(Untitled)

[zane314]

Recent math problem that's interesting enough to stick in my head but not interesting enough to solve

Comments

this isn't quite accurate, but it should be pretty close under many circumstances

[tortoise]

If I'm thinking about this correctly, you expect to add 2/5 of a nickel, 3/5 of a dime, 1.5 quarters, and 2 pennies every day. So you expect to wait 117.5 days to get 47 nickels. This means you expect to have about 235 pennies and about 176.25 quarters, and about 70.5 dimes. So the likeliest values should be integers somewhere in that range

Re: this isn't quite accurate, but it should be pretty close under many circumstances

[zane314]

Hmm. Yeah, the discreteness of the issue is a problem. Claiming "70.5" dimes satisfies me, though, since it indicates that 70 or 71 are the most likely actual values.

Ah, it becomes certain because I'm at most getting one nickel a day in that scenario. That's an interesting twist to it, I like that part. Amusingly, you're also guaranteed _not_ to have two dimes on each of those days in that case (although I think you can still have the normal distribution of the others, yeah?).

The way I'd been thinking about handling the standard deviation stuff was the Monte Carlo method, but it just wasn't ever worth writing up. Given that the values of expected coins are ratios, everything should just scale with the initial value, so the "x" nickels isn't that hard.

Re: this isn't quite accurate, but it should be pretty close under many circumstances

[tortoise]

Yeah, pennies and quarters are both independent, but there's a slight negative correlation between nickels and dimes that you'd have to take into account if you were thinking about standard deviations. (Strictly speaking, it also affects the mean, but only to second order, by the linearity of expectation.)

[dragonmudd]

"The value is any number 0-99 cents, with equal probability of any of them."

I just want to point out that this is actually not true in the real world.

[zane314]

Yeah, but it came about from seeing the huge pile of change I had and saying to myself "How far off would my value of dimes from expected have to be to convince me that the values I got were not random?" (admittedly knowing they weren't random already)

From my dad, the statistician

[pktechgirl]

The mean number of dimes is 84. To see this, let's look at the numbers between 0 and 24 (the numbers larger than that are one of those numbers plus one, two or three quarters, which we don't care about for the first question