Let a fair dice be thrown $n$ times. If we count the number of time when the sequence $123456$ appears; what is its expected value?
Clearly it is $0$ if $n<5$; and my guess is that when $n=6,7,\cdots, 11$, it should not be very difficult but when $n\geq 12$ then the question becomes more complicated as it is possible for the sequence to appear more than one times.
For $i=1$ to $n$, define Bernoulli random variables $X_i$ by $X_i=1$ if at $i$ we have the beginning of the sequence $123456$, and by $X_i=0$ otherwise. Then $Y=\sum_1^n X_i$ is the number of times the sequence $123456$ appears.
By the linearity of expectation, we have $E(Y)=\sum_1^n E(X_i)$. But for any $i$ which is not too large, $E(X_i)=\Pr(X_i=1)=\frac{1}{6^6}$. It follows that $E(Y)=\frac{n-5}{6^6}$.
Remark: This is an example of the power of the Method of Indicator Random Variables. The distribution of $Y$ is unpleasant to get at, but $E(Y)$ is not difficult to calculate.