PUMaC 2010 · 组合（A 组） · 第 4 题
PUMaC 2010 — Combinatorics (Division A) — Problem 4
题目详情
- Erick stands in the square in the 2nd row and 2nd column of a 5 by 5 chessboard. There are 5 bills in the top right and bottom left squares, as shown below. 5 E 1 Every second, Erick randomly chooses a square adjacent to the one he currently stands in (that is, a square sharing an edge with the one he currently stands in) and moves to that square. When Erick reaches a square with money on it, he takes it and quits. The expected value of Erick’s winnings in dollars is m/n , where m and n are relatively prime positive integers. Find m + n .
解析
- Erick stands in the square in the 2nd row and 2nd column of a 5 by 5 chessboard. There are 5 bills in the top right and bottom left squares, as shown below. 5 E 1 Every second, Erick randomly chooses a square adjacent to the one he currently stands in (that is, a square sharing an edge with the one he currently stands in) and moves to that square. When Erick reaches a square with money on it, he takes it and quits. The expected value of 2 Erick’s winnings in dollars is m/n , where m and n are relatively prime positive integers. Find m + n . Answer: 18 Solution: In each square, we write the expected value of Erick’s winnings starting from that square. From any square in the middle column, Erick has an equal probability of ending in the top left and top right squares, and an equal probability of ending in the bottom left and bottom right squares, by symmetry. Therefore, the total probability of Erick ending in a $1 square is the same as the total probability of Erick ending in a $5 square, so both probabilities are 1/2, and therefore Erick’s expected winnings from any square in the middle column are $3. By an analogous argument, Erick’s expected winnings from any square in the middle row is $3. Let x be the expected value of Erick’s winnings starting from the 2nd row and 2nd column, and let y be the expected value of Erick’s winnings starting from either the 1st row and 2nd column, or the 2nd row and 1st column, since these expected values are the same by symmetry. 3 $5 y x $3 3 3 $3 $3 3 $1 Then we have a system of equations: 1 1 x = · y + · 3 2 2 1 1 1 y = · x + · 1 + · 3 3 3 3 Solving yields x = 13 / 5, so the answer is 18 .