AMC 10 · 2018 · #24

Grade 9 probability

Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Region

geometric-probabilityoptimizationquadratic-equationscaseworkinterval-arithmetic convert-to-algebracaseworkextreme-principle ↑ Prerequisites: geometric-probabilityquadratic-equations

📏 Medium solution 💡 4 insights

Problem

Alice picks a number at random, spread evenly between 0 and 1. Bob picks a number at random, spread evenly between one half and two thirds. Carol hears both rules and then picks one fixed number. The winner is whoever's number lands between the other two. Find the value Carol should pick to make her chance of winning as large as possible.

Pick an answer.

(A)

$\frac{1}{2}$

(B)

$\frac{13}{24}$

(C)

$\frac{7}{12}$

(D)

$\frac{5}{8}$

(E)

$\frac{2}{3}$

How to solve

Strategy Introduce a Variable

Tool #4 (Introduce a Variable) is the move that turns a vague question into a solvable one: call Carol's number c, and "what should Carol pick?" becomes "which c maximizes the function P(c)?" Tool #1 (Draw a Diagram) converts the phrase "uniform at random" into lengths on a number line, which is how uniform probabilities are actually computed. Tool #7 (Identify Subproblems) splits the work twice: Carol wins in two mutually exclusive ways, and c itself falls into three separate ranges that need separate formulas. Tool #14 (Extreme Principle) finds the peak of the quadratic that appears in the middle range. Tool #3 (Eliminate Possibilities) closes the argument by showing the other two ranges cannot beat that peak.

1STEP 1

Name Carol's number and the win event

Split the win into two cases.

P(c) = Pr(a ＜ c ＜ b) + Pr(b ＜ c ＜ a)

2STEP 2

Turn uniform into length

For a uniform pick, probability is length.

Pr(a ＜ c)=c, Pr(b ＜ c)=6c-3 for 1/2 ≤ c ≤ 2/3

3STEP 3

Handle c outside Bob's range

Outside Bob's range is always worse.

P(c)=c ＜ 1/2 if c ＜ 1/2; P(c)=1-c ＜ 1/3 if c ＞ 2/3

4STEP 4

Build P(c) inside Bob's range

Inside, it is a downward quadratic.

P(c)=c(4-6c)+(1-c)(6c-3)=-12c²+13c-3

5STEP 5

Find the peak of the parabola

The vertex gives the maximum.

-12c²+13c-3=-12(c-13/24)²+25/48

6STEP 6

Compare the three ranges

Comparing the three ranges gives thirteen twenty-fourths.

max P = 25/48 ＞ 1/2 ＞ 1/3, attained at c=13/24

Answer

13/24

Three checks agree. First, position: 13/24 sits between 1/2 (the center of Alice's range) and 7/12=14/24 (the center of Bob's range), which is where a player who wants to be squeezed in the middle should stand. Among the five choices only 13/24 lies strictly between those two values. Second, symmetry: plugging in c=1/2 and c=7/12 both give P=1/2, and a parabola peaks exactly halfway between two points of equal height, and the midpoint of 12/24 and 14/24 is 13/24. Third, size: 25/48≈ 0.52 is a bit above the 1/2 that plain choices like 1/2 or 7/12 already earn, and well above the 1/3 a player with no information would get. A real but small edge is exactly what a tiny bit of extra positioning should buy.

💡Key takeaway

When a question asks what you should choose, name your choice with a letter, write your chance of winning as a formula in that letter, and then find where that formula peaks.

Name Carol's number and the win event
Turn uniform into length
Handle c outside Bob's range
Build P(c) inside Bob's range
Find the peak of the parabola
Compare the three ranges