I'm trying to understand the solution to the question below and it does not seem intuitive. So trying to understand how to derive the inputs
Q: A single fair-dice is rolled repeatedly. a) What is the probability that the 4th six appears on the 12th roll? b). What is the E(x) of total rolls needed to get 4 sixes?
Solution for a) suggests that if 4th six appears on 12th roll there are 16 initial failures. The answer in the book is : P(X=16) = C(19,3)((5/6)^16)((1/6)^4))
I don't understand how there are 16 initial failures if the dice is rolled 12 times..
Solution for b) is E[x] + 4.
I don't understand why the 4 is being added to the E[x] of the result.. as other problems in this category when it came to expected value did not add/subtract anything additional to the E[x] result.
We can think of the sequence of dice rolls as a sequence of independent Bernoulli trials with success (roll a six) probability $p=1/6$. Let $X$ be the minimum number of trials to get $4$ successes. Then $X$ is negative geometrically distributed. Question a) asks for $P(X=12)$. Note that $X=12$ corresponds to $4$ successes and $8$ failures with success on the last trial. Hence $$ P(X=12)=\binom{12-1}{4-1}p^4(1-p)^8 $$
where the binomial accounts for the number of ways to arrange the $3$ successes that are not the last success amongst the first $11$ trials.
For the second question, note that $X$ is equal in distribution to a sum of independent geometric random variables $W_1+\dotsb+W_4$ where $W_{i}$ is geometric (minimum number of trials to get one success) with success probability $p=1/6$. Here $W_i$ is the waiting time between the $i-1$ th success and $i$ th success. Hence $EX=4EW_1=4(6)=24$.