Probability/Math Help

[Velox]

Hello, friends! I need some help. I'm trying to figure out what the chances/odds are for certain die results, and I've hit a wall with my knowledge of probability. Any mathier people out there can help me out?

What are the chances of four of X (variable) rolled six-sided dice producing a result of "5" or "6" in one roll?
4 dice = 1.2%
5 dice = ???
6 dice = ???
etc.

I know that each die has a 1 out of 3 (or 1/3, or 33.4%) chance of producing the desired result of 5 or 6, which I'm calling a "hit," as per SR4. I know that the chances of all four dice of a four-die roll "hitting" is equal to the chance of one of them hitting times the chance of each other one hitting, i.e.
(1/3)*(1/3)*(1/3)*(1/3)

Before I start pouring over stat books, I was wondering if anyone knows this off hand. Thanks much! Peace be with you.

[sparkletwist]

Take a look at http://www.anydice.com/ , it'll help you with a lot of these kinds of problems.

[Velox]

Cool! That's handy. While I dig into that, does anyone actually know the math behind this? I know I'm basically asking for a probability lesson, but I figure if anyone is going to know this stuff it's either going to be a math professor a gamer.

[sparkletwist]

Ok, here goes.  :yumm:

What's going to help us here is a binomial distribution.

Let's start with the simplest case, 4 dice, in which we need 4 hits. As you've observed, the probability of success there is (1/3)^4, or, more generally speaking, p^h, where p represents the probability of a hit and h represents the number of hits needed.

Let's now expand this to 5 dice. The probability of success on 5 dice is simply the probability of exactly 5 hits on 5 dice plus the probability of exactly 4 hits on 5 dice. The probability of exactly 5 hits on 5 dice is p^h = (1/3)^5, much like before.

How about exactly 4 hits on 5 dice? In other words, we're calculating the probability of 4 hits and 1 miss.
This is given by
In other words, we want the probability of exactly 4 hits and 1 miss, so the probability of 4 hits and the probability of a 1 miss are multiplied. However, there are nCh different ways to get that outcome, so we multiply those two expressions by 5C4 (i.e., the binomial coefficent nCr) to get the probability.

So, the probability of (4 or 5 hits) =
I think now you can see how to expand it to 6 dice, or however many you want...  :)

[Velox]

Excellent! Thank you! Time to dig in and try to grok this... scribble scribble scribble :D

[Velox]

Dear God... I love wikipedia but it does not communicate mathematical lessons very well. I'm going with this page instead: http://www.stattrek.com/probability-distributions/binomial.aspx

Thanks again! Time to learn :D :D

[LordVreeg]

well done, Sparkle.