I haven't had to do much with probabilities since university, so please excuse if this is trivial or the question is not well specified.
Let $X$ and $Y$ be two independent, normally distributed random variables with means $\mu_x $ and $\mu_y$, respectively and standard deviations $\sigma_1 = \sigma_2 = 1$. Having observed one "draw" of $X$ and $Y$ ($x_1$, $y_1$) with $x_1 > y_1$, what is the probability that $\mu_x > \mu_y$?
In case the following question is easier to answer, an answer to it is even preferred:
Under the observation that $x_1 - y_1 = \delta$ , what is the probability that $\mu_x > \mu_y$ (as a formula of $\delta$)?
Thanks in advance.
I would have thought that unless you want to get tangled in the language of one-tailed hypothesis tests, a Bayesian approach might be best.
But even then you would need a prior distribution for $\mu_x-\mu_y$; let's suppose there is an improper prior where this difference (call it $D$) is uniformly distributed. Then given $D$, the likelihood of observing $\delta$ is proportional to $\phi\left(\frac{\delta-D}{\sqrt{2}}\right)$ and so the posterior density for $D$ is $\dfrac{\phi\left(\frac{\delta-D}{\sqrt{2}}\right)}{\int \phi\left(\frac{\delta-D}{\sqrt{2}}\right)\,dD} = \frac{1}{\sqrt{2}} \phi\left(\frac{\delta-D}{\sqrt{2}}\right)$.
So taking this prior, the posterior probability that $D \gt 0$ given $\delta$, i.e. the posterior probability that $\mu_x \gt \mu_y$ is $$\Phi\left(\frac{\delta}{\sqrt{2}}\right).$$