Finding the probability formula of minimum of 2 fair dice question Suppose there are k dice thrown. Let M denote the minimum of the k numbers rolled. 
I've learned that finding the individual probability is:
$$P(M = m) = P(M \ge m) - P(M \ge m + 1) $$
Can someone please explain this to me? I've tried plugging in values for $m = 1, ... , 6$ but it isn't clear to me how that formula is derived. 
 A: How can the minimum be $\ge m$? There are two possibilities: (i) the minimum is exactly $m$ or (ii) the minimum is greater than $m$. 
The minimum is greater than $m$ precisely if the minimum is $\ge m+1$.
The possibilities (i) and (ii) are disjoint, so $$\Pr(M\ge m)=\Pr(M=m)+\Pr(M\ge m+1).$$
From this we get immediately
$$\Pr(M=m)=\Pr(M\ge m)-\Pr(M\ge m+1).$$ 
Remark: The above deals with your specific question. But to finish your problem, note that the minimum is $\ge w$ precisely if all the values are $\ge w$. And for $w=1,2,\dots, 6$ the tosses are all $\ge w$ with probability $\left(\frac{7-w}{6}\right)^k$.
A: Note that:
$$
\Pr[M \geq m] = \sum_{i=m+1}^\infty Pr[M = i]
$$
because if $M \geq m$, then either $M=m$ or $M=m+1$ or $M=m+2$ or $M=m+3$ or ...
(by the way, the infinite sum exists because it is bounded above by 1)
Similarly, 
$$
\Pr[M \geq (m+1)] = \sum_{i=(m+1)+1}^\infty Pr[M = i]
$$
because if $M \geq (m+1)$, then either $M=m+1$ or $M=m+2$ or $M=m+3$ or ...
Now, you subtract, and what do you get?
$$
\Pr[M \geq (m+1)] - \Pr[M \geq m] = \Pr[M = m] 
$$
So that sum is easy. Even the problem is, but I'll write a separate answer for that.
