Let me elaborate a little bit more about the details of the above post, in case you want some more concrete idea.

Let [math] N(t) [/math] be the counting process, counting the aggregate number of buses arrived up to time [math] t [/math] (in minutes). One very popular model will be using (homogeneous) Poisson process to model it (with a certain mean arrival rate).

Let [math] W_i, i = 1, 2, 3, \ldots [/math] be the arrival time of the [math] i [/math]-th bus. Under the Poisson process assumption, the marginal distribution of [math] W_i [/math] is a gamma distribution.

However, as Dason pointed out, now we are conditional on [math] N(60) = 45 [/math],

and in this situation it is well known that the joint distribution of

[math] (W_1, W_2, \ldots, W_{45})|N(60)=45 [/math]

is exactly the same as

[math] (U_{(1)}, U_{(2)}, \ldots, U_{(45)}) [/math]

where [math] U_i [/math] are i.i.d. as [math] \text{Uniform}(60) [/math] and [math] U_{(i)} [/math] are the corresponding order statistics.

Assuming a deterministic waiting time of [math] \frac {1} {2} [/math] minutes in the first position of the bus stop. Then we know that the [math] i [/math]-th bus cannot enter the layby when it arrive the bus stop if and only if the previous two bus does not left yet; in math,

[math] W_i < W_{i-2} + \frac {1} {2} [/math]

(we absorbed the condition [math] N(60)=45 [/math] to shorten the notation)

The event we considered there is at least one bus facing this situation within the 60 minutes; in probability notation, it is the union of the above events:

[math] \bigcup_{i=3}^{45} \left\{W_i < W_{i-2} + \frac {1} {2}\right\} [/math]

and we are calculating the probability of this event. The way to calculate it is another issue; but at least this could be a reasonable way to model the situation you described if we did not misunderstand your post.