The probability of seeing pulsed emission from a neutron star is simply the fraction of the sky covered by the beam, i.e. the beam solid angle divided by $4\pi$ steradians.
The angle swept out on the sky by a pulsar with an emission cone of width $\rho$ turns out to be
$$\zeta=4\pi\sin^2\left(\frac{\rho}{2}\right)$$
covering a fraction of the sky
$$f=\frac{\zeta}{4\pi}=\sin^2\left(\frac{\rho}{2}\right)$$
The opening angle $\rho$ can often be deduced from the pulsar's period. Many long-period pulsars obey the power-law model $\rho\propto P^{-1/2}$; the proportionality constant is sometimes described piecewise. However, millisecond pulsars tend to deviate downwards from the $P^{-1/2}$ relation by a factor of a few, as shown in Fig. 12 of Kramer et al. (1998):

If you wanted to choose a representative opening angle to determine the probability that a particular pulsar will sweep its beam across Earth, it might be best to pick an angle calculated from the pulsar's period. On the other hand, if you care about a population of pulsars with a random period distribution, you would be better off simply looking up a mean value of $\rho$. Picking $\rho=40^{\circ}$, for example, gives $f\approx0.12$, as tuomas quoted in their answer, which is a reasonable value.