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The probability of not getting a five is $(\frac56)^3$, and I figure the probability of getting at least one 5 is $1-(\frac56)^3$, but I don't know how to figure out if it is rolled at least twice. Thoughts? Thanks in advance!

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    $\begingroup$ While it does not help at all in figuring out the formula or reasoning behind the result (and that's one of several reasons why I'm not suggesting it as an answer) anydice.com is an excellent tool for either getting immediate result data or exploring the solution space of probability questions like this. Here's a link to your specific question: anydice.com/program/1090 - it shows the raw probabilities but also the cumulative probabilities and other related statistics. $\endgroup$
    – Ben
    Jan 20, 2015 at 13:30

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The long winded brute force method would be to add up the probabilities of the four outcomes that would give you the desired result. $$5,5,5\\5,5,x\\5,x,5\\x,5,5\\\text{Where x is any number other than 5.}$$

So, these outcomes would have the probabilities of $$\frac{1}{6}^3\\\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}\\\frac{1}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}\\\frac{5}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}$$

And you would add up these probabilities.

The more general way to look at it would be to, look into the binomial distribution, as suggested by a different answer to this question. Also, @robjohn gives a good explanation on how to find the probabilities for three and two 5's.

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  • $\begingroup$ That's really clear! Thanks! I learned about the binomial distribution last year, and I knew it had something to do with it, but I didn't know how to use it. $\endgroup$ Jan 20, 2015 at 2:42
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The number of ways to arrange $k$ fives out of $3$ dice is $\displaystyle\binom{3}{k}$

The probability of each arrangement, $k$ fives and $3-k$ others, is $\displaystyle\left(\frac16\right)^k\left(\frac56\right)^{3-k}$

Thus, the probability of getting exactly $k$ fives out of $3$ dice is $\displaystyle\binom{3}{k}\left(\frac16\right)^k\left(\frac56\right)^{3-k}$

So the probability of getting at least $2$ fives would be $$ \sum_{k=2}^3\binom{3}{k}\left(\frac16\right)^k\left(\frac56\right)^{3-k} $$

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There's a famous distribution that answers the exact question you pose; namely the binomial distribution.

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  • $\begingroup$ Bernoulli distribution is meant for a single trial only and takes values of 0 or 1. Binomial distribution is what you mean to say since it is for several trials. $\endgroup$
    – JMoravitz
    Jan 20, 2015 at 2:26
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Since you're performing multiple experiments each with exactly two outcomes (the dice is a five vs. the dice is not a five), this is a classic example of a Bernoulli trial.

The standard formula to calculate the probability is then to use (from the Binomial distribution):

$$ P\{\text{that A occurs exactly k times in n trials}\} = \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) p^k (1 - p)^{n-k} $$

where event $A$ is the trial and $p$ is the probability for the success of a single trial ($\frac{1}{6}$).

In your case, you want to know the probability of it happening at least twice, so you need to sum the chances for it happening exactly twice and exactly three times:

$$ P\{\text{that A occurs exactly 2 times in 3 trials}\} + P\{\text{that A occurs exactly 3 times in 3 trials}\} $$

Therefore, $$ \begin{align} P&= \left( {\begin{array}{*{20}c} 3 \\ 2 \\ \end{array}} \right) p^2 (1 - p)^{(3-2)} + \left( {\begin{array}{*{20}c} 3 \\ 3 \\ \end{array}} \right) p^3 (1 - p)^{(3-3)} \\ &= \left( {\begin{array}{*{20}c} 3 \\ 2 \\ \end{array}} \right) \cdot \frac{1}{6}^2 \cdot \frac{5}{6}^{1} + \left( {\begin{array}{*{20}c} 3 \\ 3 \\ \end{array}} \right) \cdot \frac{1}{6}^3 \cdot \frac{5}{6}^{0}\\ &= 3 \cdot \frac{1}{36} \cdot \frac{5}{6} + 1 \cdot \frac{1}{216} \cdot 1\\ &= \frac{15}{216} + \frac{1}{216}\\ &= \frac{16}{216}\\ &= \frac{2}{27} = 0,074... = 7,41 \% \\ \end{align} $$

As you can see, this method of calculation can unfortunately become very tedious for larger examples, but it has the advantage of being a consistent approach for any problems surrounding Bernoulli trials. The intuitive approaches mentioned in some of the other answers can be quicker for small cases where all the possible outcomes can be manually considered.

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While Lee's answer is correct, if you do not know about the Binomial distribution, you can try this:

1) The complementary event is that you get $0$ or $1$ fives.

2) Probability of getting $0$ fives has been computed by you.

3) Let's compute probability of $1$ five: It can occur in $1st$, $2nd$ or $3rd$ try: each of them have the same probability, namely: $\dfrac{1}{6}\dfrac{5}{6}\dfrac{5}{6}$.

So the total probability for this case is: $3\dfrac{1}{6}\dfrac{5}{6}\dfrac{5}{6}$.

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  • $\begingroup$ I think you missed the case wheren three fives are rolled, $\endgroup$
    – Jasen
    Jan 20, 2015 at 10:35
  • $\begingroup$ Well it is correct, but cryptic. The numerator (over $6^3=216$) is $6^3-5^3-3\times 5^2=216-125-75=16$. $\endgroup$ Jan 20, 2015 at 10:55
  • $\begingroup$ Ah, now i see. thanks. $\endgroup$
    – Jasen
    Jan 20, 2015 at 11:07
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$\displaystyle\frac{2}{27}$

There are $\displaystyle6^3$ ways (which is 216) to roll three dice (or to roll a dice three times, all equally probable) There are 15 ways to get only a pair of fives and one way to get three fives, there are no other ways to get at-least two fives so the answer is

$\displaystyle\frac{16}{216}$ which is $\displaystyle\frac{2}{27}$

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  • $\begingroup$ Welcome to our site! $\endgroup$ Jan 20, 2015 at 10:31
  • $\begingroup$ @MarcvanLeeuwen thank you. $\endgroup$
    – Jasen
    Jan 20, 2015 at 11:04

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