Average amount of money received from pulling three random coins

Question

Let's say I have a bag of coins, which contains $1$ quarter, $2$ dimes, $3$ nickles and $4$ pennies. If I were to randomly pull out $3$ coins, on average, how much money would I get?

Answer 1

Each coin is pulled 30% of the time. So on average you will get 30% of 64 cents=19.2 cents

Answer 2

Ross Millikan's answer is entirely correct, but perhaps it deserves slightly more elaboration.

The expected amount of money you receive is the sum of the expected amounts of money that each coin gives you, which is the value of the coin times the probability that you draw it. (Linearity of expectation!) The probability that you draw each coin is the same: it is one minus the probability that you do not draw it, or $1 - \frac{ {9 \choose 3} }{ {10 \choose 3} } = \frac{3}{10}$. So the expected amount of money you receive is $\frac{3}{10}$ of the total value of the coins.

Note that I am basically treating coins with the same value as distinguishable (pretend they have the same value but look different; it can't affect the answer to the question). The question of distinguishability is a red herring.

Answer 3

Ross Millikan has already given a very elegant solution, but if you're still in doubt, this problem is small enough that you can easily verify it by letting a computer enumerate all the possibilities. Here's a Python program to do it:

coins = [25, 10, 10, 5, 5, 5, 1, 1, 1, 1]
total = count = 0
for k in range(10):
    for l in range(10):
        for m in range(10):
            if k!=l and k!=m and l!=m:
                total += coins[k] + coins[l] + coins[m]
                count += 1
print "Got", total, "cents in", count, "draws; on average", total*1.0/count, "cents."

The answer comes instantly: "Got 13824 cents in 720 draws; on average 19.2 cents."