Probaability of a neverending random process

Question

Roll a fair die forever. What is the relative frequency of rolling exactly three consecutive sixes, i.e., 3 consecutive sixes and a non-six for the first four rolls, then subsequently a non-six, three sixes, and a non-six? Note that this process is not the same as rolling five times with the same criterion for success. It has no trials because the isolated three sixes can occur anywhere in the infinite sequence of rolls, not just in five trial intervals. I don't know how to deal with a stochastic process that has no trials. How do we calculate the limit of the ratio of successes to number or rolls? Is it zero? Am I wrong in thinking that it's different from a five roll trial with its very simple probability?

Answer 1

Let have a dice rolling forever. At each time, $t$ discrete, we have a dice result $k$.

At time $t$, the probability of having the sequence $abbba$ from time $t+1$ to $t+5$ is: $$ p(t)={5\cdot 1\cdot 1\cdot 1\cdot 5\over 6\cdot 6\cdot 6\cdot 6\cdot 6}={5^2\over 6^5} $$

At any point, the probability of having the whole sequence having just a part completed is then given by the probabilities of the events. Hence: $$ p(t|a)={1\cdot 1\cdot 1\cdot 5\over 6\cdot 6\cdot 6\cdot 6}={5\over 6^4}=p_4\\ p(t|ab)={1\cdot 1\cdot 5\over 6\cdot 6\cdot 6}={5\over 6^3}=p_3\\ p(t|abb)={1\cdot 5\over 6\cdot 6}={5\over 6^2}=p_2\\ p(t|abbb)={5\over 6}=p_1\\ p(t|abbba)=p(t|a)=p_4 $$

Hence this is a discrete machine with 5 relevant states, and other complementary states, defining a matrix of transition probabilities between each states. When an $abbba$ happens, a counter state is increased, and when $a$ happens, the machine returns to state $p_4$. Hence a markov state space model describe the process fairly well.

From here you can do almost anything you could want, such as calculating the probability of having each state when $t \to \infty$.

Answer 2

Let $X_i$ be the random variable defined as $1$ if the $i$'th throw is the first of the streak "non-6, 6, 6, 6, non-6", and $0$ otherwise. Then $E(X_i) \approx 0.0032$. The number of times that we get a streak of exactly three sixes when throwing the die $n$ times is $\epsilon + \sum_{i = 1}^nX_i$. The error correction term $\epsilon$ is there because we haven't taken into account that we might start with three sixes, and we haven't taken into account that we might cut off in the middle of a streak. Note, however, that $\epsilon$ is only ever $0$ or $\pm 1$.

The expected value of this sum is $$ E\left(\epsilon + \sum_{i = 1}^nX_i\right) = E(\epsilon) \sum_{i = 1}^nE(X_i) \approx \sum_{i = 1}^n0.0032 = 0.0032n + E(\epsilon) $$ The expected relative density is therefore given by dividing this by $n$, which gives $0.0032 + \frac{E(\epsilon)}{n}$. As $n\to \infty$, the error correction term goes to $0$, and we're left with just $0.0032$.