$3$ identical Dice are tossed simultaneously, The probability that all dice shows same number. 
If $3$ identical Dice are tossed simultaneously, The find probability that all dice shows
same number.

$\bf{My\; Try::}$ Let $A$ be the event in which upper face of all dice shows same number
and $S$ be the sample space
Now Here we have $3$ identical dice.
So $x_{1}$ be the number of times in which dice shows number $1$ on upper face.
Similarly  $x_{2}$ be the number of times in which dice shows number $2$ on upper face.
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$x_{6}$ be the number of times in which dice shows number $6$ on upper face.
So here $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6} = 3,$ Where $x_{1},x_{2},x_{3},x_{4},x_{5},x_{6}\geq 0$
So we get $\displaystyle n(S) = \binom{8}{2}=56$ and $n(A) = 6$
So required probability $\displaystyle P(A) = \frac{6}{56}$
although i have solve that question using that post
How many are the possible outcomes from throwing $n$ (identical) dice
but i did not understand what is the difference between identical dice and simple dice.
and why answer can not be equal to $\displaystyle \frac{6}{256}$
bcz whether dice are identical or not total number of possible outcome will remain same.)(Its my assumption.)
plz explain me, Thanks
 A: Note that, as the outcomes of the three dice are independent, throwing them simultaneously is equivalent to throwing them after each other. Also note that, for all numbers to be equal, there is no restriction on the outcome of the first die. Assume the outcome of the first die is $n$, with $1\leq{n}\leq6$. Then the probability that all dice show the same number is the probability you throuw $n$ with both the second and the third die, i.e.
$$
\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}.
$$
A: If the dice are indistinguishable (I suppose that ''identical'' means this) than all the outcomes that have the same tree numbers are identified, in the sense that we cannot distinguish, as example:
$ 1,4,3$ from $1,3,4$ or $4,1,3$ and so on.
This means that the space of all possible outcomes is formed from the ''Combination with repetition'' of $6$ elements of order $3$. As you can see here, such $3-$multicombinations of $6$ elements are:
$$
\binom{6+3-1}{3}=\binom{8}{3}=\binom{8}{5}=56
$$
since the outcomes with three same numbers are $6$ it seams that the  answer is $6/56$, but this is wrong (as noted by @trueblueanil), because the $56$ outcomes are not equiprobable, so the correct answer is $1/36$.
A: For this problem, the dice being distinguishable or not is immaterial,
since they are all to show the same number.
The first die can show anything. The second and third must show the same number,
thus $Pr = \dfrac16\cdot\dfrac16 = \dfrac1{36}$
Even if they were distinguishable (say red, blue, yellow) the red die could show anything,
and the blue and yellow ones would have to show the same number with $Pr = \dfrac16$ each

Added explanation
When $3$ dice are rolled, or one die is rolled $3$ times, there will be $216$ outcomes, as under:
$3$ of a kind: $\binom61 = 6$,  only $1$ permutation $\to 6$ ways
$2-1$ of a kind: $\binom61\binom51 = 30, \frac{3!}{2!} = 3$ permutations $\to 90$ ways
$1-1-1$ of a kind: $\binom63 = 20, 3! = 6$ permutations $\to 120$ ways
Combinations with repetitions counts $6+30+20 = 56$, ignoring frequency of occurrence,
and hence are inappropriate for computing probabilities.   
