AMC 10 · 2010 · #15

Grade 9 probability

Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count

probability-basiccombinations-basicquadratic-equationscompleting-the-square convert-to-algebrasymmetry-argument ↑ Prerequisites: probability-basic

📏 Long solution 💡 3 insights

Problem

A biased coin flipped four times gives an even split with a known small chance. Find its head probability.

Pick an answer.

(A)

$\frac{\sqrt{15}-3}{6}$

(B)

$\frac{6-\sqrt{6\sqrt{6}+2}}{12}$

(C)

$\frac{\sqrt{2}-1}{2}$

(D)

$\frac{3-\sqrt{3}}{6}$

(E)

$\frac{\sqrt{3}-1}{2}$

How to solve

Strategy Introduce a Variable

Tool #4 (Introduce a Variable): the coin is described only by the number I am asked for, so naming it p and writing tails as 1-p puts the entire problem on one letter. Tool #13 (Convert to Algebra): the sentence "the probability of an equal split is 1/6" becomes a single equation once I count how many even splits there are. Tool #15 (Organize Information in More Ways): the equation never mentions p alone — only the product p(1-p) — so I re-measure p by its distance below 1/2, which turns that product into a difference of squares and makes the square root fall out in one line. Tool #3 (Eliminate Possibilities): taking a square root opens a ±, and the fact that p and 1-p are both positive closes one branch. Tool #14 (Extreme Principle): before trusting any algebra I check the largest value p(1-p) can reach, which proves such a coin exists at all and that only one such coin has p ＜ 1/2.

1STEP 1

Name the head probability

One even split has probability built from both faces.

P(HHTT) = p · p · (1-p) · (1-p) = p²(1-p)²

2STEP 2

Count the even splits, then build the equation

There are six such splits in four flips.

6p²(1-p)² = 1/6

3STEP 3

Fold the two squares into one

The two squares fold into a single square.

6(p(1-p))² = 1/6 ⟹ (p(1-p))² = 1/36

4STEP 4

Take the root, drop the impossible sign

The bias kills the negative root.

p(1-p) = ±1/6, 0 ＜ p ＜ 1/2 → p(1-p) ＞ 0 ⟹ p(1-p) = 1/6

5STEP 5

Check such a coin exists, and only one

Such a coin exists, and only one does.

p(1-p) = 1/4 - (p-1/2)² ↑ on [0,1/2]; 0 ＜ 1/6 ＜ 3/8

6STEP 6

Measure p from one half and finish

Measuring from one half finishes it, choice (D).

1/4 - d² = 1/6 → d² = 1/12 → d = √(3)/6 → p = 1/2 - √(3)/6 = (3-√(3))/6 → (D)

Answer

(3-√(3))/6

Substitute back exactly, not in decimals. With p = (3-√(3))/6 we get 1-p = (3+√(3))/6, so p(1-p) = (3-√(3))(3+√(3))/36 = (9-3)/36 = 1/6, and therefore 6p²(1-p)² = 6(1/6)² = 1/6 — the given probability, on the nose. Also p = (3-√(3))/6 ≈ 0.2113 ＜ 1/2, as required, and it lies between 0 and 1 so it is a legal probability. Two traps are worth naming. First, the mirror value (3+√(3))/6 ≈ 0.7887 satisfies the probability equation just as well — a coin biased toward heads by the same amount produces the same even-split chance — which is exactly why the problem bothers to say "less than 1/2"; that clause is the tie-breaker, not decoration. Second, choice (C) (√(2)-1)/2 ≈ 0.2071 sits only 0.004 away from the true value, so a decimal check carried to two places cannot tell (C) from (D). Test it exactly: for (C), p(1-p) = (4√(2)-5)/4 ≈ 0.16421, and 6(0.16421)² ≈ 0.1618 ≠ 1/6 ≈ 0.16667. It fails. Working in exact radicals, not decimals, is what separates the two.

💡Key takeaway

When an equation only ever uses p times 1-p, stop tracking p and track how far it sits from one half — the product becomes 1/4 minus that distance squared, and one square root finishes the job.

Name the head probability
Count the even splits, then build the equation
Fold the two squares into one
Take the root, drop the impossible sign
Check such a coin exists, and only one
Measure p from one half and finish