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.

  • $\begingroup$ 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

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.


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?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.