HMMT 二月 2008 · COMB 赛 · 第 6 题
HMMT February 2008 — COMB Round — Problem 6
题目详情
- [ 5 ] A Sudoku matrix is defined as a 9 × 9 array with entries from { 1 , 2 , . . . , 9 } 1 and with the constraint that each row, each column, and each of the nine 2 3 × 3 boxes that tile the array contains each digit from 1 to 9 exactly once. ? A Sudoku matrix is chosen at random (so that every Sudoku matrix has equal probability of being chosen). We know two of squares in this matrix, as shown. What is the probability that the square marked by ? contains the digit 3?
解析
- [ 5 ] A Sudoku matrix is defined as a 9 × 9 array with entries from { 1 , 2 , . . . , 9 } 1 and with the constraint that each row, each column, and each of the nine 2 3 × 3 boxes that tile the array contains each digit from 1 to 9 exactly once. ? A Sudoku matrix is chosen at random (so that every Sudoku matrix has equal probability of being chosen). We know two of squares in this matrix, as shown. What is the probability that the square marked by ? contains the digit 3? 2 Answer: The third row must contain the digit 1, and it cannot appear in the leftmost three 21 squares. Therefore, the digit 1 must fall into one of the six squares shown below that are marked with ? . By symmetry, each starred square has an equal probability of containing the digit 1 (To see this more precisely, note that swapping columns 4 and 5 gives another Sudoku matrix, so the probability that the 4th column ? has the 1 is the same as the probability that the 5th column ? has the 1. Similarly, switching the 4-5-6th columns with the 7-8-9th columns yields another Sudoku matrix, which implies in particular that the probability that the 4th column ? has the 1 is the same as the probability that the 7th column ? has the 1. The rest of the argument follows analogously.) Therefore, the probability that the ? square contains 1 is 1 / 6. 1 2 ? ? ? ? ? ? Similarly the probability that the digit 2 appears at ? is also 1 / 6. By symmetry, the square ? has equal ( ) 1 1 probability of containing the digits 3 , 4 , 5 , 6 , 7 , 8 , 9. It follows that this probability is 1 − − / 7 = 6 6 2 . 21