With a 6 sided die, what is the probability of 3 rolls being increasing?
I understand from reading other questions that the answer is $\frac{\binom{6}{2}}{6^3}$. The $6^3$ on the denominator makes sense. Can someone please explain to me why it's choose 2, and what that part means?
On
There are $\binom63=20$ sets of $3$ distinct numbers chosen from $\{1,2,3,4,5,6\}$, and each of those can be arranged in increasing order in exactly one way, so there are $\binom63$ strictly increasing sequences of $3$ die rolls. The probability of getting one of them is therefore $\dfrac{20}{6^3}=\dfrac5{54}$, not $\dfrac{\binom62}{6^3}=\dfrac{15}{216}=\dfrac5{72}$.
If you allow all non-decreasing sequences, there are in addition $6$ constant sequences, and such sequences as $\langle 1,3,3\rangle$ and $\langle 1,1,3\rangle$ containing two distinct entries. There are $\binom62$ ways to choose which $2$ numbers from $\{1,2,3,4,5,6\}$ are represented, and then there $2$ ways to choose which one appears twice, so there are $2\binom62$ sequences of this type. Thus, the number of non-decreasing sequences is
$$\binom63+6+2\binom62=20+6+30=56\;,$$
and the probability of getting one of them is $\dfrac{56}{216}=\dfrac7{27}$.
There are $\binom{6}{3}$ ways to choose $3$ distinct numbers. Any such choice can be arranged in order in only one way.
Thus the numerator should be $\binom{6}{3}$.