Probability of getting 6 heads in a row from 200 flips and intuition about this high value

Here is how to calculate the exact answer. Consider a Markov chain $X_0,X_1,\ldots,X_{200}$, taking integer values in the range $0\le X_n\le 6$, with transition matrix (with row and column indices in the range $0\le i,j\le6$) $$M=\pmatrix{\frac12&\frac12&0&0&0&0&0&\\ \frac12&0&\frac12&0&0&0&0&\\ \frac12&0&0&\frac12&0&0&0&\\ \frac12&0&0&0&\frac12&0&0&\\ \frac12&0&0&0&0&\frac12&0\\ \frac12&0&0&0&0&0&\frac12\\ 0&0&0&0&0&0&1}$$ Here the idea is that $X_n$ represents the number of consecutive heads ending at flip $n$ (with the conventional courtesy value $X_0=0$) so that the flip sequence HTHH would cause $X_0=0$, $X_1=1$, $X_2=0$, $X_3=1$, $X_4=2$, and so on. Except the value $X_n=6$ means something different: either $X_{n-1}=6$ or the $n$-th flip was H and $X_{n-1}=5$. The chain is started with the value $X_0=0$; what is sought is the probability that $X_{200}=6$. This is the $(0,6)$-th entry in the matrix $M^{200}$. When I do these calculations I get a value very close to $.8$.

Here is a way to visualize this chain. There is a small spider that aspires to climb to the top of a $6$ segment pipe. It starts at the bottom, and does the following $200$ times:

  • if it is at the top, it stays at the top,
  • otherwise it flips a coin and drops all the way to the bottom if the coin shows T, and climbs up more more segment if the coin shows H.