With an <absorbing boundary condition for a diffusion> on each vertical side, the <splitting probability> $u(x)$ of hitting the left side before the right is harmonic for the <Kolmogorov backward equation>:
$$
u''+a_1u'=0,\qquad
u(-1)=1,\quad u(1)=0.
$$
For $a_1\neq0$,
$$
u(x)=\frac{e^{-a_1x}-e^{-a_1}}
{e^{a_1}-e^{-a_1}}.
$$
The particle starts at $x=0$, so
$$
\boxed{
g(a_1)=u(0)
=\frac{1-e^{-a_1}}{e^{a_1}-e^{-a_1}}
=\frac1{1+e^{a_1}}.
}
$$
The continuous limit at zero is $g(0)=1/2$, as required by reflection symmetry. A large positive drift drives the particle toward the right and gives $g(a_1)\to0$; a large negative drift drives it toward the left and gives $g(a_1)\to1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 356 / 2 / d / Solution

A splitting probability is the probability that a stochastic process reaches one target before competing targets. For a diffusion it is harmonic for the <Kolmogorov backward equation>, with boundary values one on the selected target and zero on the others.

= Absorption probability
{synonym}

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 Incoming links: Splitting probability