Problem 21 · 2007 AMC 8
Medium
Counting & Probability
fix-first-card
Two cards are dealt from a deck of four red cards labeled A, B, C, D and four green cards labeled A, B, C, D. A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?
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Answer: D — 4/7.
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Hint 1 of 2
Don't worry about which card comes first — whatever it is, the picture from the second card's point of view is identical. So freeze the first card and just ask: which of the 7 leftover cards pairs with it?
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Hint 2 of 2
Fix one outcome to kill the symmetry: by symmetry the first card's identity doesn't matter, so condition on it and count favorable second cards out of the rest.
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Approach: fix the first card, count winning seconds
- Hold the first card fixed; 7 cards remain. Count which of them make a winning pair with it.
- Same color (different letters): the 3 other cards of that color — all winners.
- Same letter (different color): exactly 1 card — also a winner.
- So 3 + 1 = 4 of the 7 remaining cards win: probability 4/7.
- Why no double-count: a same-color card has a different letter and vice versa, so the two winning sets never overlap — you can just add them.
Another way — count via the losing pairs instead:
- A pair loses only when it's different color AND different letter. Fix the first card: of the 7 left, 3 share its color and 1 shares its letter, so 7 − 4 = 3 are total mismatches (losers).
- P(lose) = 3/7, so P(win) = 1 − 3/7 = 4/7.
Mark:
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