Expected value of number of throws of a dice to get element $1$ four times

Question

Here is the question:

Find the expected value and variance of the number of times one must throw a dice until the element $1$ has been obtained $4$ times.

My attempt:

The minimum number of throws required to get $4$ ones is $4$ when each of the subsequent throws gives a result of $1$. Using binomial, the probability of this case is ${4 \choose 4}(\frac{1}{6})^4(\frac{5}{6})^0$.

Similarly, for the case when we throw the dice $5$ times and obtain $1$ four out of the five times, the probability becomes ${4 \choose 4}(\frac{1}{6})^4(\frac{5}{6})^1$.

I was able to generalize the expectation as: $$E(n) = \sum_{n=0}^{\infty}(n+4){n+4 \choose 4}(\frac{1}{6})^{4}(\frac{5}{6})^n$$

I'm stuck here and can't think of any way to go forward.

Answer 1

An easy way to get the expected number of trials is to use linearity of expectation. Let $X_1$ be the number of trials until the first $1$ is rolled, $X_2$ be the number of trials after the first $1$ until the second $1$, and define $X_3$ and $X_4$ similarly. Then $X=X_1+X_2+X_3+X_4$ is the number of rolls until the fourth $1$. We have $E(X)=4E(X_1)=4\cdot6=24$ since the number of Bernoulli trials with probability $p$ until the first success is $1/p.$

Also, since the $X_i$ are independent, the variance of the sum is the sum of the variances, and $Var(X)=4Var(X_1).$ I leave it to you to continue from here.

Answer 2

Notice that this is a negative binomial random variable for which we want to find the probability that it takes $n$ trials to get $k$ successes. Let's denote this by $X$. The pdf of $X$ is of the form

$$P(X=n)={n-1 \choose k-1}p^k (1-p)^{n-k}$$

since we want to get $k-1$ successes on the first $n-1$ trials and then another success of the $n$-th trial. Here, $k$ is fixed at $4$, and $p=\frac{1}{6}$ so we get

$$P(X=n)={n-1 \choose 3}\left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^{n-4}$$

Then the expected value would just be

$$E(X)=\sum_{n=4}^{\infty} n\cdot{n-1 \choose 3}\left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^{n-4}=24$$

since as you said, the support of $X$ is $(4,5,...,)$

Can you go from here to obtain the variance?