Let $Z \sim\text{Geometric}(0.5)$. Let's say we're using $Z$ to model the number of times we need to flip a fair coin (independently of previous flips) to get heads. Let's say we know that the first coin flip is tails, and we are now interested in the number of flips after the first until getting heads. Let's call $Y$ the number of flips after the first until getting heads.
Clearly, for a given value of $Z$, the value of $Y$ will be one less (since it doesn't include the first flip). Then it seems to make sense that $Y = Z - 1\Rightarrow \mathbb{E}[Y]=\mathbb{E}[Z-1]=\mathbb{E}[Z]-1=2-1=1$.
On the other hand, the memoryless property of the geometric distribution seems to suggest that we have $Y \sim\text{Geometric}(0.5)$, which implies $\mathbb{E}[Y]=2$.
Why do these two lines of reasoning disagree, and which is correct?
We have: $$Z=1-T+T(1+Y)=1+TY$$ where $T$ corresponds with tails at the first flip.
This with $T\sim\mathsf{Bernoulli}(0.5)$ and $Y\sim \mathsf{Geometric}(0.5)$ where $T$ and $Y$ are independent.
Then: $$\mathsf EZ=\mathsf E(1+TY)=1+\mathsf ET\mathsf EY=1+0.5\cdot2=2$$
Next to that we have:$$\mathsf E [Z\mid T=1]=\mathsf E (1+Y)=1+\mathsf EY=1+2=3$$
You came up with $\mathsf EY=\mathsf EZ-1$. This is wrong and should have been $\mathsf EY=\mathsf E [Z\mid T=1]-1$.