# Thread: Given 5 dice, how many ways can you roll exactly two 6s?

1. ## Given 5 dice, how many ways can you roll exactly two 6s?

Given 5 dice, how many ways can you roll exactly two 6s?

2. ## Re: Given 5 dice, how many ways can you roll exactly two 6s?

$\dbinom{5}{2}5^3$

you'll never learn this stuff if you don't start working it out on your own.

3. ## Re: Given 5 dice, how many ways can you roll exactly two 6s?

Originally Posted by Unrated063
1.How many possible sums can you roll if you have the choice to use at most 3 dice? You must use at least one die.

2.If you were to roll 5 dice each 5 times, how many possible outcomes are there?
Originally Posted by Unrated063
Given 5 dice, how many ways can you roll exactly two 6s?
@Unrated063, You posted both of the above. The second is almost trivial, while the first is advanced(difficult) in the extreme.
Is it true that you meant the second when you posted the first? Please respond.

4. ## Re: Given 5 dice, how many ways can you roll exactly two 6s?

1. With 1 die you can roll 1 to 6. 2 dice = 2 to 12. 3 dice = 3 to 18. So with 1 to 3 dice you can roll any of 18 sums.

5. ## Re: Given 5 dice, how many ways can you roll exactly two 6s?

Here is the original post (the question) in this thread.
Originally Posted by Unrated063
Given 5 dice, how many ways can you roll exactly two 6s?
Originally Posted by SlipEternal
1. With 1 die you can roll 1 to 6. 2 dice = 2 to 12. 3 dice = 3 to 18. So with 1 to 3 dice you can roll any of 18 sums.

6. ## Re: Given 5 dice, how many ways can you roll exactly two 6s?

Originally Posted by Plato
Here is the original post (the question) in this thread.

what OP actually meant was how many unique unordered 5 tuples there are using 6 sided dice

but I don't understand why that previous post has crept into this thread.

The question at the beginning of this thread was straightforward enough and I answered it.

7. ## Re: Given 5 dice, how many ways can you roll exactly two 6s?

Originally Posted by romsek
but I don't understand why that previous post has crept into this thread.
The question at the beginning of this thread was straightforward enough and I answered it.
Yes indeed romsek. I agree with you completely. In fact, that was my whole point in asking for a reply.
I suspect that Unrated063 really has no idea about any of this.
I asked Unrated063 for a reply, giving the exact wording of the original question.

8. ## Re: Given 5 dice, how many ways can you roll exactly two 6s?

Originally Posted by Plato
Here is the original post (the question) in this thread.

I just saw the question you posted that had a question 1 and not its context. Yes, that was my attempt at an answer. I thought you were commenting on the question's ambiguity (it was poorly worded). Now that I reread the context I see you weren't looking for an interpretation that was easily answered. I was answering:

1.How many possible sums can you roll if you have the choice to use at most 3 dice? You must use at least one die.
You mentioned that it was an extremely advanced question, but because of how poorly it is worded, I offered an easily answered interpretation. Again, I just misread the context. Sorry about that.

9. ## Re: Given 5 dice, how many ways can you roll exactly two 6s?

$Binomial{5 2}{\{fac{1}{6}}raised to 5$

10. ## Re: Given 5 dice, how many ways can you roll exactly two 6s?

Originally Posted by Vinod
$Binomial{5 2}{\fac{1}{6}^5$
If we toss five dice, then the number of ways to have exactly two $6's$ appear is $\dbinom{5}{2}\cdot(5^3)$

Now what is the question?

11. ## Re: Given 5 dice, how many ways can you roll exactly two 6s?

Originally Posted by Vinod
$Binomial{5 2}{\{fac{1}{6}}raised to 5$
aside from being nearly indecipherable this is incorrect.

Post #2 has the correct expression.

12. ## Re: Given 5 dice, how many ways can you roll exactly two 6s?

A die has six outcomes: $1,~2,~3,~4,~5,~6$. Toss $n$ dice. Then there are $6^n$ different outcomes. We can model this with n-tuples.
If $0\le k\le n$ we can ask how many of those contain exactly $k$ $6's$? ANSWER: $\dbinom{n}{k}\cdot 5^{n-k}$.
We choose $k$ places to put a $6$, then there are $n-k$ places left to put any one of five other numbers.