I know there a lot of math folks here that play HD, and maybe one or more of them can help me out.

For a game I am creating I need to know the probability of coming up with at least 2 '1s' when rolling multiple 6 sided dice. I know that for 2d the probability is 1/36. What is the probability if 3d are rolled? 4d? 5d? It is the "at least" 2 "1s" keeping me from figuring this out myself... I've tried just counting all the possibilities, but I am certain I am wrong with what I got so I need a formula...

I've found a site that discusses the probability of rolling a Yatzhee. They address various multiple dice rolls. The site shows that the probability of rolling any pair with 4 dice is 936/1296, while the probability of rolling any pair on 5 dice is 7056/7776. 

If I simply divide these values by 6 (the probability of rolling any pair vs the probability of rolling a specific pair) would that do the trick?

I could be wrong on this but here is my thought:

P of rolling 2 1s with n dice = nC2*(1/6)^2 * (5/6)^(n-2)

That looks better.

So this is my logic. You are playing with combinations here. Suppose you have 3 dice, you can have 1,1,x    1,x,1, or x,1,1. So it's clearly a combination problem. 

You want 2 dice to be 1, with a total of 3 dice, 4 dice...n dice, so it's n C 2 (n choose 2) total combinations. You then times it by the (1/6)^2. 1/6 is the probability of getting you 1, to the power of 2, because you want it to happen 2x. 5/6 is the prob of not getting a 1, and you want it to happen (n-2) times. 

And obviously you can modify the formula for any number of dice to be 1. If you want 3 dice to be 1, it's nC3(1/6)^3 * (5/6)^(n-3).

hate not being able to figure this out...what is the C?

Font size on this probably screwed it up. Explanation is added on. 

That's true, tianyi, but that doesn't take into account that if you have more than 2 dice you could theoretically have more than 2 ones, which takes into account more possibilities than just rolling exactly 2 ones.  The easiest way is to look at the space (what you want to define, namely, 2 or more ones) as the probability of not rolling either 0 or 1 ones on n dice.  That would look a little like this:

1- (5/6)^n - nC1*(1/6)^1*(5/6)^(n-1).  

C, by the way, is a combination.  If you have a handheld calculator that does anything more than 4 functions it's probably on there.  

I think it does with (1/6)^2, and (5/6)^(n-2). That limits the outcome to only 2 1s, with the rest not being 1s. 

Right, but according to what I'm reading back in the initial post, that's not what dacj501 is asking.  What you just listed is the probability of rolling exactly 2 ones with n dice.  I think he's trying to figure out the probability of rolling at least two ones with n dice.  For instance, if he's got 6 dice, nC2*(1/6)^2 *(5/6)^(n-2) gives you the probability of rolling exactly two ones in those six dice.  The formula I listed above will give you the probability of rolling either 2, 3, 4, 5, or 6 ones (all put together- not each one individually) in those 6 dice.  

I read that he wants exactly 2 1s. 

For at least 2 1s, your way of thinking is right. Start with overall sample space and subtract all combinations that doesn't give you at least 2 1s. 

Edit: your formula is right. I forgot that 11x = 5 combinations.

I came up with this formula, which is the same as yours:
Overall sample space = 1
Prob of rolling no 1s = nCn (5/6)^n * (1/6)^0
Prob of rolling only one 1 = nC(n-1) (5/6)^(n-1) * (1/6)^1

End formula = 1 - [nCn (5/6)^n * (1/6)^0] - [nC(n-1) (5/6)^(n-1) * (1/6)^1]

They are the same formula. 

Posted by dacj501 on 9/11/2012 7:32:00 AM (view original):
I know there a lot of math folks here that play HD, and maybe one or more of them can help me out.

For a game I am creating I need to know the probability of coming up with at least 2 '1s' when rolling multiple 6 sided dice. I know that for 2d the probability is 1/36. What is the probability if 3d are rolled? 4d? 5d? It is the "at least" 2 "1s" keeping me from figuring this out myself... I've tried just counting all the possibilities, but I am certain I am wrong with what I got so I need a formula...

Says here at least 2.  

As for the formula, I think I have it.  1 is the overall probability space.  Subtract out the probability of rolling no ones (only one way to do that if you're looking at this as a binomial - all rolls are not ones), which would be nC0 (which is 1) * (1/6)^0 (again, 1, which is why I didn't include those) * (5/6)^n.  Then subtract out the probability of rolling exactly 1 one, which is nC1 (choose 1 roll to be the roll that comes up as one) *(1/6)^1 * (5/6)^(n-1).  

Not sure if that clarifies or confuses the issue.  

Posted by tianyi7886 on 9/11/2012 11:08:00 AM (view original):
I read that he wants exactly 2 1s. 

For at least 2 1s, your way of thinking is right. Start with overall sample space and subtract all combinations that doesn't give you at least 2 1s. 

Edit: your formula is right. I forgot that 11x = 5 combinations.

I came up with this formula, which is the same as yours:
Overall sample space = 1
Prob of rolling no 1s = nCn (5/6)^n * (1/6)^0
Prob of rolling only one 1 = nC(n-1) (5/6)^(n-1) * (1/6)^1

End formula = 1 - [nCn (5/6)^n * (1/6)^0] - [nC(n-1) (5/6)^(n-1) * (1/6)^1]

They are the same formula. 

Yup- looks like we're on the same page.  I just tried to simplify it a little, especially since we know some of the values and some of it works out to 1.  

Well at dacj got more than he bargained for. He now has ways of finding exactly 2 1s and at least 2 1s. 

thx both of you - ti, fwiw gumbercules is right, I need it to be for at least 2 1s. 

dacj, if you have any more probability questions you can sitemail me if you want.