AMC 10 · 2005 · #18
Grade 7 probabilityTeam A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Teams $A$ and $B$ play a best-of-five series (first to $3$ game wins takes the series). Every game is a $50\text{-}50$ 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 $1$.
Givens: First team to reach $3$ game wins takes the series (at most $5$ games); Each single game is won by $A$ or $B$ with probability $\tfrac{1}{2}$ each, games independent; We are told (conditioned on) two things: $B$ won game $2$, and $A$ won the whole series; Answer choices: (A) $\tfrac{1}{5}$, (B) $\tfrac{1}{4}$, (C) $\tfrac{1}{3}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{2}{3}$
Unknowns: The probability that team $B$ won game $1$, given that $B$ won game $2$ and $A$ won the series
Understand
Restated: Teams $A$ and $B$ play a best-of-five series (first to $3$ game wins takes the series). Every game is a $50\text{-}50$ 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 $1$.
Givens: First team to reach $3$ game wins takes the series (at most $5$ games); Each single game is won by $A$ or $B$ with probability $\tfrac{1}{2}$ each, games independent; We are told (conditioned on) two things: $B$ won game $2$, and $A$ won the whole series; Answer choices: (A) $\tfrac{1}{5}$, (B) $\tfrac{1}{4}$, (C) $\tfrac{1}{3}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{2}{3}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #15 Organize Information in More Ways, #7 Identify Subproblems
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 $1$.
Execute — Answer: A
7.SP.C.7 Step 1 Make every outcome equally likely
- A best-of-five series can end in $4$ or $5$ games, and a $4$-game series is more probable than any single specific $5$-game one, so the finished series are not equally likely — counting them directly would be wrong.
- Fix this by pretending the two teams always play all $5$ games, even after the series is decided (the extra games are just coin flips that do not change who already won).
- Now every outcome is a string of $5$ letters, each $A$ or $B$, and each of the $2^5=32$ strings has the same probability $\tfrac{1}{32}$.
- In this full-length picture, $A$ wins the series exactly when $A$ wins at least $3$ of the $5$ games.
💡 Padding every series to a fixed $5$ games makes all outcomes equally likely, so we can count instead of weigh.
7.SP.C.8 Step 2 Count outcomes that meet the condition
- Now impose the condition: game $2$ is a $B$, and $A$ wins the series (i.e.
- $A$ wins at least $3$ of the $5$ games).
- Game $2$ is a $B$, so it is not one of $A$'s wins.
- That means $A$ must collect its $3$-or-more wins from the other four games — games $1,3,4,5$.
- Count the strings where $A$ wins at least $3$ of those four games: exactly $3$ of $4$ gives $\binom{4}{3}=4$ strings, and all $4$ of $4$ gives $\binom{4}{4}=1$ string.
- That is $4+1=5$ equally likely outcomes in which $B$ won game $2$ and $A$ won the series.
💡 With game $2$ fixed as a $B$, $A$ must win $3$ or more of the remaining four games, so just count those winning strings.
7.SP.C.8 Step 3 Count how many also have B winning game 1
- Among those $5$ outcomes, which ones also have $B$ winning game $1$?
- If game $1$ is a $B$, then games $1$ and $2$ are both $B$, so neither can be an $A$ win.
- Then $A$ must win at least $3$ games out of only the three games left — games $3,4,5$ — which forces $A$ to win all three.
- That is the single string $B\,B\,A\,A\,A$.
- So exactly $1$ of the $5$ qualifying outcomes has $B$ winning game $1$.
💡 If $B$ takes both of the first two games, the only way $A$ can still win is to sweep the last three.
7.SP.C.8 Step 4 Form the conditional probability
- The conditional probability is the number of outcomes with $B$ winning game $1$ divided by the number of outcomes meeting the condition, since all these outcomes are equally likely.
- That is $\tfrac{1}{5}$.
- So the probability that $B$ won game $1$ is $\tfrac{1}{5}$, which is choice (A).
💡 Favorable equally-likely outcomes over all equally-likely outcomes that meet the condition gives the conditional probability.
7.SP.C.7 A best-of-five series can end in $4$ or $5$ games, and a $4$-game series is more 7.SP.C.8 Now impose the condition: game $2$ is a $B$, and $A$ wins the series (i.e. $A$ w 7.SP.C.8 Among those $5$ outcomes, which ones also have $B$ winning game $1$? If game $1$ 7.SP.C.8 The conditional probability is the number of outcomes with $B$ winning game $1$ Review
Reasonableness: The answer $\tfrac{1}{5}$ is small, which fits: $B$ winning game $1$ (on top of game $2$) puts $A$ in a deep $0\text{-}2$ 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 $0\text{-}2$ comeback $BBAAA$ needs all $5$ games, chance $\tfrac{1}{32}$; the other qualifying series (where $A$ won game $1$) total $\tfrac{4}{32}$ of the chance, so the condition has probability $\tfrac{5}{32}$ and the answer is $\tfrac{1/32}{5/32}=\tfrac{1}{5}$. The equal-weight count and the true-probability computation match. The trap answer $\tfrac{1}{4}$ 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.
Alternative: Skip the padding and weight the real finished series by their true probabilities. If $B$ wins game $1$: the only series where $A$ still wins is $BBAAA$, probability $(\tfrac{1}{2})^5=\tfrac{1}{32}$. If $A$ wins game $1$ (with $B$ still winning game $2$): the $A$-winning series are $ABAA$ with probability $(\tfrac{1}{2})^4=\tfrac{1}{16}=\tfrac{2}{32}$, plus $ABABA$ and $ABBAA$ each $(\tfrac{1}{2})^5=\tfrac{1}{32}$, totaling $\tfrac{4}{32}$. The condition's total probability is $\tfrac{1}{32}+\tfrac{4}{32}=\tfrac{5}{32}$, so the answer is $\dfrac{1/32}{5/32}=\tfrac{1}{5}$ — the same (A).
CCSS standards used (min grade 7)
7.SP.C.7Develop a probability model and use it to find probabilities of events (Recognizing that variable-length series are not equally likely, and rebuilding the model by padding every series to a full $5$ games so all $32$ outcomes carry the equal probability $\tfrac{1}{32}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Counting the $5$ equally-likely outcomes that meet the condition ($B$ wins game $2$, $A$ wins series) and the $1$ that also has $B$ winning game $1$, then forming the ratio $\tfrac{1}{5}$.)
⭐ 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.
⭐ 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.
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