A fair die is rolled four times. The probability that each of final three rolls is atleast as large as the roll preceding it is $\frac{m}{n}$ , where $m\&n$  are relatively prime positive integers, then $m$  is

If we take any combination of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from $\left\{1,2,3,4,5,6\right\}$ . We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of $\left\{1,2,3,4,5,6\right\}$ ), or dividing them amongst 5 separators. Thus, there are $\left(\begin{array}{c}9\\ 4\end{array}\right)=126$  outcomes of four dice.

The solution is therefore $\frac{126}{6^4}=\frac{7}{72}$ , and

So, $m=7$