AMC 10 · 2024 · #17

Grade 8 probabilityalgebra

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicsystematic-enumerationlinear-equations-one-var systematic-enumerationcaseworkconvert-to-algebra ↑ Prerequisites: probability-basicfraction-arithmetic

📏 Medium solution 💡 3 insights

Problem

In a best-of-three playoff, Team A hosts game 1 (win probability

\frac{2}{3}

) and Team B hosts games 2 and 3 (Team A's away win probability p). Game outcomes are independent and the playoff ends as soon as a team wins two games. Given that Team A's overall chance of winning the playoff is

\frac{1}{2}

, and p can be written as

\frac{1}{2}

(m - √(n)) for positive integers m, n, find m + n.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make an Organized List

A best-of-three has very few outcomes, so Tool #2 (Make an Organized List) is the right opening move: write down every game-by-game sequence in which A finishes with 2 wins before B does, and stop. The list is short enough to enumerate by hand — exactly three patterns: AA, ABA, BAA. Once each sequence is given a probability in terms of p, Tool #13 (Convert to Algebra) takes over: add the three case probabilities, set the sum equal to $\frac{1}{2}$ , and solve the resulting quadratic in p. The quadratic has two roots; Tool #3 (Eliminate Possibilities) discards the one that lies outside [0,1], leaving a unique probability p = $\frac{1}{2}$ (4 - √(10)) that matches the requested form.

1STEP 1

The series stops at a team's second win, so Team A finishes in exactly three mutually exclusive orders: AA, ABA, BAA.

Winning sequences for A: {AA, ABA, BAA}

2STEP 2

Game 1 is home ( $\frac{2}{3}$ ), games 2 and 3 are away (p); independence multiplies them into 2p/3, 2p(1-p)/3, p²/3.

P(AA) =

\frac{2}{3}

· p = 2p/3, P(ABA) =

\frac{2}{3}

·(1-p)· p = (2p(1-p))/3, P(BAA) =

\frac{1}{3}

· p · p = p²/3

3STEP 3

Sum the three cases, set the total equal to $\frac{1}{2}$ , and clear the denominators to get 2p² - 8p + 3 = 0.

2p/3 + (2p(1-p))/3 + p²/3 =

\frac{1}{2}

→ 4p + 4p - 4p² + 2p² = 3 → 2p² - 8p + 3 = 0

4STEP 4

Apply the quadratic formula; √40 = 2√10 collapses the radical to give p = (4 ± √(10))/2.

p = (8 ± √(64 - 24))/4 = (8 ± √(40))/4 = (8 ± 2√(10))/4 = (4 ± √(10))/2

5STEP 5

The plus root exceeds 1, so only p = (4 - √(10))/2 is a valid probability, giving m = 4, n = 10.

p = (4 - √(10))/2 =

\frac{1}{2}

(4 - √(10)) → m = 4, n = 10

6STEP 6

Matching p to $\frac{1}{2}$ (m - √(n)) gives m = 4 and n = 10, so m + n = 14.

m + n = 4 + 10 = 14 → (E)

Answer

Plug p = (4 - √(10))/2 back into the case sum. Numerically p ≈ 0.4189, so P(AA) ≈

\frac{2}{3}

(0.4189) ≈ 0.2793, P(ABA) ≈

\frac{2}{3}

(0.4189)(0.5811) ≈ 0.1623, and P(BAA) ≈

\frac{1}{3}

(0.4189)² ≈ 0.0585. Their sum is 0.2793 + 0.1623 + 0.0585 ≈ 0.5001 ≈

\frac{1}{2}

. The condition holds. Sign check:

\frac{2}{3}

＞

\frac{1}{2}

, so Team A's home edge alone beats 50%, meaning the away probability p has to be noticeably less than

\frac{1}{2}

to drag the overall chance back down to

\frac{1}{2}

— and 0.42 fits that picture. Answer (E) 14 is consistent.

💡Key takeaway

A short series has a short list of outcomes — write them all down, add their probabilities, and the unknown probability falls out of one quadratic equation. The [0,1] rule for probability picks the right root.

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