Coherent beams produced by lasers follow Poissonian statistics, that is, the number of photons passing through the cross-section of the beam per unit time can be described by the Poisson distribution. As such, the waiting time, $T$, between photons follows an exponential distribution.
Squeezed light, on the other hand, has "sub-exponential" waiting times. We may define a squeezed version of the exponential distribution as follows. Let $H,s>0$, $X\sim\operatorname{Gamma}(1/s,H)$ (shape-rate parameterization), and $T=X^{1/s}$. Then $T$ follows a squeezed exponential distribution $$ f_T(t)=\frac{H^{1/s}}{\Gamma(1+1/s)}e^{-Ht^s}. $$ Note that when $s=1$, we recover the exponential density function.
Now let $N$ denote the number of photons passing through the cross-section of a squeezed light beam over a a time period $t_0$. The probability of observing zero photons is easy to work out: $$ \mathsf P(N=0)=\mathsf P(T>t_0)=\tilde\Gamma(1/s,Ht_0^s), $$ where $\tilde\Gamma$ is the regularized upper incomplete gamma function $$ \tilde\Gamma(a,z)=\frac{1}{\Gamma(a)}\int_z^\infty t^{a-1}e^{-t}\mathrm dt. $$
However, the probability of observing exactly one photon during the time period $t\in[0,t_0]$ is $$ \mathsf P(N=1)=\mathsf P(T_1\leq t_0,T_2>t_0-T_1)=\int_0^{t_0}\frac{H^{1/s}}{\Gamma(1+1/s)}e^{-Hx^s}\tilde\Gamma(1/s,H(t_0-x)^s)\ \mathrm dx, $$ which is not so nice.
Probabilities for larger $N$ can be further written in terms of double, triple, quadruple, etc., integrals. This of course is not very useful for computation. I wonder if a closed-form solution (a solution in terms of known special functions) to the integral defining $\mathsf P(N=1)$ can be achieved as it may shed light on how to obtain solutions to the integrals for larger $N$.