Roll a fair 6−sided die until a 6 appears.Given that the first 6 occurs before the first 5, find the expected number of times the die was rolled

Question

I am trying to solve this problem but my answer is coming out to be 1.5 where the answer is 3. I even ran a simulation to verify it and its correct. I am not really sure whats wrong in my method and how can I correct it.

I considered for getting my first 6 on Nth trials, without any 5's till now, I have probability of $$P(A^N) = \left(\frac{4}{6}\right)^{N-1} \times \frac{1}{6}$$

Now to get the expected number of rolls,

$$\sum_{N=1}^{\infty} N \times \left(\frac{4}{6}\right)^{N-1} \times \frac{1}{6}$$

So, If we bring 1/6 outside, and consider x = 4/6 then the above equation is the 1st derivative of x/(1-x)

$$\frac{d}{dx} \left( \frac{x}{1 - x} \right) = \frac{1}{{(1 - x)}^2}$$

Putting back the values and multiplying by 1/6 we get 1.5 (9/6). Where am I going wrong?

Answer 1

Roll a fair 6−sided die until a 6 appears.Given that the first 6 occurs before the first 5, find the expected number of times the die was rolled

Let $N$ be the number of rolls until you stop, and $\mathcal C$ be the condition.

• $(1/6)$ is the probability for rolling a 6 (which is not a 5, of course).
• $(4/6)$ is the probability for rolling neither a 5 nor a 6.
• $(4/6)^{n-1}(1/6)$ is the probability for never rolling a 5 and the first 6 being on roll n. IE: $\mathsf P(N{\,=\,}n\cap\mathcal C)$
• Thus $\color{blue}{\sum\limits_{n=1}^\infty n(4/6)^{n-1}(1/6)}$ is the expected count of rolls until the first 6 and we never roll a 5 before stopping.
  • This is $\mathsf E(N\mathbf 1_\mathcal C)$. Yes this is $1.5$, but this is not what we wanted to calculate.
  • We actually want $\mathsf E(N\mid\mathcal C)$, the conditional expectation of the count for rolls until we stop on a 6 when given that we never rolled a 5 before stopping.

$$\begin{align}\mathsf E(N\mid\mathcal C) &= \dfrac{\mathsf E(N\mathbf 1_\mathcal C)}{\mathsf P(\mathcal C)} \\[1ex]&= \dfrac{\sum\limits_{n=1}^\infty n (4/6)^{n-1}(1/6)}{1/2}\\[1ex] &= 3\end{align}$$