Problem:
Amelia has a coin that lands on heads with probability , and Blaine has a coin that lands on heads with probability . Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is , where and are relatively prime positive integers. What is ?
Answer Choices:
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Solution:
Let be the probability that Amelia wins. Then , because either Amelia wins on the first toss, or, if she and Blaine both get tails, then the chance of her winning from that point onward is also . Solving this equation gives . The requested difference is .
The probability that Amelia wins on the first toss is , the probability that Amelia wins on the second toss is , and so on. Therefore the probability that Amelia wins is
Let be the greatest integer such that Amelia and Blaine each toss tails times in a row. Then the game will end in the next round of tosses, either because Amelia tosses a head, which will occur with probability , or because Amelia tosses a tail and Blaine tosses a head, which will occur with probability . The probability that it is Amelia who wins is therefore
The problems on this page are the property of the MAA's American Mathematics Competitions