Problem 21 · 2000 AMC 8
Stretch
Counting & Probability
casework
Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
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Answer: B — 3/8.
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Hint 1 of 2
Keiko has only one coin, so she can only get 0 or 1 head — that means there are just TWO ways to 'match,' and you only ever have to consider those two head-counts.
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Hint 2 of 2
Independent people ⇒ multiply their probabilities within each matching case, then ADD across the cases. So total = P(both 0) + P(both 1).
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Approach: split into the only two matchable head-counts
- Keiko's single coin: 1 head or 0 heads, each probability ½. Ephraim's two coins: 0 heads (TT) with probability ¼, 1 head (HT or TH) with probability ½, 2 heads (HH) with probability ¼. His '2 heads' can never match Keiko, so ignore it.
- Match at 1 head: ½ · ½ = ¼. Match at 0 heads: ½ · ¼ = ⅛.
- These cases are exclusive, so add: ¼ + ⅛ = ⅜.
- The principle: 'same outcome for two players' = sum over each shared value of P(A gets it)·P(B gets it). Limiting to the value the *smaller* experiment can produce keeps the casework tiny.
Another way — count equally-likely outcomes directly:
- Keiko has 2 equally likely results (H, T); Ephraim has 4 (HH, HT, TH, TT). Together that's 2 × 4 = 8 equally likely combined outcomes.
- Matching ones: Keiko H pairs with Ephraim's 1-head results HT, TH (2 ways); Keiko T pairs with Ephraim's TT (1 way). That's 3 favorable out of 8.
- Probability = 3/8 = ⅜.
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