AMC 10 · 2019 · #21
Grade 7 probabilityProblem
Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head？
Pick an answer.
AMC 10 2019 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: Debra flips a fair coin until she gets either two heads in a row (HH) or two tails in a row (TT). What is the probability that the run ends with HH but along the way she has already seen a second T before she ever saw a second H?
Givens: Fair coin: each flip is H or T with probability $\tfrac{1}{2}$, independent; Stopping rule: stop the first time we see HH or TT; Target event: the run ends with HH, and the second T appeared before the second H; Answer choices: (A) $\tfrac{1}{36}$, (B) $\tfrac{1}{24}$, (C) $\tfrac{1}{18}$, (D) $\tfrac{1}{12}$, (E) $\tfrac{1}{6}$
Unknowns: The probability of the target event
Understand
Restated: Debra flips a fair coin until she gets either two heads in a row (HH) or two tails in a row (TT). What is the probability that the run ends with HH but along the way she has already seen a second T before she ever saw a second H?
Givens: Fair coin: each flip is H or T with probability $\tfrac{1}{2}$, independent; Stopping rule: stop the first time we see HH or TT; Target event: the run ends with HH, and the second T appeared before the second H; Answer choices: (A) $\tfrac{1}{36}$, (B) $\tfrac{1}{24}$, (C) $\tfrac{1}{18}$, (D) $\tfrac{1}{12}$, (E) $\tfrac{1}{6}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #9 Solve an Easier Related Problem, #5 Look for a Pattern, #3 Eliminate Possibilities
Tool #2 (Systematic List): list the very short winning sequences first (THH, THTHH, THTHTHH, ...) — they form an obvious family. Tool #3 (Eliminate): the first flip cannot be H (any opening H either ends with HH on flip 2 with no second T yet, or forces a TT loss before HH). Tool #5 (Pattern): each winning sequence is T followed by some number of HT pairs followed by HH — a geometric pattern. Tool #9 (Easier Problem): the resulting infinite probability sum is a single geometric series, easy to total.
Execute — Answer: B
7.SP.C.7 Step 1 - Eliminate impossible openings.
- If flip 1 is H: flip 2 is H (game ends HH but only 1 head total before HH — the 'second T before second H' clause is impossible) or flip 2 is T (now we have HT and the game continues, but the only way for HH to happen later requires another H first; however before two H's can occur in a row, a T might appear, and tracking shows the game ends with TT before HH in the relevant subcases — bottom line, no winning string starts with H).
- So flip 1 is T.
💡 An opening H locks us out of the target event, so the run must begin with T.
7.SP.C.7 Step 2 - After the opening T, flip 2 cannot be T (that would end the game as TT immediately, no HH ever — and we still need HH to happen).
- So flip 2 = H.
- Now we have TH.
💡 TT after the first T ends the game the wrong way, so the second flip is forced to H.
7.SP.C.7 Step 3 - From state TH, the game continues.
- Flip 3 is either H (game ends with HH — winning, with one T and two H's, satisfying 'second T appears before second H' vacuously?
- No: we need the second T to appear before the second H, but here there is no second T at all).
- So flip 3 = H actually FAILS the 'second T before second H' condition.
- So flip 3 must be T.
- Now we have THT.
💡 Ending with HH on flip 3 gives only 1 T total, so the second T never showed up first — must keep flipping.
7.SP.C.7 Step 4 - From state THT, flip 4 cannot be T (that's TT, game over wrong way).
- So flip 4 = H.
- Now we have THTH.
- From here, flip 5 can be H (game ends with HH — winning, and we now have 2 T's and 2 H's, with the second T appearing on flip 3 and the second H on flip 5, so 'second T before second H' holds) OR flip 5 = T (we have THTHT, now flip 6 must be H, then we are at THTHTH and the same choice repeats).
💡 The shortest winner has 5 flips: $THTHH$.
7.SP.C.8 Step 5 - Pattern: every winning sequence is $T(HT)^k HH$ for some $k \ge 1$.
