Dice= probability

Here is my explanation please tell if it is correct, First i will find a general formula for the problem if a = 1 then p(b>a)= $\frac{n-1}{n}$ if a=2 then P(b>a)= $\frac{n-2}{n}$ . . . if a = k then P(b>a)= $\frac{n-k}{n}$ so our net probability turns out to be = $\frac{n-1+n-2+n-3...n-m}{nm}$ =1- $\frac{n-1}{2m}$ =P(b>a) hence the P(b</=a)= $\frac{n-1}{2m}$ =0.1 ergo,m=5(n-1) ergo m is a multiple of 5.

There are a few issues with this problem.

1) You've put no constraints on $M, N$ to be integers. If $M$ is allowed to be any arbitrary positive number, it's impossible to conclude that $M$ needs to be integral, much less a multiple of $5$ .

2) By constraining the random numbers to be at least $1$ , you have broken the problem. Consider this: if I let $M=2$ and $N=6$ , and use uniform distributions for both $A$ and $B$ , then my distribution satisfies the requirements. But $N$ is not a multiple of $5$ . What happened? It's clear that you want the distributions to range from $0$ to $M$ , $N$ respectively. And then, if you have uniform distributions and $P(B \leq A) = 0.1$ , it follows that $M = 5N$ .

3) My pet peeve: please don't use "random" to mean "random and uniform" when discussing probability distributions. Unless you constrain these distributions to be uniform, there really is no way to solve this problem.