# If a die is rolled thrice, what are the possible different outcomes.

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.

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Try making a list of outcomes for the die being rolled twice. Use a two-column table in which you list the outcome of the first roll in the first column and the outcome of the second roll in the second column. Do NOT stop when you are tired; make sure you list them all. Then count the number of rows in your table. See if you can generalize your answer to three rolls. –  Dilip Sarwate Dec 14 '11 at 12:34
would it be meaning to same to say that each dice is rolled twice and two dice are rolled once together? –  Nikhil Mulley Dec 14 '11 at 12:40
You have a single die. You are rolling it three times. So there is a number that shows on the first roll, a number that shows on the second roll, and a number that shows on the third roll. Why don't you ignore the third roll for now, and just make a list of all possible things that can happen on the first two rolls as I suggested in my previous comment? If you are too lazy to do so, look at David Mitra's answer to this question –  Dilip Sarwate Dec 14 '11 at 12:47
@Nikhil: You are correct in saying that there is no difference between rolling one die twice and rolling two dice together (the dice are identical). However, this doesn't simplify the problem at all. Following the advice given to you above should help you to see why –  Daniel Freedman Dec 14 '11 at 13:15
@DanielFreedman "there is no difference between rolling one die twice and rolling two dice together (the dice are identical)." Yes indeed. However, beginners often confuse identical as meaning indistinguishable and David Mitra's answer to the question I cited previously discusses this issue at some length. My personal belief is that it is best to resolve the simpler "roll a die twice" first before getting into "roll two identical dice once" –  Dilip Sarwate Dec 14 '11 at 13:31
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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.

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