AMC 10 · 2019 · #21

Grade 7 probability

probability-basicsequences-geometricpattern-recognitionsystematic-enumeration pattern-recognitionsystematic-enumeration ↑ Prerequisites: probability-basicsequences-geometric

📏 Medium solution 💡 3 insights

Problem

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？

$\textbf{(A) } \frac{1}{36} \qquad \textbf{(B) } \frac{1}{24} \qquad \textbf{(C) } \frac{1}{18} \qquad \textbf{(D) } \frac{1}{12} \qquad \textbf{(E) } \frac{1}{6}$

Pick an answer.

(A)

$\frac{1}{36}$

(B)

$\frac{1}{24}$

(C)

$\frac{1}{18}$

(D)

$\frac{1}{12}$

(E)

$\frac{1}{6}$

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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

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

#3 Eliminate Possibilities 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.

$$\text{first flip} = T$$

💡 An opening H locks us out of the target event, so the run must begin with T.

#3 Eliminate Possibilities 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.

$$\text{flips 1-2} = TH$$

💡 TT after the first T ends the game the wrong way, so the second flip is forced to H.

#3 Eliminate Possibilities 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.

$$\text{flips 1-3} = THT$$

💡 Ending with HH on flip 3 gives only 1 T total, so the second T never showed up first — must keep flipping.

#2 Make a Systematic List 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).

$$\text{shortest winning sequence} = THTHH$$

💡 The shortest winner has 5 flips: $THTHH$.

#5 Look for a Pattern 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 $(1/2)^{2k+3}$.

$$P(T(HT)^k HH) = \left(\tfrac{1}{2}\right)^{2k+3} = \tfrac{1}{8} \cdot \left(\tfrac{1}{4}\right)^k$$

💡 Each fixed flip sequence has probability $1/2$ per flip; multiply for the whole string.

#9 Solve an Easier Related Problem 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 $= \tfrac{a}{1 - r} = \tfrac{1/32}{3/4} = \tfrac{1}{32} \cdot \tfrac{4}{3} = \tfrac{1}{24}$.

$$P = \sum_{k=1}^{\infty} \tfrac{1}{8} \left(\tfrac{1}{4}\right)^k = \tfrac{1/32}{1 - 1/4} = \tfrac{1}{24}$$

💡 Reduce infinite probability sum to a single geometric-series formula.

#3 Eliminate Possibilities 7.SP.C.7 Step 7

The answer is (B) $\tfrac{1}{24}$.

$$\boxed{\tfrac{1}{24}}$$

💡 Match the sum to one of the answer choices.

[1] #3 7.SP.C.7 Eliminate impossible openings. If flip 1 is H: flip 2 is H (game ends HH but onl

[2] #3 7.SP.C.7 After the opening T, flip 2 cannot be T (that would end the game as TT immediate

[3] #3 7.SP.C.7 From state TH, the game continues. Flip 3 is either H (game ends with HH — winni

[4] #2 7.SP.C.7 From state THT, flip 4 cannot be T (that's TT, game over wrong way). So flip 4 =

[5] #5 7.SP.C.8 Pattern: every winning sequence is $T(HT)^k HH$ for some $k \ge 1$. Length $= 1

[6] #9 7.RP.A.3 Sum the geometric series over $k = 1, 2, 3, \ldots$. First term $a = \tfrac{1}{8

[7] #3 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.7 Develop 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.8 Find 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.3 Use 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 $\tfrac{1}{24}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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