1
$\begingroup$

This is probably (...) a very basic probability question, but I'd like to be sure about it.

I have 6-sided dice. When rolled, I count one success per dice if the number on the dice is 5 or more.

for each dice, I have a probability of 1/3, but what's the probability of having at least 1 success when rolling n dice at a time?

$\endgroup$
2
  • 1
    $\begingroup$ This is a generalization of the question asked at stats.stackexchange.com/questions/53154, which is directed at the cases $n=1$ and $n=2$. One of the answers gives a formula for all $n$ and any types of dice (not just six-sided): see "Solution 2" at stats.stackexchange.com/a/53160. $\endgroup$ – whuber Jan 15 '15 at 20:19
  • 1
    $\begingroup$ The usual trick on "at least one success" type problems is to work out the probability of the complementary event ("no successes"), which is usually simple enough to write down immediately. $\endgroup$ – Glen_b Jan 16 '15 at 1:21
4
$\begingroup$

You are describing the binomial distribution. The binomial distribution specifies the probabilities of $x$ number of successes in $n$ independent trials each with probability of success $p$. In your case, $x=1$ and $p=1/3$.

You can answer this question with the following calculator: http://stattrek.com/online-calculator/binomial.aspx. You can also do this in R with the function pbinom.

If you are interested in learning more about the binomial distribution and want a good textbook I recommend A First Course in Probability by Sheldon Ross. If you want a less in-depth look, I recommend this video: https://www.khanacademy.org/math/probability/random-variables-topic/binomial_distribution/v/binomial-distribution

$\endgroup$
2
  • $\begingroup$ Great, thanks! would you have an idea of the formula (for P(X >= 1)) used on the website? I need to program it from scratch. $\endgroup$ – Lucien S. Jan 15 '15 at 18:51
  • 1
    $\begingroup$ $P(X >= 1) = 1 - P(X=0) = 1 - (1-p)^n$; I highly recommend the Khan Academy video. $\endgroup$ – TrynnaDoStat Jan 15 '15 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.