AMC 10 · 2025 · #18

Grade 12 probability

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

expected-valueprobability-basicrecursive-sequence identify-subproblemseasier-related-problem ↑ Prerequisites: probability-basic

📏 Medium solution 💡 3 insights

Problem

Awnik plays the same game over and over. Each game is a win with probability 1/3 and a loss with probability 2/3, and the outcomes are independent. He keeps playing until his record contains at least one win and at least one loss. Find the expected number of games he plays.

Pick an answer.

(A)

$\frac{5}{2}$

(B)

(C)

$\frac{16}{5}$

(D)

$\frac{7}{2}$

(E)

$\frac{15}{4}$

How to solve

Strategy Identify Subproblems

The stopping rule asks for two different things at once, which is what makes a direct count of outcomes messy. Tool #7 (Identify Subproblems) cuts the run at the first game: that game can never finish the job, and once it is played exactly one outcome is still missing, so the rest of the run is a single-target wait. Tool #4 (Introduce a Variable) names the two possible leftover waits, E_W and E_L, so they can be handled symbolically instead of case by case. Tool #9 (Solve an Easier Related Problem) then answers the general question "how many independent trials until the first success of probability p?" once, giving 1/p, which both waits are instances of. Tool #16 (Change Focus / Count the Complement) supplies the independent cross-check in the review: instead of tracking when the run ends, count the chance it has not ended yet and sum those tail probabilities.

1STEP 1

The first game cannot finish it

After game one you wait for the opposite result.

N = 1 + M, where M is the wait for the outcome opposite to game 1

2STEP 2

Name the two possible waits

There are only two waits.

E_W = E[games until the first loss], E_L = E[games until the first win]

3STEP 3

Expected wait for a first success

The wait is one over the probability.

E = 1 + (1 - p)E → pE = 1 → E = 1/p

4STEP 4

Fill in both waits

They are three halves and three.

E_W = 1/2/3 = 3/2, E_L = 1/1/3 = 3

5STEP 5

Weight the two branches

Take the weighted average by the first game.

E[N] = 1 + 1/3E_W + 2/3E_L

6STEP 6

Do the arithmetic

The arithmetic gives seven halves.

E[N] = 1 + 1/3 · 3/2 + 2/3 · 3 = 1 + 1/2 + 2 = 7/2 → (D)

Answer

7/2

Both a win and a loss are required, so N ≥ 2 always and no value at or below 2 could be right; 7/2 = 3.5 clears that floor. A tighter bound comes from the branches themselves: the total is 1 + M where M is either E_W = 3/2 or E_L = 3 on average, so E[N] must lie between 1 + 3/2 = 5/2 and 1 + 3 = 4. The slower branch, waiting for a win, is the one entered with the larger probability 2/3, so the value has to sit nearer 4 than 5/2 — and 7/2 does. The two nearby choices are exactly the classic slips: (A) 5/2 is what you get by assuming he always ends up waiting for a loss, and (B) 3 is the win-wait E_L reported on its own without the opening game or the other branch.

💡Key takeaway

The first game only decides which outcome you are still missing, and an outcome with probability p takes 1/p games on average to show up.

The first game cannot finish it
Name the two possible waits
Expected wait for a first success
Fill in both waits
Weight the two branches
Do the arithmetic