AMC 10 · 2012 · #11

Grade 11 probability
Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count
probability-basicindependent-eventscombinations-basicfundamental-counting-principle identify-subproblemsconvert-to-algebra ↑ Prerequisites: probability-basiccombinations-basic
📏 Medium solution 💡 3 insights
Problem
Independent rounds each have one winner, with one player's chance tied to another's. Find the chance of a given split.

Pick an answer.

(A)
$\frac{5}{72}$
(B)
$\frac{5}{36}$
(C)
$\frac{1}{6}$
(D)
$\frac{1}{3}$
(E)
1
How to solve
Strategy Identify Subproblems

The question splits cleanly into three subproblems (tool #7): what is one round's probability table, what is the probability of one specific 6-round sequence of winners, and how many sequences produce the required tally. Tool #4 (Introduce a Variable) handles the first: name Mel's and Chelsea's per-round chances m and c, then turn "a single winner" and "twice as likely" into two equations. Tool #2 (Make a Systematic List) handles the third: think of the answer as a 6-letter word made of three A's, two M's, and one C, and count those words by deciding which rounds go to whom. The step most solutions skip is the bridge between subproblems two and three: multiplying "one sequence's probability" by "the number of sequences" is only legal because every qualifying sequence has the same probability (products do not depend on the order of their factors) and because distinct sequences are disjoint events. Both facts are proved below rather than assumed. Tool #15 (Organize Information in More Ways) supplies the independent cross-check: re-slice the same experiment as Alex-versus-not-Alex first, then Mel-versus-Chelsea inside the rounds Alex lost, and confirm the two routes land on the same number.

1STEP 1

Pin down one round's probabilities

One round's three probabilities are forced.

1/2+m+c=1, m=2c ⟹ 3c=1/2 ⟹ c=1/6, m=1/3
2STEP 2

Price one full six-round sequence

Any one qualifying sequence has the same price.

P(one such sequence)=(1/2)³(1/3)²(1/6)¹=1/(8 · 9 · 6)=1/432
3STEP 3

Count the qualifying sequences

There are 60 such sequences.

C(6, 3)C(3, 2)C(1, 1)=20 · 3 · 1=60=6!/3! 2! 1!
4STEP 4

Add up the equal pieces

Multiplying gives 5/36, choice (B).

60·1/432=60/432=5/36 ⟹ (B)
Answer
5/36
Recompute by a structurally different route that never writes a multinomial coefficient. First slice each round only as "Alex" or "not Alex": that is 6 independent trials with success probability 1/2, so P(Alex wins exactly 3)=C(6, 3)(1/2)⁶=20/64=5/16. Now condition on the 3 rounds Alex lost. Within a round Alex loses, the conditional probability Mel won is (1/3)/(1/2)=2/3 and Chelsea won is (1/6)/(1/2)=1/3, independently across those rounds. So P(exactly 2 Mel among the 3)=C(3, 2)(2/3)²(1/3)=3·4/9·1/3=4/9. Multiplying, 5/16·4/9=20/144=5/36 — the same value, reached without ever assuming that all 60 orders share a probability. Bounds also agree: the answer must be strictly smaller than P(Alex wins exactly 3)=5/16=0.3125, and 5/36≈ 0.139 is. Distractors are diagnosable: (D) 1/3 is just Mel's per-round probability, (C) 1/6 is just Chelsea's, (E) 1 is impossible since many other tallies (for instance Alex 6) are also possible, and (A) 5/72 is exactly what comes out if Mel's and Chelsea's probabilities are swapped: 60(1/2)³(1/6)²(1/3)=60/864=5/72. Getting 5/36 rather than 5/72 therefore confirms that m=1/3 was assigned to the player with two wins, not one.
💡Key takeaway

"A single winner each round" forces 1/2+m+c=1, which turns "Mel is twice Chelsea" into m=1/3 and c=1/6; then every one of the 60 orderings of three Alex wins, two Mel wins and one Chelsea win carries the same 1/432, so the answer is 60/432=5/36.

  • Pin down one round's probabilities
  • Price one full six-round sequence
  • Count the qualifying sequences
  • Add up the equal pieces

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