AMC 10 · 2015 · #9

Grade 11 probability

Archetype Geometric Series Share └ Sub-archetype Cyclic Participant Share

probability-basicgeometric-series-infiniteindependent-events systematic-enumerationcomplementary-counting ↑ Prerequisites: probability-basicgeometric-series-infinite

📏 Medium solution 💡 2 insights

Problem

Two players alternate attempts and the first success wins. Find the first player's chance.

Pick an answer.

(A)

$\dfrac{1}{2}$

(B)

$\dfrac{3}{5}$

(C)

$\dfrac{2}{3}$

(D)

$\dfrac{3}{4}$

(E)

$\dfrac{4}{5}$

How to solve

Strategy Make a Systematic List

There is no upper limit on how long the game runs, so I cannot list outcomes throw by throw forever. Instead I sort every outcome by one number: which throw ends the game. Those cases never overlap, Larry's cases are exactly the odd ones, and their probabilities follow a clean pattern, so a systematic list turns the whole game into an addition problem. The one thing a list of endless cases cannot do by itself is prove that nothing important got left out, so I will add only finitely many cases at a time and separately measure how much is still unaccounted for. When that leftover shrinks to nothing, the running total is forced to be the exact answer.

1STEP 1

Name the throw that ends it

The winning attempt's number decides the winner.

Larry throws 1, 3, 5, … Julius throws 2, 4, 6, … Larry wins ⇔ N is odd

2STEP 2

Probability the game ends on throw n

Ending on a given attempt has a simple chance.

P(N = n) = 1/2 · 1/2 … 1/2_n-1 misses · 1/2 = (1/2)ⁿ

3STEP 3

List Larry's cases and add

The first player's cases are the odd ones.

S_m = P(N=1) + P(N=3) + … + P(N=2m-1) = 1/2 + 1/8 + 1/32 + … + (1/2)²m-1

4STEP 4

Sum the first m cases exactly

Any finite stretch adds exactly.

S_m = (1/2(1 - (1/4)^m))/(1 - 1/4) = 2/3(1 - 1/4^m)

5STEP 5

Measure what is left out

What is left out shrinks to nothing.

P(first 2m throws all miss) = (1/2)²m = 1/4^m → S_m ≤ P(Larry wins) ≤ S_m + 1/4^m

6STEP 6

Squeeze to the exact value

Squeezing gives 2/3, choice (E).

2/3(1 - 1/4^m) ≤ P(Larry wins) ≤ 2/3(1 - 1/4^m) + 1/4^m for all m → P(Larry wins) = 2/3

Answer

2/3

Larry throws first, so his probability must beat 1/2 — that rules out (A) — and it must fall short of 1, since Julius clearly wins whenever Larry's first two throws both miss and Julius's first throw hits, an event of probability 1/8. The value 2/3 sits between them. A stronger check: redo the work with hit probability p instead of 1/2. The sum becomes p/(1-(1-p)²) = 1/(2-p). This behaves the way it should — it is near 1/2 when p is tiny (a near-endless game is nearly fair) and equals 1 when p = 1 (Larry ends it immediately) — and at p = 1/2 it gives 1/(2-1/2) = 2/3, matching choice (C). Note also that 3/4 and 4/5 would require p = 2/3 and p = 3/4, so those distractors answer a different question.

💡Key takeaway

Sort every outcome by which throw ends the game, add up Larry's odd-numbered ones, and make sure the part you have not added yet shrinks to nothing.

Name the throw that ends it
Probability the game ends on throw n
List Larry's cases and add
Sum the first m cases exactly
Measure what is left out
Squeeze to the exact value