2015 AMC 10B Problem 18

Think about one coin at a time.

1. When does a coin end up tails? Only if it shows tails on the first toss, tails again on the second toss, and tails again on the third toss. Any heads along the way stops the tossing with heads showing.
2. Chance of that. Each toss is tails with probability $\frac12$, so three tails in a row has probability
$$\frac12 \cdot \frac12 \cdot \frac12 = \frac18.$$
3. Chance of heads. Every other outcome ends in heads, so one coin ends heads with probability $1 - \frac18 = \frac78$.
4. All 64 coins. The expected number of heads is the number of coins times the chance for each one:
$$64 \cdot \frac78 = 56.$$

The answer is $\boxed{\textbf{(D)}\ 56}$.

The expected number of coins with some property is just the number of coins times the chance that one coin has it. Also, the chance of heads is easier to get from the opposite event: "never heads in three tries" is the single outcome tails, tails, tails. Any time something is retried a fixed number of times, find the chance that every try fails and subtract it from $1$.

Follow the coins round by round, using the expected halves. On the first toss about $32$ of the $64$ land heads and $32$ land tails. The $32$ tails are tossed again: about $16$ turn to heads. The $16$ still tails are tossed a third time: about $8$ turn to heads. In total $32 + 16 + 8 = 56$.