## Odd odds

Last night I paid for some goods in cash. They cost $17.83. I gave a $20, then dug into my pocket for change to see if I could unload some of it. I happened to have exactly 83 cents in my pocket.

I’m wondering what the odds are of doing that. That is: for any random transaction, what are the odds you have exactly the right amount of coins on you to exactly match the fractional portion (the cents) of the price? I’m not a statistician, but this seems like a hard thing to figure out. I’m not even sure what information you’d actually need to solve it. In the first place, over all transactions are the 100 possible values for the fractional portion evenly distributed? How about the number of coins you are likely to have in your pocket, much less their denominations?

A possibly related, but probably easier problem: if you wanted to be able to always pay exact change in US currency, you need to carry four pennies, a nickel, two dimes and three quarters. If you just grabbed a gob of *x* coins from a jar, how big does *x* need to be give you a 95% chance of having at least this combination?

1sama Says:

October 11th, 2006 at 1:34 pm

let’s see, I would figure you should be able to always have exact change by having just $0.99, whereas you have over $1.

I would use two quarters, three dimes, three nickles and 4 pennies.

Now I would attempt to answer your actual question, but that would take too much time.