Q: Does the number of times a sequence repeats in a row influence its probability? Let's say I want to know what is the chance of a certain sequence of events occurring. For example I want to know the chance that the following four events will occur in a row: A - B - C - A After calculating the chance of A, then I would multiply the result with the chance of B (in case that A happened) and multiply the result with the chance of C (in case that A and B happened). My question is: if the four events happen many times in a row does the chance of their occurring increase or decrease? Say for example that we have 100 A's in a row, then would the chance of 100 B's afterwards be 100 times the chance of an A? A: If you don't expect your experiment to be biased, you can just consider each event separately and their occurrence probability can be taken as the product of the corresponding occurrence probabilities: For instance, if you want to know the probability that a dice rolls 4 heads in a row, you can try a dice (each event will have the same probability $\frac12$ since you can choose any of the 6 faces to start from) and see what is the probability of getting 4 heads in a row. Since a die will have at least one face, the chance of getting 4 heads is just the chance of getting at least one heads (you cannot have no heads nor tails), so it will be $\frac56$. In the other hand, if you want to know the probability of $4$ dice that have been rolled $100$ times in a row, all of them with different faces, then you need to sum (in the sense of event) all the possible ways to get $4$ heads (since each dice will contribute with the same probability $\frac12$ to the probability, and there will be $\binom42$ different ways to get exactly $4$ heads). This means that the chance of getting $4$ heads will be equal to $$ \frac{\binom42}6=2 $$ A: The way the question is framed, a certain sequence of events occurring means exactly one sequence. So, in the example, the chance of B being the next event after any particular A is $1\over3$. The situation you describe could occur if, say, the chance of B occurring is $1/2$ after A and $1/2$ after not A. In that case the chance of A happening twice is $\frac12\times\frac12= \frac14$, then followed by B being the next event, so the chance of B happening twice is $$1\over3\times \frac14=\frac1{12}$$ Similarly, if the chance of B occurring is $1/3$ after A and $1/3$ after not A, the chance of A being followed by B is $1\over3\times \frac13=\frac1{9}$. A: If the occurrence probabilities are independent of each other, then the chance that the sequence ABCA occurs is simply the product of the probability of getting each one: $$\mathrm{Prob}(A) \times \mathrm{Prob}(B) \times \mathrm{Prob}(C) \times \mathrm{Prob}(A)$$ However, if they are not independent of each other, then the probability is more complicated and depends upon the joint probabilities of getting both A and then B and then C. If you knew the entire joint probabilities, then you could just multiply them all together to get the exact answer for the probability of that specific sequence. Let's see what happens if we use the naive (or "uniform") model for these joint probabilities. Suppose that all the event $A,B,C$ are equally likely. Then $\mathrm{Prob}(A) = \mathrm{Prob}(B) = \mathrm{Prob}(C) = 1/3$. Then, what is the chance of getting each of the sequences in question: $A, B, A$: If we get $A$ at some point in the sequence, we are guaranteed to get either $B$ or $C$ and so $A,B,A$ occurs. If we don't get $A$ at all, then the sequence is guaranteed to end in $A$, but since the first three events are disjoint, there's no chance of going from $A$ to $C$. So then the only sequence that can occur is $A,B,A$. This gives the probability of $1/3 \times 1/2 \times 1/3 = 1/18$. $B,C,A$: Similarly, this has probability $1/18$. $B,A,B$: This also has probability $1/18$. Finally, there are $\binom{3}{2} = 3$ different sequences of $A, B, A$ that can be formed from the three events. Since they are all equally likely, the probability of getting $A,B,A$ exactly once is then $$1/18 \times 3 = 1/6$$ so the probability of having A repeated is $1/6$ and the probability of having either $A,B,A$ or $B,C,A$ is $2/18$. Let's now see what happens if we consider the true joint probabilities of the events. Let's write out all possible combinations of occurrences of $A,B,C$ that could make it into a sequence like $A,B,C,A$: $$ \begin{array}{c|cccccc} &A&B&C&A&B&C\\ \hline A&1/18&0&1/18&0&0&1/18\\ B&0&1/18&0&1/18&1/18&0\\ C&1/18&0&1/18&0&0&1/18\\ A&0&1/18&0&1/18&0&1/18\\ B&1/18&0&1/18&1/18&1/18&0\\ C&0&1/18&0&1/18&0&1/18 \end{array} $$ From this, it's possible to work out that the actual probabilities of sequences like $A,B,C,A$ are $$ \begin{array}{c|cccccc} &A&B&C&A&B&C\\ \hline A&1/6&1/6&1/6&1/6&1/6&1/6\\ B&1/6&0&1/6&0&1/6&1/6\\ C&1/6&1/6&0&0&0&1/6\\ A&0&0&0&0&0&0\\ B&0&0&0&0&0&0\\ C&0&0&0&0&0&0 \end{array} $$ so we can indeed get a probability of $1/6$ for all three sequences.