Bounding Bernoulli trials by the standard Bernoulli process Suppose we have a Bernoulli-like process $P$. At each step a coin is tossed and the outcome ("success", "failure") is recorded. What differentiate $P$ from the standard Bernoulli process, is that we pick a probability of "success" uniformly at random in range $(1/2, 1)$ at each step before we toss the coin.

I'm interested in finding an upper bound on the expected number of trials 
  until the first "success" is tossed.

What I have thought, if the probability of "success" is at least $1/2$, then at each step $P$ is more probable to stop than a standard Bernoulli process, therefore an expectation of a standard geometrically distributed variable bounds from above the expectation of steps until the first "success".

How can I make this claim formal?


While, kimchi's answer is an answer to the problem as I have stated it in the first place ...
I want to share an approach, that tackles directly the expectation bound.
Suppose, we have a series of independent Bernoulli trials $X_i$ each with a probability of success $p_i \geq \frac{1}{2}$. And a series of standard i.i.d. Bernoulli trials $Y_i$ with a probability of success $p = \frac{1}{2}$.
Define by $X$ - the index of the first success in the series $X_i$, and by $Y$ - an index of the first success in the series $Y_i$.
We have that $Y \sim Geom(\frac{1}{2})$, and $\mathbb{E}(Y) = 2$.
We ask, what is $\mathbb{P}(X > k)$ ?
In other words what is the probability that the first success in series $X_i$ happens after the $k^{th}$ trial. The answer can be calculated in a straight-forward way:
$$\mathbb{P}(X > k) = \prod\limits_{i=1}^k(1-p_i),$$ as all trials before and including $k^{th}$ should fail.
We further use the fact that $p_i \geq p$ to show that
$$\mathbb{P}(X > k) = \prod\limits_{i=1}^k(1-p_i) \leq \prod\limits_{i=1}^k(1-p) = \mathbb{P}(Y > k)$$
Recall, that for a discrete variable $Z$(such as $X$ and $Y$) taking values in $\{1, 2, \ldots \} \cup \{ +\infty\}$
$$\mathbb{E}(Z) = \sum\limits_{k = 1}^{\infty}\mathbb{P}(Z > k)$$
Now sum the probabilities to get the expectations:
$$\mathbb{E}(X) = \sum\limits_{k = 1}^{\infty}\mathbb{P}(X > k) \leq \sum\limits_{k = 1}^{\infty}\mathbb{P}(Y > k) = \mathbb{E}(Y)$$
Thus a direct upper bound is shown.
