Let $B(t)$ be a standard Brownian Bridge, where $T=1$ and $B(T)=0$.
Finding the time $t^\ast$ that maximizes $P\{B(t)>b\}$, where $b>0$, is straightforward. We simply want to minimize: $$\Phi\left(\frac{b}{t(1-t)}\right),$$ so we choose the time at which the process has the greatest variance, which is $t^\ast = 0.5$.
Now, suppose I want to find $t^\ast \in (0,t_0)$ that maximizes $$y(t) \equiv P\{B(t)>b|B(t_0)<b\}.$$
This is equivalent to maximizing
$$\int^b_{-\infty} P\{B(t)>b|B(t_0)=x\}P\{B(t_0)\in[x,x+dx]\},$$
which in turn is equal to:
$$\Phi\left(\frac{b}{t_0(1-t_0)}\right) - \int^b_{-\infty} \Phi\left(\frac{b-\frac{t}{t_0}x}{\sqrt{\frac{t}{t_0}(t_0-t)}}\right) \frac{\varphi\left(\frac{x}{\sqrt{t_0(1-t_0)}}\right)}{\sqrt{t_0(1-t_0)}} dx.$$
If we take the derivative with respect to $t$, we obtain:
$$y'(t) = \int^b_{-\infty} \left(\frac{b-(2b-x)\frac{t}{t_0}}{2(\frac{t}{t_0}(t_0-t))^{1.5}}\right) \varphi\left(\frac{b-\frac{t}{t_0}x}{\sqrt{\frac{t}{t_0}(t_0-t)}}\right) \frac{\varphi\left(\frac{x}{\sqrt{t_0(1-t_0)}}\right)}{\sqrt{t_0(1-t_0)}} dx.$$
This makes intuitive sense because if we knew instead that $B(t_0) = x$, then it would be the case that $$t^\ast = \frac{b}{2b-x} t_0,$$ as it makes the above integrand vanish. If $x=0$, then of course $t^\ast = 0.5 t_0$; if $x=b$, then $t^\ast = t_0$; and as $x \to -\infty$, $t^\ast \to 0$.
Now, let's try to calculate the above integral. I complete the square and obtain:
$$y'(t) = \left(\frac{1}{2\frac{t}{t_0}(t_0-t)\sqrt{1-x_0}}\right) \left(\frac{1}{\sqrt{t}}\varphi\left(\frac{x}{\sqrt{t}}\right) \right) \int^b_{-\infty} \left(b-(2b-x)\frac{t}{t_0}\right) \frac{\varphi\left(\frac{x - b}{\sqrt{t_0-t}}\right)}{\sqrt{t_0-t}} dx.$$
The integral in the above expression is equal to:
$$\Phi(0)\left[\left(b - 2b\frac{t}{t_0}\right) + \frac{t}{t_0}E[Z|Z<b]\right],$$
where $Z$ is a normal random variable with mean $b$ and variance $t_0-t$. Since we have that
$$E[Z|Z<b] = b - \sqrt{t_0 - t} \frac{\varphi(0)}{\Phi(0)},$$ the integral must equal: $$0.5b\left(1 - \frac{t}{t_0}\right) - \frac{t}{t_0}\sqrt{\frac{t_0 - t}{2\pi}}.$$
If we set the above equal to $0$, we should get the maximizer $t^\ast$. The solution (via Mathematica) I get is:
$$t^\ast = 0.25 \left(\sqrt{\pi b^2 (\pi b^2 + 8t_0)} - \pi b^2\right) \in (0,t_0).$$
I'm very confused by this solution however. My intuition is that if $t_0$ is very close to $1$, then $t^\ast$ should be very close to $0.5$, since we learn very little information given that $B(1)<b$. But $t^\ast \neq 0.5$ if $t_0 = 1$. I've omitted some steps in my calculations, but I fear I may have gone astray somewhere.
I was hoping someone could help me either correct my calculations or clarify why the solution isn't approximately equal to $0.5$ when $t_0$ is near $1$.