Which is more likely with perfect six-side dice? Find the probability of the following 
$(i)$ to roll a $1$,then a $2$ and then a $3$ on consecutive roll of a single dice,
$(ii)$to roll three identical dice and have the outcome be a $1$,a $2$ and a $3$.
Which is more likely with perfect six-side dice?

My Attempt:
Probability to roll a $1$,then a $2$ and then a $3$ on consecutive roll of a single dice$=\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}+\frac{5}{6}\times(\frac{1}{6})^3+(\frac{5}{6})^2\times(\frac{1}{6})^3+(\frac{5}{6})^3\times(\frac{1}{6})^3+.....$
$=\frac{1}{36}$
Probability to roll three identical dice and have the outcome be a $1$,a $2$ and a $3=\frac{6}{216}=\frac{1}{36}$
There are 216 sample points in the sample space when we throw three dices and record the outcomes.Out of them six are favourable.$(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)$
So the two events are equally likely.
But the book answer says $(ii)$ event is more likely.I dont know how.Is some mistake i have made.Please help me.
 A: The first one should just be ${1\over6}\times{1\over6}\times{1\over6}$ and nothing else if you assume you only have three chances of rolling (which is much more natural to think in my opinion and is probably what the intention of the question).
However, based on your formula you seem to assume you have infinitely many chances of rolling.
But you are still wrong because you did not count the cases for, for example $4,5,1,2,5,1,2,3$ in which "1" appears but "123" does not fully appear. If you really had infinite number of chances the probability would converge to $1$.
A: Let's rephrase the question:
Which is more likely with perfect six-sided dice?


*

*Roll $1$ die $3$ times and get 1, then 2, then 3

*Roll $3$ dice $1$ time and get 1, 2, 3


Obviously, without even computing the probabilities, the second option is more likely.
This is because in the first option the order matters and in the second option it doesn't.

That said, let's compute the probabilities:

Roll $1$ die $3$ times and get 1, then 2, then 3:

$$\frac{1}{6^3}=\frac{1}{216}$$

Roll $3$ dice $1$ time and get 1, 2, 3:

$$\frac{3!}{6^3}=\frac{6}{216}$$
