My question
Lets say we have a fair 6-sided die that we roll continuously. We roll this same dice 10 times in a row. What is the probability that we get a 3 after our 4th roll but before our 10th roll?
In other words, given the events
$$R_1,R_2,R_3,R_4,R_5,R_6,R_7,R_8,R_9,R_{10}$$
What is the probability that
$$R_5,R_6,R_7,R_8,R_9$$
We get our first 3?
My attempt
$R_1,R_2,R_3,R_4$ cannot be 3s. This means that they have probabilities $\frac{5}{6}$.
The first 3 appearing somewhere here $R_5,R_6,R_7,R_8,R_9$ is probably something like $(5 \text{ choose } 1) * 6^4$ (I have no reasoning, just intuition).
I have no idea how to handle the last part of the problem with "before $R_{10}$"…
Best Answer
I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)
To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.
$$ X \sim Bin\Big(n=5, p=\frac{1}{6}\Big) $$
(Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $\frac{1}{6}$)
We want the probability:
$$ Pr(X > 0) = 1- Pr(X=0) = 1 - \Big(\frac{5}{6}\Big)^5 $$
So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result: $$ P(\text{No 3's in the first four rolls} \cap \text{at least one 3 in rolls 5 to 9} ) = \Big(\frac{5}{6}\Big)^4 \Big( 1 - \Big(\frac{5}{6}\Big)^5 \Big) $$
Which I believe works out to be $\approx0.288$