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?
Please simplify ; )
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$\begingroup$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
$\endgroup$ $\begingroup$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 $$P = P(1) + P(2) + P(3)$$
$\endgroup$ $\begingroup$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.
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