- Length $= 1 + 2k + 2 = 2k + 3$.
- Each specific sequence has probability $(1/2)^{2k+3}$.
💡 Each fixed flip sequence has probability $1/2$ per flip; multiply for the whole string.
7.RP.A.3 Step 6 - Sum the geometric series over $k = 1, 2, 3, \ldots$.
- First term $a = \tfrac{1}{8} \cdot \tfrac{1}{4} = \tfrac{1}{32}$ (the $k=1$ string $THTHH$).
- Common ratio $r = \tfrac{1}{4}$.
- Sum $= \tfrac{a}{1 - r} = \tfrac{1/32}{3/4} = \tfrac{1}{32} \cdot \tfrac{4}{3} = \tfrac{1}{24}$.
💡 Reduce infinite probability sum to a single geometric-series formula.
7.SP.C.7 Step 7 The answer is (B) $\tfrac{1}{24}$.
💡 Match the sum to one of the answer choices.
7.SP.C.7 Eliminate impossible openings. If flip 1 is H: flip 2 is H (game ends HH but onl 7.SP.C.7 After the opening T, flip 2 cannot be T (that would end the game as TT immediate 7.SP.C.7 From state TH, the game continues. Flip 3 is either H (game ends with HH — winni 7.SP.C.7 From state THT, flip 4 cannot be T (that's TT, game over wrong way). So flip 4 = 7.SP.C.8 Pattern: every winning sequence is $T(HT)^k HH$ for some $k \ge 1$. Length $= 1 7.RP.A.3 Sum the geometric series over $k = 1, 2, 3, \ldots$. First term $a = \tfrac{1}{8 7.SP.C.7 The answer is (B) $\tfrac{1}{24}$. Review
Reasonableness: Sanity check the shortest winner $THTHH$: probability $(1/2)^5 = 1/32 \approx 0.0313$. The full target probability $1/24 \approx 0.0417$ is only slightly larger than this dominant term — exactly what we expect, since the longer winning strings ($THTHTHH$ at $1/128$, $THTHTHTHH$ at $1/512$, ...) contribute small geometric corrections. Also: the four 'second-event' AMC clauses (2 heads in a row, 2 tails in a row, second tail first, second head first) partition the sample space into 4 symmetric-by-relabeling cases plus a 'still running' tail of probability 0, so each of the 4 outcomes has total probability roughly $1/4$. Our target is one of the 4 corners of the 'HH-ends' half — about $1/4 \cdot \text{(fraction with second-T-first)}$ — which is small but plausibly $\tfrac{1}{24}$.
Alternative: Tool #11 (Work Backwards) with a state machine: label states $S_0$ (start), $S_T$ (last flip T, no streak yet), $S_H$ (last flip H, no streak yet), $S_{TT}, S_{HH}$ (absorbing), tracking also whether the second T has already been seen. From $S_0$, transition to $S_T$ or $S_H$ with prob $\tfrac{1}{2}$ each. Set up the probability of reaching 'win' from each state and solve. This recovers $\tfrac{1}{24}$ algebraically without summing a series.
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Modeling each fair coin flip as an independent $\tfrac{1}{2}$ event and tracking which sequences satisfy the target.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Computing the probability of each specific winning sequence by multiplying $\tfrac{1}{2}$ across flips.)7.RP.A.3Use proportional relationships to solve multi-step ratio and percent problems (Summing the infinite geometric series of probabilities $\sum (1/4)^k$ via the ratio $a/(1-r)$.)
⭐ This AMC 10 problem only needs Grade 7 probability — list the few winning patterns ($THTHH$, $THTHTHH$, ...), notice each is $1/4$ as likely as the last, and add up the infinite list to get $\tfrac{1}{24}$.
⭐ This AMC 10 problem only needs Grade 7 probability — list the few winning patterns ($THTHH$, $THTHTHH$, ...), notice each is $1/4$ as likely as the last, and add up the infinite list to get $\tfrac{1}{24}$.
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