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What are the odds that there's exactly €10 in a jar, filled with 1 cent coins?
Hi,
Like everyone, I have a large jar full of loose change in my house. So today I tipped all the coins out and started counting them. The jar was full of all the different denominations of coins (i.e, 1c, 2c, 5c, 10c, 20c, 50c, €1, €2).
Just by guessing I'd say the jar was about 1/3 full of 1 cent coins (with all the other denominations making up the other 2/3).
After counting them it turned out I had exactly €10.01 in 1 cent coins (i.e, I had exactly 1001 one cent coins in the jar).
My question is:
Is it possible to calculate the odds of having exactly €10, in 1 cent coins, in a jar filled with all the denominations of coins?
Or it the only way of determining the amount of one particular coin by guessing?
The volume of a 1 cent is roughly = 3.46 x 10^-7 m^3,
The volume of the jar is = 7.03×10^-4 m^3
EDIT: I don't know if this is relevant but I thought I should add it:
All the coins are at arbitrary angles to each other but as they are all thin, flat disks I'd guess they have a fairly high packing efficiency... My guess would be somewhere in the order of 90%.
2 Answers
- SqdancefanLv 76 years agoFavorite Answer
The volume of the jar is about 703/0.346 ≈ 2030 times the volume of a 1-cent coin. The best packing factor of coins in a plane is π/(2√3), about 0.907, so your packing factor is unlikely to be as high as 90%. Coins packed in a rectilinear array have a packing factor of π/4, about 0.785. If you assume the volumetric packing factor is about 85%, then the jar may hold about €17 in 1 cent coins. It seems very unlikely that your 1/3 full jar will hold more than €6 in 1-cent coins.
- Barry GLv 76 years ago
No.
I think you need to define the conditions of your question more precisely. If the only coins in the jar are 1c coins, you can estimate how many there are using the information you have given. If there are other coins of different denominations as well as 1c, and these are different sizes, but all you know is the total volume of the coins, it becomes a very difficult problem. Firstly because if the coins are different sizes the packing fraction will be different for each ratio of denominations. Secondly, even if you know the total number of coins (or total weight, and the weight of each denomination), but not the numbers of each denomination, it is still a very difficult problem, but it can be solved.
It would be much easier just to count the number of coins in each denomination, as you did. The fact that it came out close to 10 euro is just a coincidence. If it were close of 20 euros, or 5 euros, or if it were 365 cents (days in the year) or 1815 cents (Battle of Waterloo) or 1234 cents, then you might be just as surprised.
