I have a single die, and it is rolled thrice. What could be the total possible different outcomes, I guess if I have the number of possible outcomes for each rolled die, then I would use it for other roll outcomes to since it is the single die that is being rolled every time.
Similarily, what could be the total possible different outcomes when the number of dice is increased, to say two and three.
Thanks.
Best Answer
The answer to this question depends on what is a "different" outcome.
Interpretation 1: The possible outcomes are the multisets $\{a,b,c\}$ such that $a,b,c \in \{1,2,\ldots,6\}$. This is justified since dice are identical objects, and so e.g. the outcome $\{1,1,2\}$ is counted as the same thing as $\{1,2,1\}$ and $\{2,1,1\}$.
In general, using a stars-and-bars argument, the number of such multisets is ${n+5} \choose 5$. For example, $\{1,4,4,6\}$ is counted by $\underbrace{\star}_{\text{one } 1} | \; | \; | \underbrace{\star \star}_{\text{two } 4\text{'s}} | \; | \underbrace{\star}_{\text{one } 6}$ (i.e. we generate a string with $n$ stars and $5$ bars, and the position of the bars determines the number of copies of each element in the multiset).
In GAP, this is implemented as
NrUnorderedTuples([1,2,3,4,5,6],3);. It is also described by Sloane's A000389.Interpretation 2: The possible outcomes are the sums $a+b+c$ of the multisets $\{a,b,c\}$ such that $a,b,c \in \{1,2,\ldots,6\}$. In general, anything from $n$ to $6n$ is a possible outcome. Hence the number of possible outcomes is $5n+1$.
You also have correctly surmised that rolling $n$ dice and rolling the same die $n$ times are equivalent.