2015 AMC 10B Problem 18

Look at one coin. It finishes as heads unless it shows tails on the first toss, tails again on the second, and tails again on the third. Those three tosses are independent, so
$$P(\text{ends tails}) = \frac12 \cdot \frac12 \cdot \frac12 = \frac18, \qquad P(\text{ends heads}) = 1 - \frac18 = \frac78.$$

The expected number of heads is the sum over the $64$ coins of the probability that each is heads (the coins need not be independent for this to hold, though they are):
$$64 \cdot \frac78 = 56.$$

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

Linearity of expectation turns "expected number of coins with a property" into $n$ times the probability one coin has it, with no need to consider the joint distribution. The complement is easier than the event itself: "heads at some point in up to three tries" is the opposite of "tails, tails, tails." Any process that retries on failure a fixed number of times has success probability $1 - p^k$.

Track the expected counts round by round: about $32$ heads after toss one, $16$ more after the $32$ tails are retossed, then $8$ more from the last $16$. Total $32 + 16 + 8 = 56$.

• Stopping after two tosses (probability 3/4 heads, giving 48) or adding 1/2 + 1/4 + 1/8 without recognizing it equals 7/8.
• Retossing the heads as well as the tails, or thinking that a coin flipped three times has a $\frac12$ chance of heads overall, giving $32$.