Probability no two dice are the same How to show that when a poker dice is played simultaneously by rolling 5 dice, that:
The probability of no two alike is equal to 0.0926.
If I take the sample space by $6^5$ = 7776, given that each roll has a $\frac{1}{6}$ chance,
I know that the probability that no two are alike when two dice are rolled is: $\frac{30}{36}$, so I thought of implementing this as: $\frac{30}{7726}$, however it's not the answer. I would really appreciate some help on approaching this!
 A: By combinatorics: $$(6/6)(5/6)(4/6)(3/6)(2/6) = {}_6P_5/6^5 = 6!/6^5 = 0.09259259.$$
But you should also investigate the connection with the coupon-collector's problem.
By simulation in R, a million iterations gives $0.092627\pm 0.0006.$
die = 1:6
set.seed(2021)
nr.faces = replicate(10^6, length(unique(sample(die,5,rep=T))))
mean(nr.faces==5)
[1] 0.092627           # aprx 0.09259
2*sd(nr.faces==5)/1000
[1] 0.0005798183       # aprx 95% margin of sim err
factorial(6)/6^5
[1] 0.09259259         # exact

Procedure sample(die, 5, rep=T) simulates rolling 5 fair dice.
Vector nr.faces has a million elements, each a number 1 through 5; logical vector nr.faces==5
has a million TRUEs and FALSEs, and its mean is the proportion
of its TRUEs.
cutp=0:5 + .5
hist(nr.faces, prob=T, ylim=c(0.,.5),
  br=cutp, label=T, col="skyblue2")


Can you argue that the probability all five dice show the same face is
$4/6^5 = 0.0005144033 \approx 0.001$ in the figure? Any others?
A: If we assume that each outcome is from a fair die then the outcomes are IID uniform random variables on the $m=6$ possible outcomes.  Suppose we roll $n$ dice and let $K_n$ be the number of distinct outcomes (called the "occupancy number") from all dice.  It is well-known that this random variable is distributed according to the classical occupancy distribution (see e.g., O'Neill 2019), with mass function:
$$\mathbb{P}(K_mn=k) = \text{Occ}(k|n,m) = \frac{(m)_k \cdot S(n,k)}{m^n}.$$
(The values $(n)_k = \prod_{i=1}^k (n-i+1)$ are the falling factorials and the values $S(n,k)$ are the Stirling numbers of the second kind.)  With $m=6$ and $n=5$, the probability that no two dice are alike is equivalent to the event $K_n=n$, so the probability is:
$$\begin{align}
\mathbb{P}(K_5=5) = \text{Occ}(5|5,6) 
&= \frac{(6)_5 \cdot S(5,5)}{6^5} \\[6pt]
&= \frac{720 \cdot 1}{7776} \\[6pt]
&= \frac{5}{54} \\[12pt]
&= 0.09259259. \\[6pt]
\end{align}$$
