# Dice probability over multiple rolls.

What is the probability of rolling one or more 6's using 3 six sided dice (1...6) that are rolled three times?

How does multiple rolls influence the probability, is it simply 3 times the probability of a single roll?

• No its not 3 times one roll as if you get a 6 on the the first roll the other rolls are irrelevant and could generate more 6's which you don't want to count as additional wins. As pointed out in another answer its useful to workout the probability of no 6's. Aug 17, 2013 at 17:42

I find it easier to consider the problem of not rolling a 6 with any of the three dice (the same problem with a different definition).

The chance of not rolling a 6 is 5/6. The chance of not rolling a six on any of your dice is therefor 5/6 * 5/6 * 5/6.

The chance of rolling at least one six is therefore the opposite of this:

1 - (5/6)^3 = 0.421

using bernoilles trials , if an event is repeated $n$ times and if probability of success is $p$ and $q$ the probability of failure . then the event happening $r$ times in $n$ trials is $$P(r) = ^{n}C_{r}p^{r}q^{n-r}$$ here $r =1,2,or3$. as you are requiring consecutive terms $p= \frac {1}{6}$ and $q =\frac{5}{6}$. $n =3$ then $$P = P(1) + P(2) + P(3)$$

Hint: We have a total of $$9$$ independent dice rolls. First find the probability of no $$6$$.

Remark: In direct answer to your question, no, that is not correct, there is no reason to multiply by $$3$$, and it will not give the right answer.

But you might find the probability of no $$6$$ on one roll of the three dice, and then find the probability this happens three times in a row, However, the approach suggested in the hint is a little shorter.

• Please let me know if I'm on the right track.. 5^9 / 6^9 = .194 (approx) 1 - .194 = .80 (approx) Aug 17, 2013 at 17:56
• Yes, the probability of no $6$ is $(5/6)^9$. So the probability of $1$ or more $6$ is $1-(5/6)^9$. Aug 17, 2013 at 17:58