins like copper are probably the most commonly circulated and easiest to come by.

So something like a pirate's buried treasure chest, or a rich merchant's hidden emergency stash, is likely to be mostly gold and maybe some silver to fill up the space. Meanwhile, a store of money in active use, like a lord's treasury or a trader's coin purse, will likely have quite a few copper coins in it, because they'll keep getting more in trade as fast as they can trade them away.

And as for a dragon's hoard or a bandit camp's loot, who knows? Did they indiscriminately gather all the coins and other loot they could get (implying lots of silver and copper), or did they just pick the most valuable and easy-to-carry stuff and leave the rest behind (implying little if any copper at all)? And, come to think of it, how exactly does a dragon transport coins to their lair anyway?

Anyway, if you've already decided on specific percentages of coins by number, as in your question, you can also convert between value percentages and coin percentages.

The easiest way to do that is a two-step process. For instance, to convert from value percentages to coin percentages:

  1. First multiply each percentage by the number of the corresponding coins that are worth one gold piece (i.e. 1 for gold, 10 for silver, 200 for copper).

    This first step will give you the correct proportions of the value in each type of coin, but these proportions will generally not sum up to 100%.

  2. Thus, the second step is to add the scaled percentages from the previous step together, divide each scaled percentage by this sum and multiply it by 100%.

So for example, in my example hoard above, gold, silver and copper coins appear in proportions of 75 gold coins to 20 × 10 = 200 silver coins to 5 × 200 = 1000 copper coins. These proportions sum up to 75 + 200 + 1000 = 1275, so we need to divide them by 12.75 to obtain the normalized percentages of 75 / 12.75 ≈ 5.88% gold coins, 200 / 12.75 ≈ 15.69% silver coins and 1000 / 12.75 ≈ 78.43% copper coins.

(Of course, we can round those to, say, 5% gold coins, 15% silver coins and 80% copper coins for convenience.)

Conversely, to convert from coin percentages to value percentages you can use the same two steps, but divide instead of multiplying in the first step.

(Or, equivalently, you can instead multiply each percentage by the number of copper pieces the corresponding coin is worth, i.e. 200 for gold, 20 for silver and 1 for copper. This will give numbers that are 200 times larger than what you'd get using the division method, but their relative proportions will be the same, and the second step will normalize them to the same percentages anyway.)

For instance, for your original example hoard of 13% gold coins, 60% silver coins, and 27% copper coins, we find that the value is distributed in proportions of 13 gp in gold to 60 / 10 = 6 gp in silver to 27 / 200 = 0.135 gp in copper.

Dividing those proportions by their sum of 19.135 gp and multiplying by 100%, we find that the value of the hoard is made up of approximately 67.94% gold, 31.36% silver and 0.7055% copper.

You can then multiply these percentages with any total value of the hoard in gp and convert the resulting share of the value to the appropriate type of coins.

For example, with a total value of 873 gp, this means that your hoard contains:

  • 873 × 67.94% / 100% ≈ 593 gp in gold coins,
  • 873 × 31.36% / 100% ≈ 273.8 gp silver, for a total of 2738 silver coins, and
  • 873 × 0.7055% / 100% ≈ 6.16 gp in copper, for a total of 1232 copper coins.

Also note that, as seen in the examples above, the number of copper coins in a hoard mixed with silver and gold typically makes very little difference to the total value.

So, as long as you don't insist on getting the total value exactly right down to the last cp, one way to simplify the math is to first figure out the approximate number of gold and silver coins in the hoard and then just add whatever number of copper coins feels appropriate to you. As long as it's not more than, say, 20 times the number of gold coins, it won't change the value of the hoard significantly.

(And if the number of copper coins in the hoard is more than 20 times the number of gold coins, that means there's either very little gold or a huge amount of copper in the hoard. In the latter case, transporting all that heavy copper out of whatever godforsaken dungeon it was found it in might become a valid concern — unless of course your players just decide to leave it behind and only take the more valuable silver and gold!)

  • \$\begingroup\$ I feel where you're coming from with this. However, it makes sense that the OP has a concrete visual goal for a ratio of gold-silver-copper coins flashing in a treasure pile. It might not be apparent to every DM exactly how little value the copper adds -- e.g., the very low 5% you picked here. Better to make that an explicit part of the calculating formula, rather than assume every DM would just intuit that up front (or rely on fudging in later passes, as you say). \$\endgroup\$ Nov 27, 2021 at 15:58
  • 1
    \$\begingroup\$ @DanielR.Collins: That's a good point. I added an explicit note about that, and also took the opportunity to generally reorganize the latter half of the answer while I was at it. \$\endgroup\$ Nov 27, 2021 at 18:09
  • \$\begingroup\$ Dragons just scoop out the innards of any conveniently handy large mammal (bull, horse, or man) and use them as a carry pouch. \$\endgroup\$
    – Arluin
    Nov 29, 2021 at 17:38

This is actually fairly straight forward, if you're used to the right thinking.

