AMC 10 · 2010 · #16

Grade 7 probability
probability-basiccombinations-basiccaseworksymmetry-argument complementary-countingcasework ↑ Prerequisites: probability-basic
📏 Long solution 💡 3 insights
Problem
Two players each draw three numbers from nearly the same pool and arrange them into a number. Find the chance the first wins.

Pick an answer.

(A)
$\frac{47}{72}$
(B)
$\frac{37}{56}$
(C)
$\frac{2}{3}$
(D)
$\frac{49}{72}$
(E)
$\frac{39}{56}$
How to solve
Strategy Change Focus / Count the Complement

Tool #16 (Change Focus / Count the Complement): the clever move is to notice that once Bernardo has no 9 he is drawing from the exact same pool as Silvia, so the chance he wins equals the chance she wins — and the only leftover is a tie, which I remove and split. Tool #7 (Identify Subproblems): I break the event into two clean cases, Bernardo picks a 9 versus he does not, because the 9 is the single thing that makes his set different from Silvia's. Tool #2 (Make a Systematic List): counting choices with combinations tells me how likely the 9 case is and how likely a tie is.

1STEP 1

A chosen set is one number

Forced ordering makes a chosen set one number.

C(9, 3)=84, C(8, 3)=56
2STEP 2

Case 1: Bernardo picks a 9

Drawing the extra number is an instant win.

P(9 chosen)=(C(8, 2))/(C(9, 3))=28/84=1/3
3STEP 3

Case 2: no 9 means a fair fight

Without it the two draws are perfectly symmetric.

P(no 9)=1-1/3=2/3; P(B > S)=P(S > B)
4STEP 4

Remove the tie, split the rest

Removing ties splits the rest in half.

P(tie)=1/56; P(B > S ∣ no 9)=1/2(1-1/56)=55/112
5STEP 5

Weight each case

Each case carries its own weight.

1/3 · 1=56/168, 2/3·55/112=55/168
6STEP 6

Add the cases

Adding them gives 37/56, choice (B).

56/168+55/168=111/168=37/56 → (B)
Answer
37/56
The answer 37/56≈ 0.661 should be a bit more than 1/2, and it is: Bernardo's advantage is only the extra number 9, so he should win a little more than half the time, not overwhelmingly. Check the fraction is valid: 0 < 37/56 < 1. Sanity-check the pieces — the guaranteed-win case alone gives 1/3≈ 0.33, and the fair-fight case adds another 55/168≈ 0.33, totaling about 0.66, matching. The distractors 47/72 and 49/72 come from mishandling the tie, and 2/3 ignores the tie entirely; our count keeps the tie explicit.
💡Key takeaway

When both players draw the same way from the same pool, neither is favored — so remove the ties, split the rest in half, and the only real edge comes from Bernardo's extra 9.

  • A chosen set is one number
  • Case 1: Bernardo picks a 9
  • Case 2: no 9 means a fair fight
  • Remove the tie, split the rest
  • Weight each case
  • Add the cases