1995 Putnam A6

Suppose that each of $n$ people writes down the numbers

1,2,3 in random order in one column of a $3 \times n$ matrix, with

all orders equally likely and with the orders for different columns

independent of each other. Let the row sums $a,b,c$ of the resulting

matrix be rearranged (if necessary) so that $a \leq b \leq c$. Show

that for some $n \geq 1995$, it is at least four times as likely that

both $b=a+1$ and $c=a+2$ as that $a=b=c$.

假设$n$个人都写下数字

1,2,3 在 $3 \times n$ 矩阵的一列中以随机顺序排列，其中

所有订单的可能性相同，并且不同列的订单

彼此独立。令该行对结果的 $a,b,c$ 求和

重新排列矩阵(如有必要)，使得 $a \leq b \leq c$。显示

对于某些 $n \geq 1995$，其可能性至少是其四倍

$b=a+1$ 和 $c=a+2$ 都是 $a=b=c$。