Need Help With Change

the idea of OldSkoolMadSkilz would not really work, because once the coins are counted, they would be the banks. If you can, use donuts to hire local kids to count the coins.

What I would suggest is measuring the box to find its area. It might even label its capacity in quarts or gallons. Type in "x quarts in m^3" and it will give the capacity of the chest to you in cubic meters. Let's say it's a 50.00-quart ice chest therefore 0.0473 m^3.

Cent: 8305 kg/m^3 (copper) and 7180 kg/m^3 (zinc) AVERAGE: 7743 kg/m^3
Nickel: 8946 kg/m^3
Dime: 6669 kg/m^3
Quarter: 8950 kg /m^3

If the chest contained only pennies it would have approximately 366.2 kg of pennies. This is calculated by multiplying the density of pennies (in kg/m^3) by the capacity of the ice chest (in m^3).

If each penny weighs (average the copper and zinc pennies) 2.805g, then this means there are 130,553 pennies therefore $1,305.53.

If all nickels, then 423.1kg of nickels at 5.000g per nickel = 84,620 nickels therefore $4,231.
If all dimes, then 315.4kg of dimes at 2.268g per dime = 139,085 dimes therefore $13,908.50
If all quarters, then 423.335kg of quarters at 5.670g per quarter = 74,662 quarters therefore $18,665.50

If you take the maximum number of coins needed in any given transaction (4 pennies because 5 pennies makes a nickel, 1 nickel because 2 would make a dime, etc.) you have:

4 pennies
1 nickel
2 dimes
3 quarters

4+1+2+3=10, so out of every 10 coins, on average, 4 should be pennies, 1 should be a nickel, 2 should be dimes, and 3 should be quarters. This translates to 40%, 10%, 20%, and 30% respectively. We must now average the dollar values calculated above using a weighted average.

(1,305.53x40%) + (4,231x10%) + (13,908.50x20%) + (18,665.50x30%) = 9326.66
9326.66/4 = $2331.66

A 50-quart ice chest filled with U.S. coins (assuming there are no halves, dollars, or silver coinage) should contain approximately $2,331.66. Round down to the nearest $100 and offer him that, therefore: $2,300.

EDIT: To be on the safe side, assume that the $2,331.66 figure could be 5% off in either direction so it could contain as little as $2,215.08 or as much as $2,448.24. Barter using the $2,215.08 figure in mind, so I would say start by offering $2,000.00 (in this hypothetical scenario where the ice-chest is completely full and is 50 quart capacity.)

I've already done the basic calculation, so once you find how big the chest is you just have to do a simple proportion. So if it's a 25-quart chest instead of a 50-quart one, just halve the $2,300 figure I came to (assuming the chest is filled to the brim).

Also, when you go to offer him money, figure that the dollar amount you eventually come to could be 5% off in either direction. So take 5% off the figure, THEN round down from there to a reasonable starting offer.

DISCLAIMER: If my math is wrong, or there is some variable I'm not seeing, and my estimation is off, then I am abrogating myself of all responsibility. You ultimately have to make the call.

I'd go by weight. There could be huge variations in the density based on how they are packed. The weight will never change.

Just dump a huge scoop on the table. Toss the pennies to the left, nickels to the right and the rest towards you. Bet it will take a couple hours and your hands will feel nasty when done.

Pennies $1.81 per pound
Nickels $4.53 per pound
Dimes, quarters, halves and dollars $19.97 per pound

Just to be sure count out 200 dimes, 80 quarters, 40 halves and 20 dollars. Should weight a tad over a pound each.

I've counted a lot of change in my life and being old school ...I want to know what I'm paying for. From what you say 5 or 6 hundred dollars is not that much. Unless he is in a hurry I would take all the halves, quarters, silver dollars and golden dollars out ...pay him for that. Then go with the dimes and pennies. Thats what I would do.

hugemistake200 you must be a statistician.