Let's say our total value is \$V\$. Then

$$ V = V_g n_g + V_s n_s + V_c n_c $$

where \$V_g\$ and \$n_g\$ is the value of each gold coin and number thereof. Same for the other coinages.

We also need the fraction of each coin, let's use \$\rho\$ which is then

$$ \rho_g = \frac{n_g}{n_T} $$

where \$n_T\$ is total number of coins. And again, we have the same for the other coinages. We can then write:

$$ n_g = \rho_g n_T $$

which gives

$$ V = (V_g \rho_g + V_s \rho_s + V_c \rho_c) n_T $$

$$ n_T = \frac{V}{V_g \rho_g + V_s \rho_s + V_c \rho_c} $$

which inserts into our expression for \$n_g\$ above to give

$$ n_g = \rho_g \frac{V}{V_g \rho_g + V_s \rho_s + V_c \rho_c} $$

This last formula gives the number of a given denomination based purely on the total value, coin values, and the fraction of each coin (which is what we wanted). Subbing in the appropriate \$\rho\$ in front gives the given coin count. (The fraction itself will be constant for a given hoard, so you may prefer to just calculate that once, and multiply by the fractions.) You'll want to make sure the same scale for all the values, whether you base it on gold coin equivalents or copper coins (or silver for that matter). Also make sure the fraction is between 0 and 1, and not expressed as a percentage.

If we choose the value based on number of copper, we get \$V_c = 1\$, \$V_s = 20\$, \$V_g = 200\$ and \$V = 212 \times 200 = 42400\$ for the second example. And we have \$\rho_g = 0.17\$, \$\rho_s = 0.54\$, and \$\rho_c = 0.29\$. Churning the numbers spits out coin counts of \$159.86, 507.78, 272.699\$ for gold, silver and copper respectivly. You'll need to round off obviously, but that's due to the percentages not being exact.

And as a final note, the general form (for any number of different coins) is

$$ n_i = \rho_i \frac{V}{\sum V_j \rho_j} $$

  • 1
    \$\begingroup\$ Coming at this as a reader: I'm not sure which formula I should actually use and what I should put into it to get a clean answer. It's clear you're deriving a few different formulae as you go, but it's not clear where the progress ends and the end begins. It would be helpful to devote a section to "here, use this equation, here's how to use it." I'd consider making the first section, and then use the rest to explain how you got there. \$\endgroup\$ Nov 26, 2021 at 17:51
  • \$\begingroup\$ @doppelgreener How does that look? I think it flows fairly well between getting the final formula and the example (when I just remember to put it in...) \$\endgroup\$
    – Someone_Evil
    Nov 26, 2021 at 18:12
  • \$\begingroup\$ Recommend that you explicitly work out the first example (instead of just the second), so we can more easily compare to other answers. \$\endgroup\$ Nov 27, 2021 at 15:59

873 gp [...]
value of 1 gp = 10 sp [...]
the value of 1 sp = 20 cp [...]
13% gold coins, 60% silver coins [...]

... and percentage of copper is irrelevant because thats simply what is left over.

Copper is the smallest coin - so calculate how much copper you got:

 873 gold * 10 silver/gold * 20 copper / silver = 174600 copper

Calculate how much of that is in gold and silver:

174600 / 100 = 1746 copper == 1 % of the copper value

13% in gold ==> 1746 * 13 =  22698 copper, 200 copper = 1 gold   => 113.49 coins
60% in gold ==> 1746 * 60 = 104760 copper,  20 copper = 1 silver => 5238 coins

Remaining ==> 174600 - 113 * 200 - 5238 * 20 => 47240 copper coins

You have to round down/up for gold or silver if they do not divide without remainder - I rounded down.

If you happen to have access to python (f.e. online here ) this calculates it for you:

def hoardCalc(totalGold: int, 
            goldPercent: int, 
            silverPercent: int, 
            silverToGold:int = 10, 
            copperToSilver:int = 10):
    """Calculates coin amounts given a gold value of a hoard, 
    percentages of gold and silver coins and optionally the 
    exchangevalues of silverToGold and copperToSilver. If 
    optional not provided 1g == 10s == 100c is assumed."""

    goldToCopper = silverToGold * copperToSilver

    totalCopper = totalGold * goldToCopper

    gold = int((totalCopper / 100 / goldToCopper  * goldPercent) // 1)
    silver = int((totalCopper / 100 / copperToSilver * silverPercent) // 1)
    copper = totalCopper - gold*goldToCopper - silver*copperToSilver

