The probability that 901 coins have total value $100

Here's a cute little problem from Reddit: Tough question for you guys. Let's say you have 901 coins that come out to exactly $100. What are the odds? (Also here.)

Everyone there who gets a solution is assuming that all the possible coins are equally likely, which isn't a reasonable assumption. Years ago I looked at the density of money, where I used a model in which I get back from each transaction n cents with probability 0.01, for n = 0, 1, ... 99; furthermore I always get back the smallest possible number of coins. The only coins allowed are pennies, nickels, dimes, and quarters (worth 1, 5, 10, and 25 cents respectively).

As I calculated before, if I! make 100 transactions, and I get each number of cents back exactly once, I'll get 200 pennies, 40 nickels, 80 dimes, and 150 quarters. This is a total of 470 coins, and worth $49.50. Thus the "average coin" is worth 495/47 = 10.53 cents; 901 coins are "on average" worth $94.89. The value $100 isn't that unreasonable.

So consider a jar with 901 coins, which are independent; they each have probability 20/47 of being a penny, 4/47 of being a nickel, 8/47 of being a dime, and 15/47 of being a quarter. The mean value of a coin is 495/47 = 10.53 cents; the variance is 238840/2309 = 108.12 "square cents".

The mean value of 901 coins, then, is 9489 cents; the variance is 93198 "square cents", so the standard deviation is 305 cents. (Everything here is rounded to the nearest integer.)

Invoking the central limit theorem, then, we say that the value of 901 randomly chosen coins is normally distributed with this mean and standard deviation. Th! e probability of having value exactly 10,000 cents is approxim! ated by the probability density function of this variable at 10,000; that's 0.000322, or 1 in 3101.

An exact answer is feasible -- but not worth computing, I'd say, because the error in the central limit theorem is surely much smaller than the error from the fact that this isn't a realistic model of what actually ends up in your change jar.

probability coins