 # Estimating the value of a mass of US coins

So I was in my cousin’s office this weekend and I saw a 5 gallon water bottle that had been emptied of water and filled with spare change. My first thought was that it must be terribly heavy, and when I tried to lift it I discovered I was correct – it’s extremely heavy. I was able to tip it to get my hands under it, but I was unable to lift it. Perhaps with help from another person I could even carry it around.

The bottle is 18.93 liters, give or take, which is .667 cubic feet. Combined with this page I have what I consider to be a lower bound for the value of the bottle: \$328, if it’s full of pennies and not packed optimally. I could probably compute an upper bound as well, assuming it’s made of the highest value-per-volume coin (I think that’s going to be dimes, but I’m not sure).

If I were able to obtain an accurate weight for the full jug (and a reasonably precise estimate for the mass of the empty jug), how closely could I estimate the value of all of the coins inside? I’d like to think that simple linear algebra would do the trick, but I can only think of three equations – one for total mass of the coins, one for volume, and one for total value – and I know I’ve got at least four unknowns (number of each type of coin).

So:
.25q + .10d + .05n + .01p = \$(some large number)
Mqq + Mdd + Mnn + Mp * p = (some large number of kilograms), where M(x) is the mass, in kilograms, of coin x.
Vq
q + Vd * d + Vn * n + Vp * p = .667 cubic feet, where V(x) is the volume, in cubic feet, of coin x.

M(x) and V(x) are almost certainly public information, but that still leaves me with three equations and five unknowns, even if I do know the mass to within a few grams. I think my only hope is to find an equation governing the frequency with which each coin is circulated relative to the others. Assuming that’s possible, could one estimate the value of the coins inside without knowing the mass of the full jug at all?

Since you know the volume, you could calculate the number of coins of each denomination that fit in that volume, given a less than optimal packing. You could then take the weighted average based on the frequency of each coin. It’s not necessarily a great estimate, but as a first approximation, it should be OK.

I don’t think you could do it even with a known mass. You have 4 non-uniform parts (coins) with unknown proportions that don’t fit together nicely.

I’ve been wondering the same thing. I have a big pepsi bottle full of coins. It is about 3 feet high. I know that I put more than 100 dollars worth each year and have been doing so for many years. I’m guessing it is more than a thousand dollars worth. (I have a separate place for pennies, so it is all quarter, dimes and nickels.

If you want to cash it in, interesting choice:

A) Spend an entire weekend rolling the coins, or

B) Brave the 10-15% vigorish that those coin machines typically take.

or C) go to your bank where (if it’s a good bank) they’ll let you use the coin counting machine for free.

In our area there is a bank that will let you use their change counter for free. You don’t even have to be a customer. Those small things are what will make me go to them if I ever need a bank again (currently I use the credit union.)

I’d be willing to bet that the dollars-per-unit-volume of the average american pocket change bucket is pretty constant, especially if you aren’t dealing with someone who intentionally biases their coin collecting habits (e.g. someone who does coin-op laundry habitually seperating out their quarters). Now the question is how to figure that out. It would probably be interesting to know the actual distribution of US coins, so that you could predict the value of a randomly chosen sample from that pool, but it’s important to note that the contents of a pocket change bucket are not randomly chosen (resulting as they are from transactions where there are 99 core permutations that make the most sense to get), so the “value density” (warning: invented term) of the average bucket is likely different from that of the US coin circulation as a whole.

A glass jar full of coins has another characteristic as well, that should be considered. It is under a very large tension from the “fluid” pressure of the coins. It can sit for decades, with no hint of problems, and then the drop of a single coin, or the torsion of trying to lift it will put some small flaw over the edge.

The failure is catastrophic, startling, and one hell of a mess to clean up.

Or, it can be one of those silly twist of fate things. A good friend of mine had a glass carboy full of coins. Someone opened the window just above his desk, and a very cold wind blew in. (VPI, dead of winter, from steam heat to plateau wind chill) There was a sharp crack. He looked at the carboy, and there was a neat crack all the way around the very bottom. He made the mistake of checking it.
Then the coins squirted out around the base faster and wider than you would ever imagine.

