Rolling a die, binomial distribution

Question

I didn't get the exact thing that this question wants. Is the concept about the binomial distribution? I am not sure if I should calculate the probability or expected value. Also, this question means that after two 6 I should expect 1 or no.

A fair die is rolled sequentially. What is the probability of finding two consecutive 6s before finding a 1? In other words, think of the problem as a game that ends when you either (a) roll two consecutive 6s or (b) roll a single 1. You must account for all possible numbers of rolls, i.e., it could take one million rolls before two consecutive 6’s are rolled.

Answer 1

As mentioned, it's not a Binomial Problem since you can't presume two exclusive events of success/failure from the first experiment or die roll. It's helpful to consider the trivial case of losing on the first roll, how do you do form probabilities for the outcomes? Define three states:

• $S_0$: Initial state,
• $S_1$: Rolled a 6,
• $S_2$: Rolled a 1 (losing state).

We aim to find $P_0$, the probability of rolling two consecutive 6s before a 1 from state $S_0$, and $P_1$, the probability of the same after rolling one 6.

From state $S_0$: $$ P_0 = \frac{1}{6}P_1 + \frac{1}{6}(0) + \frac{4}{6}P_0. $$

From state $S_1$: $$ P_1 = \frac{1}{6}(1) + \frac{1}{6}(0) + \frac{4}{6}P_0. $$

Solving these gives: $$P_1 = \frac{1}{6} + \frac{4}{6}P_0. $$

From here you can substitute and solve.

Answer 2

There are multiple ways to find out the winning probability. One way is to treat the problem as a Markov process. Define four states as

$S_0$: a throw yields anyone of $2, 3, 4$ or $5$.

$S_1$: a throw yields a $1$.

$S_2$: a throw yields a first $6$.

$S_3$: two consecutive throws yield two $6'$s.

The state transfer matrix can be written as

$\textbf T=\begin{bmatrix} 4/6 & 1/6 & 1/6 & 0\\ 0 & 1 & 0 & 0\\ 4/6 & 1/6 & 0 & 1/6\\ 0 & 0 & 0 & 1 \end{bmatrix}$.

Start from the sate $S_0$ by defining a row vector $\textbf v=[1\ \ 0\ \ 0\ \ 0]$. The final state can be obtained by evaluating $\textbf F=\textbf{vT}^\infty$. In practice, $\textbf F$ can be obtained as $\textbf F=\textbf{vT}^n$ with a sufficiently large value of $n$, say with $n=50$, yielding $\textbf F =[0\ \ 7/8\ \ 0\ \ 1/8]$, indicating that the probabilities for loss and winning are 7/8 and 1/8 respectively.