AMC 10 · 2017 · #17

Grade 7 probability
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
probability-basicindependent-eventsfraction-arithmetic identify-subproblemssystematic-enumeration ↑ Prerequisites: probability-basicindependent-events
📏 Medium solution 💡 2 insights
Problem
A coin lands heads with probability two thirds and tails with probability one third, and every toss is independent. Game A: toss three times and win if all three match. Game B: toss four times and win if the first two match each other and the last two match each other. Compare the two winning probabilities: say which game is more likely to be won, and by exactly how much.

Pick an answer.

(A)
The probability of winning Game A is } \frac{4}{81} \text{ less than the probability of winning Game B.
(B)
The probability of winning Game A is } \frac{2}{81} \text{ less than the probability of winning Game B.
(C)
The probabilities are the same.
(D)
The probability of winning Game A is } \frac{2}{81} \text{ greater than the probability of winning Game B.
(E)
The probability of winning Game A is } \frac{4}{81} \text{ greater than the probability of winning Game B.
How to solve
Strategy Identify Subproblems

Tool #7 (Identify Subproblems): the question compares two numbers, so I compute each game's probability on its own first and only compare at the end. Game B splits further into two separate subproblems, because tosses 1–2 and tosses 3–4 share no tosses and therefore do not affect each other. Tool #2 (Make a Systematic List): each game wins on a short, listable set of sequences, so I can write every winning sequence down and be sure none is missed or double-counted — and I can re-list Game B's winners a second way as a check. Tool #15 (Organize Information in More Ways): rewriting both answers over the same denominator 81 turns the comparison into subtracting two whole numbers, where the direction is impossible to misread. Tool #3 (Eliminate Possibilities): the five choices pair up as the same gap in opposite directions, so I must pin down the sign of the difference, not just its size, before naming a choice.

1STEP 1

List Game A's winning sequences

Game A wins only on all heads or all tails.

P(A) = (2/3)³ + (1/3)³ = 8/27+1/27 = 9/27 = 1/3
2STEP 2

Split Game B into two pairs

Game B splits into two independent pairs.

P(pair matches) = 4/9+1/9 = 5/9, P(B) = (5/9)² = 25/81
3STEP 3

Re-list Game B a second way

Listing the four cases gives the same value.

16/81+4/81+4/81+1/81 = 25/81
4STEP 4

Rewrite both over 81

Over a denominator of eighty-one the comparison is immediate.

P(A) - P(B) = 27/81 - 25/81 = +2/81
5STEP 5

Match the sign to a choice

Game A wins by two eighty-firsts.

P(A) = 27/81 ＞ 25/81 = P(B) → (D)
Answer
The probability of winning Game A is 2/81 greater than the probability of winning Game B.
As decimals, P(A) = 1/3 ≈ 0.3333 and P(B) = 25/81 ≈ 0.3086, so both sit sensibly between 0 and 1 and the gap is about 0.0247, which equals 2/81. The order is also believable without any arithmetic: Game A needs its three tosses to agree with each other, while Game B needs only two independent two-toss agreements, so neither game is obviously easier, and the answer should be a small gap rather than a lopsided one. The sign survives a stronger test too. Let p+q=1 and t = pq. Then P(A) = p³+q³ = 1-3t and P(B) = (p²+q²)² = (1-2t)², so P(A)-P(B) = (1-3t)-(1-4t+4t²) = t(1-4t). Since pq ≤ 1/4 whenever p+q=1, this difference is never negative: Game A is at least as good as Game B for any coin, and ties only for a fair coin. So a choice claiming Game A is smaller is impossible here, and t = 2/9 gives 2/9(1-8/9) = 2/81, matching exactly.
💡Key takeaway

Game A wins with probability 27/81 and Game B with 25/81, so Game A is better by 2/81 — and since two choices offer that same gap in opposite directions, the sign of the subtraction is what picks (D).

  • List Game A's winning sequences
  • Split Game B into two pairs
  • Re-list Game B a second way
  • Rewrite both over 81
  • Match the sign to a choice

More like this

Same archetype — closest grade level first.