Expected number of tosses to observe a fixed number $k$ times in a row = 6 * expected number of tosses to observe any number $k$ times in a row.

Question

By "fixed," I just mean, say, observing $k$ consecutive 6s when tossing a fair die.

The latter case has the solution $E[N] = \frac{6^k - 1}{5}$, shown Expected number of rolls until a number appears $k$ consecutive times, and the former has a solution $E[N_{fixed}] = \frac{6(6^k - 1)}{5}$.

Could someone provide intuition how to go from one to the other? I think there should be a very easy transition from one to the other but it's not obvious to me what it is, but there's definitely some strong symmetry in the problem.

I also imagine the factor of $6$ generalizes to an $n$-sided die, where the factor will become $n$.

So I figured this out, but I definitely want to learn about other viewpoints.

Here's what I came up with.

Let $g,f$ denote the events that we have $k$ consecutive 6s and $k$ consecutive ANY number, respectively. Suppose $f$ occurs. It has a $1/6$ probability that it is the event $g$, i.e. $f=g$. There is thus a $5/6$ probability that $f \neq g$. In this case, we have effectively started over in our request to get to event $g$, so to get form $f$ to $g$, we require $g$ tosses, on average. And note that $g$ is a subset of $f$. So if we have that $g$ has occured, then $f$ has occured as well.

So we have

$$ E[N_{fixed}] = 1/6(E[N] + 0) + 5/6(E[N] + E[N_{fixed}]) \\ \implies E[N_{fixed}] = 6E[N] $$

Answer 1

The combinatorial argument is that you throw until you get $k$ consecutive of any type; if that is of the number you want then stop, and if not then start again.

So if $A$ is the expected number of throws until you get $k$ consecutive of any type and $B$ is the expected number of throws until you get $k$ consecutive of the type you want then $$B=A +\frac1n 0+ \frac{n-1}{n}B$$ implying $$B = nA$$ and here $n=6$

Answer 2

Let $E_1$ be the expected number of trials to see $k$ in a row of any face, and let $E$ be the expected number of trials to see $k$ sixes in a row.

First, wait until you see $k$ of anything. That is expected to take $E_1$ trials. At the end of that, two things are possible. Either you got lucky and saw $k$ sixes in a row (probability $\frac 16$) or you didn't and you have to start over (probability $\frac 56$). It follows that $$E=\frac 16\times E_1+\frac 56\times (E+E_1)\implies 6E=E_1+5E+5E_1\implies E=6E_1$$ as desired.

As you surmise, there is no difficulty generalizing this to an $n-$sided die.