Y ~ n(mu, 1)
Ybar ~ n(mu, 1/n)
mu ~ n(0, 1/tau)

mu | Ybar ~ n(a, b^2)

b^2 = 1 / (tau + n)
a = n * Ybar * b^2
Post prob mu > 0 = 1 - Prob(mu <= 0 | Ybar)
 = 1 - Pr((mu - a) / b < (0 - a) / b)
 = 1 - Phi(-a / b) = Phi(a / b)


If tau=0:
b^2 = 1/n
a = Ybar
Post prob mu > 0 = 1 - Prob(mu <= 0 | Ybar)
 = 1 - Pr((mu - Ybar) / 1/sqrt(n) < (0 - Ybar) / 1/sqrt(n))
 = 1 - Phi(- Ybar * sqrt(n)) = Phi(Ybar * sqrt(n))


m * Ybar * b^2 / b = Ybar * sqrt(n)
m * Ybar * b = Ybar * sqrt(n)
m * b = sqrt(n)     # check: tau=0 b=1/sqrt(m)
m^2 b^2 = n
m^2 / (tau + m) = n
m^2 = n(tau + m) = n*tau + n*m
m^2 - n*m - n*tau = 0

m = 0.5 * (n + sqrt(n^2 + 4 *n * tau))

Check: tau=0: m = .5 * (n + n)

z <- list()
n <- seq(1, 100, by=2)
for(v in c(.05, .1, .25, .5, 1, 4, 100))
  z[[paste0('v=', v)]] <-
	list(x=n, y=0.5 * (n + sqrt(n^2 + 4 * n / v)) - n)

labcurve(z, pl=TRUE, xlab='Sample Size With No Skepticism',
         ylab='Extra Subjects Needed Due to Skepticism')

Caption: Effect of discounting by a skeptical prior with mean zero and
variance $v$.  This is in the increase needed in the sample size in
order to achieve the same posterior probability of $\mu > 0$ as with
the flat (non-informative) prior.