For example, If I roll 4 dice (set size 4, range of possible values 1-6), what is the probability of getting at least 2 of 6s.
Formula for probability that a set of size S contains at least X objects of value V, given common range R of possible values for each object
0
$\begingroup$
probability
-
1Ideally, the body of the question should be self-contained without relying on the title; in the present case, the two seem to contradict each other. Rolling $4$ dice doesn't generate a permutation. Also, the concept of "value of an object" isn't explained in the body. If, as the example seems to suggest, your question is what the probability is of getting a particular value $V$ at least $X$ times drawn from $S$ independent uniform distributions over $R$ items, the answer is $\sum_{n=X}^S\binom{S}{n}p^n(1-p)^{S-n}$, where $p=1/R$. – 2011-06-16
-
0@joriki: sorry for the semantic errors, I'm not too familiar with math jargon. But how does rolling 4 dice not generate a permutation? – 2011-06-16
-
0See http://en.wikipedia.org/wiki/Permutation. A permutation is an arrangement of different elements without repetition; rolling $4$ dice doesn't generate all of the $6$ possible values and may lead to repetitions. An example of a permutation of the values $1$ to $6$ would be $2,5,4,3,1,6$; an example of the result of rolling $4$ dice would be $2,3,6,3$. – 2011-06-16
-
0@joriki: Ah, ok. My definition of the word permutations was clearly way off. – 2011-06-16
1 Answers
1
For your example, the probability is $$ \sum_{n=2}^4\binom{4}{n}\left(\frac{1}{6}\right)^n\left(\frac{5}{6}\right)^{4-n} $$ as this is the sum of the probabilities that you roll exactly $2$ sixes, exactly $3$ sixes, or exactly $4$ sixes in four rolls using the binomial distribution. In general, the formula will given by $$ \sum_{n=X}^S\binom{S}{n}p^nq^{S-n} $$ assuming there is a probability $p=\frac{1}{R}$ of the object having value $V$, and probability $q=1-p$ of the object not having value $V$.
-
0@yunone: I have to admit I don't quite understand the notation, and I didn't expect the answer to be this complicated. But I will figure it out. Thanks a bunch :) – 2011-06-16
-
0@Matt: To understand the notation, see http://en.wikipedia.org/wiki/Binomial_coefficient and http://en.wikipedia.org/wiki/Sigma_notation. For why this is the correct answer, see http://en.wikipedia.org/wiki/Binomial_distribution. Note that despite the similarity in appearance, in $\binom{4}{n}\left(\frac{1}{6}\right)^n\left(\frac{5}{6}\right)^{4-n}$ the first factor is a binomial coefficient whereas the other two are just fractions in parentheses. – 2011-06-16
-
0@Matt, sure thing. What notation do you not understand? It's all straightforward multiplication and addition, so it's not as bad as it may seem at first. – 2011-06-16
-
0@ joriki: So if I understand correctly then, the notation is equivalent to $$ \left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{4-2}+ \left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^{4-3}+ \left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^{4-4} $$ , which simplifies to 25/1296 + 5/1296 + 1/1296 = 31/1296. But that can't be right. I don't know what I'm supposed to do with the binomial coeffeicient. – 2011-06-16
-
0@Matt, you evaluate the binomial coefficient in each term as well. $\binom{n}{k}$ is shorthand for $n!/(k!(n-k)!)$. So for example, $\binom{4}{2}=4!/(2!(4-2)!)=4!/(2!2!)=24/(2\cdot 2)=6$. I take it you're familiar with the factorial? If not, $n!=1\cdot 2\cdot 3\cdots (n-1)\cdot n$, the product of all positive integers up to $n$. – 2011-06-16
-
0@yunone: Ah, ok. No I wasn't actually familiar with factorial, but I'm glad to know it. Out of curiosity, what is the rational behind binomial coefficient? and how does it fit (conceptualy speaking) into this formula? – 2011-06-16
-
0@Matt, $\binom{n}{k}$ counts the number of ways to "choose" $k$ flips out of $n$ flips which will give a $6$. You need to include $\binom{4}{2}$ in counting the probability that you get exactly $2$ sixes, since they could happen at different flips. You might flip sixes the first two flips only, or you might flip sixes the first and last flip only, or flip sixes the first and third flips, so you need to count all these possible combinations to get the full probability. – 2011-06-16
-
0@Matt: For the ping to work, there shouldn't be a space between the '@' and the name. (That's why I didn't respond right away.) – 2011-06-16
-
0@Matt Munson: What mathematicians call $\binom{n}{r}$ can be found on most scientific calculators as a key label that looks like ${}_nC_r$. But do calculate a number of small cases by hand! – 2011-06-16
-
0@Matt: Quite generally speaking, these sorts of things are usually well explained on Wikipedia, and it's usually worthwhile to read the corresponding article if you're not familiar with a concept. In the present case, yunone's answer was contained in the articles I'd linked to; see in particular this section of the article on the binomial distribution: http://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function. – 2011-06-16
-
0Btw, I meant to say roll not flip in my above comment. Heh, got a little mixed up. – 2011-06-16
-
0Thanks allot guys. I learned much more than I expected here. Cheers. – 2011-06-16