AMC 10 · 2005 · #18

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count

conditional-probabilityprobability-basicsystematic-enumeration systematic-enumerationcasework ↑ Prerequisites: probability-basicconditional-probabilitycombinations-basic

📏 Medium solution 💡 3 insights

Problem

Teams A and B play a best-of-five series: the first team to win 3 games takes the series. Every game is a 50-50 coin flip, independent of the others. You are told two facts about how one particular series turned out: team B won game 2, and team A won the series. Given only that, find the probability that team B also won game 1.

Pick an answer.

(A)

$\frac{1}{5}$

(B)

$\frac{1}{4}$

(C)

$\frac{1}{3}$

(D)

$\frac{1}{2}$

(E)

$\frac{2}{3}$

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

A conditional probability is just (chance of the target-and-condition) divided by (chance of the condition), so the whole job is measuring two chances built from game sequences — Tool #2 (Make a Systematic List) is the engine for laying those sequences out. The trap is that finished series have different lengths, so they are not equally likely; Tool #15 (Organize Information in More Ways) fixes this by re-imagining every series as a full 5 games, which makes all 32 outcomes equally likely and lets us count instead of juggle unequal weights. Tool #7 (Identify Subproblems) then splits the count into the two pieces a conditional probability needs: how many equally-likely outcomes meet the condition, and how many of those also have B winning game 1.

1STEP 1

Make every outcome equally likely

Pretend they always play all 5 games even after the series ends: all 32 A/B strings are equally likely, and A wins by taking 3 or more.

2⁵ = 32 equally likely 5-game strings, each 1/32

2STEP 2

Count outcomes that meet the condition

Game 2 is a B, so A must take 3 or more of games 1,3,4,5: C(4,3)+C(4,4)=5 equally likely outcomes meet the condition.

C(4, 3)+C(4, 4)=4+1=5 outcomes

3STEP 3

Count how many also have B winning game 1

If game 1 is a B too, A must sweep games 3,4,5 — only BBAAA survives, so just 1 of those 5 outcomes has B winning game 1.

B B A A A → C(3, 3)=1 outcome

4STEP 4

Form the conditional probability

All these outcomes are equally likely, so the conditional probability is favorable over qualifying: 1/5, which is choice (A).

P(B won game 1 ∣ condition) = 1/5 → (A)

Answer

1/5

The answer 1/5 is small, which fits: B winning game 1 (on top of game 2) puts A in a deep 0-2 hole, and A climbing out of that to win the series is a rare way for the condition to be satisfied, so it should be the least likely of the qualifying stories. A check by actual probabilities agrees: the single 0-2 comeback BBAAA needs all 5 games, chance 1/32; the other qualifying series (where A won game 1) total 4/32 of the chance, so the condition has probability 5/32 and the answer is 1/32/5/32=1/5. The equal-weight count and the true-probability computation match. The trap answer 1/4 comes from listing the four distinct finished series (BBAAA, ABAA, ABABA, ABBAA) and treating them as equally likely — but the 4-game series ABAA is twice as likely as a specific 5-game one, which the padding correctly accounts for.

💡Key takeaway

When a series can end in different numbers of games, imagine they always play the full five games so every outcome is equally likely — then a conditional probability is just favorable outcomes over the outcomes that fit the condition.

Make every outcome equally likely
Count outcomes that meet the condition
Count how many also have B winning game 1
Form the conditional probability

Make every outcome equally likely

Count outcomes that meet the condition

Count how many also have B winning game 1

Form the conditional probability

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