AMC 10 · 2019 · #13
Grade 8 probabilityPick an answer.
Tool #16 (Complement): direct summation Σ_g Σ_r ＞ g 2^-g · 2^-r is messy. Instead, count the tie probability P(R = G) and use symmetry. Tool #15 (Reorganize): the three events R ＞ G, R ＜ G, R = G partition the sample space — that's the right way to slice the problem. Tool #5 (Pattern): the tie probability sums a geometric series in 4^-k, an arithmetic move kids see early.
Spot the symmetry
Both directions are equally likely.
Red beating green and green beating red are mirror images — same chance.
7.SP.C.7Organize Information In More WaysSplit into three cases
Three cases cover everything.
Exactly one of three things happens — split, win, or lose for red.
7.SP.C.7Organize Information In More WaysThe same-bin probability
Write the same-bin chance as a series.
Tie cases are the easy ones to count — pick a bin, both balls land there.
7.SP.C.8Change Focus Count The ComplementSum the series
Add the geometric series.
Geometric sum with ratio 1/4 — first-term-over-(1-ratio) rule.
A sum whose every term is a fixed fraction of the one before collapses into one short formula.
▸ Why?
Each round is the previous one multiplied by the same fixed factor, which makes the list geometric.
▸ Why?
A shrinking geometric series totals its first term divided by one minus the common ratio.
Solve for the rest
Halving the remainder gives one third.
Solve a one-step equation for p — the symmetry trick paid off.
6.EE.B.7Change Focus Count The ComplementThis AMC 12 problem only needs Grade 8 exponent reasoning you already know — by symmetry, red beating green and green beating red are equally likely, and ties happen 1/3 of the time (geometric series). So P(R ＞ G) = (1 - 1/3)/2 = 1/3. The answer is (C).