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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.

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  • $\begingroup$ There's an error in your formula: to achieve a fourth success on the $n$th turn, you need three previous successes out of $n-1$ attempts (in your case, the sequence $1, 1, 1, 1, 2$ would also be counted). $\endgroup$ – jvdhooft Feb 13 at 16:43
  • $\begingroup$ Also, since there are three previous successes, the factor $\frac{5}{6}$ should not be raised to the power of $n$. $\endgroup$ – jvdhooft Feb 13 at 16:46
  • $\begingroup$ I fixed the summation. @jvdhooft $\endgroup$ – Abhyudaya Sharma Feb 13 at 16:47
  • $\begingroup$ Possible duplicate of What is the expected number of trials until x successes? $\endgroup$ – jvdhooft Feb 13 at 16:55
  • $\begingroup$ My standard kvetch: You throw a single [b]die[/b]. "Dice" is the [I]plural[/I] or "die". $\endgroup$ – user247327 Feb 13 at 16:59
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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.

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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?

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  • $\begingroup$ Thanks for your answer! It makes sense. Could you please tell how you arrived at the value of $24$? $\endgroup$ – Abhyudaya Sharma Feb 13 at 16:57
  • $\begingroup$ @AbhyudayaSharma If you require $r = 4$ successes, and the probability of success is $p = \frac{1}{6}$, we have: $E(X) = \frac{r}{p} = 24$. $\endgroup$ – jvdhooft Feb 13 at 17:01
  • $\begingroup$ @jvdhooft Thanks! I'll study up on negative binomial distribution. $\endgroup$ – Abhyudaya Sharma Feb 13 at 17:04
  • $\begingroup$ I would suggest accepting the above answer instead. $\endgroup$ – Remy Feb 13 at 17:09

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