So I was wondering about the distribution of coins in loose change. If most of loose change is a result of transactions where you get change for a random purchase that you have paid for with bills, how many of each coin are you likely to have? First, assume the amounts of the random purchases are uniformly distributed between one cent and 99 cents more than an even number of dollars. Then what I did was determine how you would receive change in quarters, dimes, nickels and pennies, assuming the change was delivered in the most efficient way, i.e., minimum coins. For example, if the purchase was $4.31, and you presented a $5 bill, you would receive your 69 cents as two quarters, one dime, one nickel and four pennies. I examined every possible amount of change from even dollars and totaled the number of each coin. The result was 150 quarters, 80 dimes, 40 nickels and 200 pennies. This reduces to 15 quarters, 8 dimes, 4 nickels and 20 pennies for a total value of $4.95. So these ratios should apply to any random collection of loose change.
Next I wanted to know the value of this change in dollars per pound. The weights of modern coins are quarter = 5.67 g., dime = 2.268 g., nickel = 5.0 g. and penny = 2.50 g. This means the weight of the $4.95 group is 173.2 g. or 0.3815 pounds and this gives the value of loose change as $12.98 per pound. So if you have a container of change that weighs 20 pounds, it probably contains about $260.
So if you throw your loose change into a jar or drawer, check it and you should find you have fewer nickels than any other coin and more quarters and pennies than any others. I have left half-dollars out of this analysis because they are so rarely seen.
