AMC 10 · 2019 · #21

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

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

📏 Medium solution 💡 3 insights

Problem

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?

Pick an answer.

(A)

$\frac{1}{36}$

(B)

$\frac{1}{24}$

(C)

$\frac{1}{18}$

(D)

$\frac{1}{12}$

(E)

$\frac{1}{6}$

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

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.

1STEP 1

If flip 1 is H, the target event becomes impossible either way, so the run must open with T.

first flip = T

2STEP 2

A second T here would end the game as TT with no HH, so flip 2 must be H — the state is now TH.

flips 1-2 = TH

3STEP 3

Ending HH on flip 3 leaves only one T, failing the second-T-first clue, so flip 3 must be T — state THT.

flips 1-3 = THT

4STEP 4

Flip 4 must be H (TT loses), giving THTH; then flip 5 = H wins with two T's and two H's, so the shortest winner is THTHH.

shortest winning sequence = THTHH

5STEP 5

Every winner is T(HT)^k HH for k ≥ 1, a fixed string of length 2k+3 with probability $\frac{1}{8}$ ·( $\frac{1}{4}$ )^k.

P(T(HT)^k HH) = (

\frac{1}{2}

)²k+3 =

\frac{1}{8}

· (

\frac{1}{4}

)^k

6STEP 6

Sum the geometric series: first term $\frac{1}{32}$ , ratio $\frac{1}{4}$ , total $\frac{\frac{1}{32}}{\frac{3}{4}}$ = $\frac{1}{24}$ .

P = Σ_k=1^∞

\frac{1}{8}

(

\frac{1}{4}

)^k =

\frac{\frac{1}{32}}{1 - \frac{1}{4}}

\frac{1}{24}

7STEP 7

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

\frac{1}{24}

Answer

\frac{1}{24}

Sanity check the shortest winner THTHH: probability (

\frac{1}{2}

)⁵ =

\frac{1}{32}

≈ 0.0313. The full target probability

\frac{1}{24}

≈ 0.0417 is only slightly larger than this dominant term — exactly what we expect, since the longer winning strings (THTHTHH at

\frac{1}{128}

, THTHTHTHH at

\frac{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

\frac{1}{4}

. Our target is one of the 4 corners of the 'HH-ends' half — about

\frac{1}{4}

· (fraction with second-T-first) — which is small but plausibly

\frac{1}{24}

💡Key takeaway

This AMC 10 problem only needs Grade 7 probability — list the few winning patterns (THTHH, THTHTHH, ...), notice each is $\frac{1}{4}$ as likely as the last, and add up the infinite list to get $\frac{1}{24}$ .

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