2010 AIME I Problems/Problem 4

Problem

Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac47$ . Jackie flips the three coins, and then Phil flips the three coins. Let $\frac {m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution

Case 1: No heads. The only possibility is TTT (the third coin being the unfair coin). The probability for this to happen to Jackie is $\frac {1}{2} \cdot \frac {1}{2} \cdot \frac {3}{7} = \frac {3}{28}$ Thus the probability for this to happen to both players is $\frac {3}{28} = \frac {9}{784}$

Case 2: One head. We can have either HTT, THT, or TTH. The first two happen to Jackie with the same $\frac {3}{28}$ chance, but the third happens $\frac {4}{28}$ of the time, since the unfair coin is heads instead of tails. With 3 possibilities for Jackie and 3 for Phil, there are a total of 9 ways for them both to have 1 head.

Multiplying and adding up all 9 ways, we have a $\frac {4(3 \cdot 3) + 4(3 \cdot 4) + 1(4 \cdot 4)}{28^{2}} = \frac {100}{784}$ overall chance for this case.

Case 3: Two heads. With HHT $\frac {3}{28}$ , HTH $\frac {4}{28}$ , and THH $\frac {4}{28}$ possible, we proceed as in Case 2, obtaining $\frac {1(3 \cdot 3) + 4(3 \cdot 4) + 4(4 \cdot 4)}{28^{2}} = \frac {121}{784}$ .

Case 4: Three heads. Similar to Case 1, we can only have HHH ( $\frac {4}{28}$ chance). Then in this case we get $\frac {16}{784}$

Finally, we take the sum: $\frac {9 + 100 + 121 + 16}{784} = \frac {246}{784} = \frac {123}{392}$ , so our answer is $123 + 392 = \fbox{515}$ .

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2010 AIME I Problems/Problem 4

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