Tris

This one’s got me curious so I did a little exercise.

Assumption 1: change will be given from a particular transaction in the most logical way, e.g. 45 cents will be given as a quarter and two dimes, not as four dimes and a nickel. Obviously, this is not always true, but is probably true enough not to affect the results too much.

Assumption 2: all sub-dollar totals are equally likely (you’re just as likely to have a transaction ending in .23 as in .99). This one is probably wrong enough to impact the results, due to retail pricing practices, but I’ll continue anyway, and maybe someone can come up with a way to correct this one and weight the results.

If you got all 99 different change amounts once each, you would have:
200 pennies, 40 nickels, 60 dimes and 150 quarters.

So given the assumptions above, the average change jar should be 44.4% pennies, 8.8% nickels, 13.3% dimes, and 33.3% quarters. Note that this is just the relative number of each coin, not the relative weight, so you need to know how many coins there are total in order to apply them. More calculations are necessary to say that a given jar filled this way is n% pennies by weight, etc. Hang tight…

On preview: wow, Tris, I wish I could have seen that. Ok, plugging those numbers and weights of coins from here into a spreadsheet, I get:

By weight, a bucket of pocket change (using my assumptions about collection from above) is 29.6% pennies, 11.9% nickels, 8.1% dimes and 50.42% quarters.

So a pound (453.6g) of said change contains:

134.5 grams of pennies (= \$0.53)
53.8 grams of nickels (= \$0.53)
36.6 grams of dimes (= \$1.61)
228.7 grams of quarters (= \$10.09)

… giving this hypothetical change mixture a value of \$12.77 per pound
Ok, anyone see anything blatantly wrong with my calculations?

My change weighs 55 lbs - no pennies.

If you are able to get an accurate total weight and empty container weight, perhaps the best way to estimate the total value is through sampling. Take out a couple to a few pounds of coins at random, count out the value of this sample and weigh it. From this you can obtain a \$/lb value that you can extrapolate to the entire container full of coins. The only assumption required is that the composition of the coin mixture is fairly uniform.

Well if you took a sample of my hypothetical standard change mixture and simply threw out all the pennies, it would be (by weight) 16.9% nickels, 11.5% dimes, and 71.7% quarters, putting its value at about \$17.39 per pound.

FWIW I keep my pennies, nickles and dimes***** in a one-quart glass milk bottle. I shake it from time to time to get the coins to settle. When it’s full I usually have \$45-\$55 in change.

*I save the quarters for laundry. When I’m home (with my washer and dryer) I just save them.

I would like to make a prediction. Next time your jar is full, if you weigh it, it will be between 8 and 10 pounds. (my spreadsheet says that without quarters, pocket change will be \$5.43 per pound) That was a great analysis.

Given galt’s earlier math, can we deduce that a quart of change with quarters would be worth (12.77/5.43)*100% more, or about 2.35 times as much? If so, then my cousin’s five gallon jug is probably worth:

5 gallons * (4 quarts / 1 gallon) * (12.77 with quarters / 5.43 without quarters) * \$50 per quart-without-quarters = about \$2,350!

Also, for anyone interested in playing around with the numbers, I’ve put together a Google Spreadsheet here that includes mass estimates for full distributions of coins, and the “no pennies” and “no quarters” mixes.

For some strange reason I’ve been tallying the coins in my coffee can before taking them to the bank. It’s how I deal with change.

2.5 years worth of coins.

10,843 coins worth \$1204.36 (I excluded 253 1 coins.) Average coin value is .1111.

Quarters at 33.9%.
Dimes at 16.5%.
Nickels at 12.2%.
Pennies at 37.4%.

I’m pretty good at emptying the pockets and rarely raid the can for prime coins. Hope this helps.

Fantastic. Earlier today I found myself wishing I had done this in the past so I could find out how close my numbers were. My prediction by coin count was:
Quarters at 33.3%
Dimes at 13.3%
Nickels at 8.8%
Pennies at 44.4%

My percentages would have come up with a prediction of \$1142.74 for 10,843 coins, which is off by 5%. I have to say, I feel like that’s not too bad.