What is the probability of rolling at least one $6$ in $12$ rolls of a dice?
This is a question in my Mathematics textbook.
The answer is $$1 – \left(\frac56\right)^{12} = 0.89$$
I do not understand how or why this method is used and why the answer is not $\frac16$, because if you roll the same dice $12$ times, does that not mean that the events are independent and that every roll has the probability of $1$ in $6$?
Best Answer
Yes, the events are independent, and every roll has a chance of $\frac{1}{6}$ to be a $6$. But that does not mean that getting a $6$ in $12$ rolls is also $\frac{1}{6}$.
Basically you are saying that it doesn't matter whether I roll the die $1$ time or a billion times, the probability of getting a $6$ is always $\frac{1}{6}$. But if you think about it, not getting a $6$ in $1$ roll is quite plausible, but not getting a $6$ in a billion rolls is outrageously unlikely. So, this should tell you that there is something wrong in your thinking about this.
OK, but what exactly is going wrong? Well, not getting a $6$ in 12 rolls means more events need to take place than just not getting a $6$ in 1 roll, and the more events need to take place, the less likely it becomes for all those events to take place, even if the probability for each of those events is still the same. So that is possibly where your confusion lies: you didn't distinguish between the probability of all of those events to take place and the probability of each of those events to take place.
Finally, having hopefully cleared up your confusion, let's explain the formula. Well, as I said, in order to not get any $6$ in $12$ rolls, $12$ events need to happen: $12$ times you need to not roll a $6$, and each of those individual events has a probability of $\frac{5}{6}$. So that means that there is a $(\frac{5}{6})^{12}$ probabiity for the overall even of not getting a $6$ at all to take place. Meaning that the chance of getting at least one $6$ is $1-(\frac{5}{6})^{12}$.