Suppose you roll a die until you roll three six. Calculate the probability that you need exactly 5 throws for this.
My solution is the following, however it supposed to be wrong since the 'correct' answer is $0.0193$ (a colleague verified this with a simulation).
Solution:
Suppose
$$G= \text{exactly five throws is needed to throw three times 6}$$
It is clear that $G$ is determined by all events where the fifth throw is a $6$ and where within the first four throws there are exactly two $6$. You can partition $G$ in two sets, namely
- a) $\omega=(6,6,x,x,6)$ where $x$ is a number from $\{1,2,3,4,5\}$, or
- b) $\omega'=(6,6,x,y,6)$ where $x,y$ are different numbers from $\{1,2,3,4,5\}$.
The probobality of an a) like event is
$$P(\omega)=\frac{1}{6}\cdot\frac{1}{6}\cdot \frac{5}{6}\cdot \frac{1}{6} \cdot \frac{1}{6}=\left(\frac{1}{6}\right)^4\cdot \frac{5}{6} $$ since the 5 throws are statistically independent.
The probablity of an b) like event is
$$P(\omega')=\left(\frac{1}{6}\right)^3\cdot \frac{5}{6}\cdot \frac{4}{6}.$$
To compute $P(G)$ we need to take the arrangements into account. With a) you have $\overline{P}_4^{2,2}=\frac{4!}{2!2!}$ different arrangements. And with b) there are $\overline{P}_4^{2,1,1}=V_4^2=\frac{4!}{2!}$ different arrangements.\
The probablity we are after is
$$P(G)=\frac{4!}{2!2!}\cdot \left(\frac{1}{6}\right)^4\cdot \frac{5}{6}+\frac{4!}{2!}\left(\frac{1}{6}\right)^3\cdot \frac{5}{6}\cdot \frac{4}{6}=0.0347 $$
Best Answer
I have called out the error in your working in the end. Throw of die will follow the binomial distribution and here is a simple way to look at this -
Of the first four throws, two are $6$ with probability $\frac{1}{6}$ for a throw and other two are any of the remaining five ($1, 2, 3, 4, 5$) with a probability $\frac{5}{6}$ for a throw.
Any of the two out of four throws can be $6$ and remaining two throws are decided automatically by that. Also the last throw has to be six and probability of that is again $\frac{1}{6}$.
So the desired probability is $ = {4 \choose 2} \times (\frac{1}{6})^2 \times (\frac{5}{6})^2 \times \frac{1}{6} = \frac{25}{1296} \approx 0.0193$
EDIT: Coming to mistake in your working - in the second part of your working, please note that you are choosing two non-six numbers by $5 \cdot 4$ which assumes both positions are distinct. For example, if the first number is $5$, four choices of second number will give you $5, 1; 5, 2; 5, 3; 5, 4$. But when the first number is say $1$, the second number can be $5$. So you get both $5, 1$ and $1, 5$. Then when you are arranging positions of first $4$ numbers, you will again rearrange $5, 1$ as $1, 5$ and vice versa. That leads to duplicates. So there are two ways to fix your working. Keep the positions unnamed by dividing $5 \cdot 4$ by $2!$ or while making arrangements, go with $\frac{4!}{2! \ 2!}$. One of $2!$ is to account for the fact that two non-six numbers do not need to be arranged within and only need to be arranged with respect to six.