Problem:
Jerry starts at on the real number line. He tosses a fair coin times. When he gets heads, he moves unit in the positive direction; when he gets tails, he moves unit in the negative direction. The probability that he reaches at some time during this process is , where and are relatively prime positive integers. What is (For example, he succeeds if his sequence of tosses is HTHHHHHH.)
Answer Choices:
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Solution:
Jerry arrives at 4 for the first time after an even number of tosses. Because Jerry tosses 8 coins, he arrives at 4 for the first time after either
4,6 , or 8 tosses. If Jerry arrives at 4 for the first time after 4 tosses, then he must have tossed HHHH. The probability of this occurring is . If Jerry arrives at 4 for the first time after 6 tosses, he must have tossed 5 heads and 1 tail among the 6 tosses, and the 1 tail must have come among the first 4 tosses. Thus, there are 4 possible sequences of valid tosses, each with probability , for a total of . If Jerry arrives at 4 for the first time after 8 tosses, then he must have tossed 6 heads and 2 tails among the 8 tosses. Both tails must occur among the first 6 tosses; otherwise Jerry would have already reached 4 before the toss. Further, at least 1 tail must occur in the first 4 tosses; otherwise Jerry would have already reached 4 after the toss. Therefore there are sequences for which Jerry first arrives at 4 after 8 tosses, each with probability , for a total of . Thus the probability that Jerry reaches 4 at some time during the process is . The requested sum is .
OR
Count the sequences of 8 heads or tails that result in Jerry arriving at 4. Any sequence with appearing fewer than 3 times results in Jerry reaching 4 . There are such sequences. If Jerry's sequence contains exactly 3 Ts, then he reaches 4 only if he does so before getting his second T. As a result, Jerry can get at most one in his first 5 tosses. This happens if the first 4 tosses are and there is exactly one in the last 4 tosses, or there is one within the first 4 tosses followed by the remaining , accounting for ways for Jerry to get to 4 with exactly 3 Ts. Finally, the only way for Jerry to get to 4 by tossing exactly 4 Ts is HHHHTTTT. Jerry cannot get to 4 by tossing fewer than . Thus there are sequences where he reaches 4 , out of equally likely ways to toss the coin 8 times. The required probability is , and the requested sum is .
The problems on this page are the property of the MAA's American Mathematics Competitions