Variance of fair die roll - two in a row

If you had a fair die roll where you rolled until you got the same number twice in a row, you'd almost get the geometic distribution.

I think you can use the geometric approach to calculate the expected number of rolls if you just add the one roll back.

For example:

$E(X)=$$(1-p)^(k-1^)p$ $\,$ for $k=1,2,3...$ would work out when you shift things to the right by 1 and start at $k=2$, so you can treat the problem after $2$ geometric and add back the one to the result. So you'd get $E(X) =$ $1/$$1 \over 6$ $\,$ because if the random variable X has the geometric distribution with parameter p, then $E(X) =$ $1 \over p$. $\quad$Ultimately, $E(X) = 6 + 1 = 7$

For the variance of a geometric distribution with paramater $p$ you'd have $(1-p)/(p^2)$ normally. How do you account for the additional roll here? Would $V(X)$ be ($1-$$1\over 6$)$/$$1 \over 6 $$^2$ $= 30$ ? Or would it be $30+1$? Or something else entirely?

You have already done the important work when you noticed that $X-1$ has the $\text{Geometric}(\frac16)$ distribution.

For expected value, you are right that $\mathbb E[X-1] = 6$ tells us that $\mathbb E[X] = 7$.

For variance, knowing $\text{Var}[X-1] = 30$ also tells us $\text{Var}[X]$, but the logic is different. Variance is not changed by adding a constant, so $\text{Var}[X]$ will also be $30$.

This can be done algebraically, by using the definition of variance: either $\text{Var}[X] = \mathbb E[(X - \mathbb E[X])^2]$ or $\text{Var}[X] = \mathbb E[X^2] - \mathbb E[X]^2$. Use one of these to simplify $\text{Var}[X-1]$, and you'll see that you get the same thing as $\text{Var}[X]$.

However, this relationship also has an intuitive explanation. Variance measures the average squared distance from the mean. In every outcome of our random experiment, the distance between $X$ and its mean of $7$ is the same as the distance between $X-1$ and its mean of $6$. So the squared distance is always the same - in particular, the average squared distance is the same. Therefore $\text{Var}[X-1] = \text{Var}[X]$.