    Total copper  {'':>20}   {totalCopper:>20}
    Gold          {gold:>20} = {gold*goldToCopper:>20}
    Silver        {silver:>20} = {silver*copperToSilver:>20}
    Copper        {'':>20}   {copper:>20}\n""")


# 873 total gold, 13% gold coins, 60% silver coins, 1:10:200 exchange

hoardCalc(873, 13, 60, 10, 20) 


    Total copper                                       174600
    Gold                           113 =                22600
    Silver                        5238 =               104760
    Copper                                              47240
  • 2
    \$\begingroup\$ Note that this is giving different results from other answers. The problem is that you've assumed the OP wanted the 13/60/27% to be percents of the value, whereas most everyone else has read that as percents of the coin numbers. \$\endgroup\$ Nov 27, 2021 at 15:50
  • 1
    \$\begingroup\$ @DanielR.Collins agreed - slight interpretation error on my side. Leaving the answer up as other may interpret it exactly like this. The logistics of carrying away mixed loots lead to some creative t(h)inkering in my adventuring life ... \$\endgroup\$ Nov 28, 2021 at 11:29

Both of the examples you gave would use the same formulas (which you'll need 3 of), so I'll put them here followed up by the math:

1-Number of Gold coins = (Hoard Gold Value)(percentage of hoard that's gold)(ratio of gold coins to gold coins)
2-Number of Silver coins = (Hoard Gold Value)(percentage of hoard that's silver)(ratio of silver coins to gold coins)
3-Number of Copper coins = (Hoard Gold Value)(percentage of hoard that's copper)(number of copper coins to gold coins)

Or (for ease of writing)
r1 = ratio of gold to gold, r2 = ratio of silver to gold, r3 = copper to gold

1- G = H(g)(r1)
2- S = H(s)(r2)
3- C = H(c)(r3)

Example 1
Assuming that gp = one gold coin, sp = one silver coin, and cp = one copper coin

G = 873(0.13)(1)
G = 113.49 gp(1)
G = 113.49 gp

S = 873(0.60)(10)
S = (523.8 gp)(10)
S = 5238 sp

C = 873(0.27)(200)
C = (235.71 gp)(200)
C = 47142 cp
So the Example 1 horde would give 113.49 gold coins, 5238 silver coins, and 47142 copper coins.

Example 2
This one is pretty much the exact same as Example 1, but with different inputs. Same assumption that gp = one gold coin, sp = one silver coin, and cp = one copper coin

G = 212(0.17)(1)
G = 36.04 gp

S = 212(0.54)(10)
S = (114.48 gold value)(10)
S = 1144.8 sp

C = 212(0.29)(200)
C = (61.48 gold value)(200)
C = 12296 cp

So the Example 2 horde gives 36.04 gold coins, 1144.8 silver coins, and 12296 copper coins

Hopefully this answered your question adequately and you were able to follow it. However, if the above assumption (that gp = one gold coin, sp = one silver coin, and cp = one copper coin) is incorrect, then you'll have to add in an additional conversion step with the actual value of said coins. Which is essentially the same as what's been shown but with an extra set of multiplication, so not too bad.

  • \$\begingroup\$ Note that you've assumed OP has set ratios for the value of the coins, whereas most other readers have understood it as number of the coins instead. \$\endgroup\$ Nov 27, 2021 at 15:54
  • \$\begingroup\$ Hmm, true, that was an assumption I uncounciously made. Didn't see any reason why not to draw my assumption but will edit my answer to show the gp value before it gets spilt into my assumed coin amount. \$\endgroup\$
    – Joe D.
    Nov 27, 2021 at 20:39
  • \$\begingroup\$ Not to be a pest, but the answer is still showing markedly different results from other answers (because of that difference in initial assumption). \$\endgroup\$ Nov 28, 2021 at 14:17
  • 1
    \$\begingroup\$ Nah, you're not being a pest. But given my assumptions, which are now stated, are my answers wrong? If they are or my assumption is resulting in an answer that doesn't accurately answer the question, then I'll definitely fix my answer to give accurate info. If they are correct and do accurately answer the question, then I don't see why a different answer due to different stated assumptions is bad. \$\endgroup\$
    – Joe D.
    Nov 28, 2021 at 18:30

Hmm... There's not too much available to vary here. Unless the percentages are a bit loose and not exactly those integer values...

If you know that there's G% gold coins, S% silver coins, and C% copper coins for a total of 100 coins, where every coin can be rendered into Copper...

Then you convert them all into copper and your total value can only be a multiple of that resultant value in copper?

You already know the percentages, so just start from the initial numbers and multiply accordingly to get the integer number of coins.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .