Consider a sequence of identically distributed (but possibly dependent) Bernoulli random variables $X_i$ for $i=1,2,\ldots$, with $P(X_i=1)=p$. Let $Y$ be the index of the first $1$, and say the $X_i$ depend on one another in such a way that $Y$ is almost surely finite and has an expectation (to rule out things like all $X_i$ being equal). For instance, if the $X_i$ turn out to be independent then $Y$ is geometrically distributed with expectation $1/p$.
Must the expectation of $Y$ be $1/p$?
I considered this question because if you toss a coin or a die looking to get a certain sequence (of length $n$, say), then the random variables $X_i$ given by "The $(n+i-1)$th toss finishes the sequence" is exactly such a sequence as described above, and it does have expectation as described above. So I'm basically wondering whether there is something nice about looking for sequences of tosses that gives a nice expectation, or whether it's inevitable.
Consider the bernoulli sequence given by:
$X_1 \sim$ Bernoulli($p)$,
$X_i = X_{i-2}$, $i$ odd.
$X_{i} \sim $Bernoulli$(p)$, $i$ even iid.
$E[Y] = p+\sum_{i=1}^{\infty}(2i) p(1-p)^{i}= p+\frac{2(1-p)}{p}$
The idea here is that $Y$ has effectively half as many goes at being $1$, since $Y \neq 2i+1, i \in \mathbb{N}$. Thus the expectation is changed.