So first let $X$ be the number of independent coin tosses until first head; $P(H)=p$. Suppose we have observed that the first coin toss is tails, what is the expectation $E(X-1|X>1)$?

So intuitively this is just equal to $E(X)$ which makes sense to me. I tried to derive this more formally:

$E(X-1|X>1)=sum_xg(x)p_{X-1|X>1}(x)=sum_x(x-1)p_{X-1|X>1}(x)$

Since a geometric random variable is memoryless, we have:

$E(X-1|X>1)=sum_x(x-1)p_{x}(x)=E(X)-1$

I seem to be off by a factor of 1, I suspect it might be something to do with the fact that I am still summing over all $x$ as denoted in $sum_x$ but I am not quite sure how to fix this.