HMMT 十一月 2010 · GEN1 赛 · 第 4 题

英文解析

4. [ 4 ] An ant starts at the point (1 , 0). Each minute, it walks from its current position to one of the fouradjacent lattice points until it reaches a point ( x, y ) with | x | + | y | ≥ 2. What is the probability that the ant ends at the point (1 , 1)?
   7 1 1
   Answer: From the starting point of (1 , 0), there is a chance we will go directly to (1 , 1), a
   24 4 2
   chance we will end at (2 , 0) or (1 , − 1), and a chance we will go to (0 , 0). Thus, if p is the probability 1
   1 14
   that we will reach (1 , 1) from (0 , 0), then the desired probability is equal to + p , so we need only
   4 4
   calculate p . Note that we can replace the condition | x | + | y | ≥ 2 by | x | + | y | = 2, since in each iterationthe quantity | x | + | y | can increase by at most 1. Thus, we only have to consider the eight points
   (2 , 0), (1 , 1), (0 , 2), ( − 1 , 1), ( − 2 , 0), ( − 1 , − 1), (0 , − 2), (1 , − 1). Let p , p , ..., p be the probability of
   1 2 8
   General Testreaching each of these points from (0 , 0), respectively. By symmetry, we see that p = p = p = p
   1 3 5 7
   and p = p = p = p . We also know that there are two paths from (0 , 0) to (1 , 1) and one path from
   2 4 6 8
   (0 , 0) to (2 , 0), thus p = 2 p . Because the sum of all probabilities is 1, we have p + p + ... + p = 1.
   2 1 1 2 8
   1 1 1
   Combining these equations, we see that 4 p +4 p = 12 p = 1, so p = and p = . Since p = p = ,
   1 2 1 1 2 2
   12 6 6
   1 1 1 7
   then the final answer is + · =
   4 4 6 24
   6 5 4 3 2