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2017 AMC 10A Problem 18

2017 AMC 10A Problem 18AMC 10 Step by Step

Name: Date: September 13, 2026

Conditional Probability & States · difficulty 3 of 5 · about 4 min · position P16-20

Amelia has a coin that lands heads with probability $\tfrac{1}{3}$ , and Blaine has a coin that lands on heads with probability $\tfrac{2}{5}$ . Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $q-p$ ?

A)
$1$
B)
$2$
C)
$3$
D)
$4$
E)
$5$

Answer

Key insight

If both first tosses fail (probability 2/5), the game restarts from the same state, so p = 1/3 + (2/5) p, giving p = 5/9.

Solution

Let $P$ be the probability that Amelia wins from the start of a round (Amelia about to toss).

In the first round, Amelia wins immediately with probability $\tfrac13$ . Otherwise (probability $\tfrac23$ ) Blaine tosses; if he gets heads (probability $\tfrac25$ ) Amelia loses, and if he gets tails (probability $\tfrac35$ ) the game is back exactly where it started, so Amelia wins from there with probability $P$ again.

Therefore
$P = \frac13 + \frac23 \cdot \frac35 \cdot P = \frac13 + \frac25 P.$
Solving, $\tfrac35 P = \tfrac13$ , so $P = \tfrac59$ .

Thus $p = 5$ , $q = 9$ , and $q - p = 4$ .

The answer is $\boxed{\textbf{(D)}\ 4}$ .

Why this works

"Repeat until someone succeeds" processes are memoryless: after a full round of failures the situation is identical to the start. That lets you write one equation in the unknown probability instead of summing an infinite series. The same trick works whenever a random process can return to its initial state.

Alternative approach

Sum the series directly. Amelia wins on round $n$ if the first $n-1$ rounds are double failures (probability $\left(\tfrac25\right)^{n-1}$ ) and she then flips heads: total $\tfrac13\left(1 + \tfrac25 + \tfrac{4}{25} + \cdots\right) = \tfrac13 \cdot \frac{1}{1 - 2/5} = \tfrac59$ .

The trap

Stopping after one round and answering 1/3, or computing Blaine's winning probability 4/9 and reporting 9 - 4 = 5.

Common mistakes

Stopping after one round and answering 1/3, or computing Blaine's winning probability 4/9 and reporting 9 - 4 = 5.
Using $\tfrac23 \cdot \tfrac25$ (Blaine wins) as the restart probability instead of $\tfrac23 \cdot \tfrac35$ (both fail).

Techniques

Define states/recurrence and iterate