Here’s a question that plagues my life, and if anyone can solve it mathematically, props to you.

Truth is, I don’t like change. Coinage, I mean. As a male, I have no place to keep loose coins, unless they jangle in my pockets all day (and fall out) or pudge out my wallet. So I start out every day without change in my pockets, and I end each day depositing the accumulated coins into my coin jar.

My coin jar now has $101.65 in pennies, nickels, dimes, and quarters. (Incidentally, I also lost $0.44 due to change falling out of my pocket in 2007.)

So here’s the puzzle:

What is the optimum coin set to begin every day with, such that if I make a single transaction that day, I will on average have the fewest coins with me? For easy math, assume that I make the transaction in the middle of the day (giving equal weight to the before and after transaction periods).

For example, I always start out with 0 coins. But if I make a transaction that costs me $X.01, I’ll have 9 coins for the rest of the day (3 quarters, 2 dimes, 4 pennies). So I average (0+9)/2 = 4.5 coins for that situation. Of course, to find the true average for the day, you’d have to consider all 100 possible transactions ($X.00 through $X.99 in change). You can assume equal probability of each situation, for simpler math.

You can also assume that I’ll pay intelligently to yield the least coins in change. For example, if the cost is $X.23 and I have a quarter and 3 pennies, I’ll pay all $0.28 so as to receive back one coin, a nickel. (Whereas if I paid stupidly the quarter alone, I would get back $0.02 and now have 5 pennies.)

A compounding challenging aspect of this puzzle is that my starting coin set might not have to be optimal. For example, starting with 6 pennies might yield a better set of change than starting with a nickel and 1 penny, because of the pennies’ flexibility to prevent extra coins in return (not sure if this case is true, it’s just an illustration). So there are far more than 100 initial coin sets to consider.

So, what is the optimal coin set to start out with, so that (considering all 100 possible transactions using that starting amount) I will have the least number of coins in my pocket on average throughout the day?

No, I don’t know the answer to this question. It’s way too complex for me to figure out.

[Solution posted here: Loose Change Solution And Loose Change Part II]