Consider a 1 second frame with two users having 2 milli-second(ms) symbol length transmission each, now each of the user can randomly select any 2 ms interval within the frame, what is the probability that both the transmissions overlap?

PS: I first assumed that the user-1 randomly selects any interval with uniform probability $\frac{2}{1000}$ and conditioned on that calculated the probability that user-2 overlaps with the given user-1's interval, is the approach right way to proceed or I'm missing out on something? Can someone help me solving this problem.

Also, the main problem is lets say, there are now N-users now what is the probability that atleast 2 users overlap in the 1 second frame?

  • $\begingroup$ Do you that intervals start at even integer number of ms? Can the interval for the first use be (in milliseconds) [505.3, 507.3]? Can it be [501, 503]? $\endgroup$
    – mihaild
    Commented May 11, 2022 at 15:19
  • $\begingroup$ @mihaild The interval for any user, say for example, for user-1, the interval be (t1) < t < (t1+2), then t1 can take any value between 0 to 998 milliseconds. Be it irrational or rational number, anything between the range of 0 to 998 ms. $\endgroup$ Commented May 12, 2022 at 8:44

1 Answer 1


I don't know how to solve it for general case, because even for two users it's a bit more complex (may be there is a way to simplify, but I don't see it). The problem is that not all intervals for the first user are equal: it's less likely that 2nd user will intersect it if the first user starts near the very beginning or end.

For simplicity, lets rescale time unit so that transmission takes $1$ unit (and so it can start at any moment from $0$ to $449$, let's denote $a = 449$)

For 2 users, formally we have $t_1$ and $t_2$ independent uniformly distributed on $[0, a]$ and we are interested in event $|t_1 - t_2| < 1$.

Depending on $t_1$, this condition can be rewritten as follow:

  1. If $t_1 < 1$, then condition is equal to $t_2 < t_1 + 1$
  2. If $t_1 > a - 1$, then condition is equal to $t_2 > t_1 - 1$.
  3. Otherwise, condition is equal to $t_2 \in (t_1 - 1, t_1 + 1)$ - $t_2$ is in some interval of length $2$.

This variants are mutually exclusive (if $a > 2$), so we can calculate probability of each and sum them up.

The third is the simplest: it's that $t_1$ gets into interval of length $a - 2$ (probability $\frac{a - 2}{a}$) and then $t_2$ gets into interval of length $2$ (probability $\frac{2}{a})$, multiplying we get $\frac{2(a - 2)}{a^2}$.

The first and the second are similar (and in fact equal, you can easily see it if draw square $[0, a]\times [0, a]$ and color corresponding parts). For the first, we have probability

$$\int_0^1\,dt_1 \int_0^{t_1 + 1}\, dt_2 \frac{1}{a^2} = \int_0^1\,dt_1 \frac{t_1 + 1}{a^2} = \frac{3}{2a^2}$$

Summing up, we get probability of intersection $\frac{2a - 1}{a^2}$.


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