Assume that you have just finished talking about one topic. Let's assume we have $n$ people.
If we model the number of seconds the $i^{th}$ person typically takes to start a new conversation as a continuous random variable $S_i$, what we need to compute is the probability that, in a sample of these n random variables, the first two happen within 1 second of each other.
A natural assumption is that each of the $S_i$ is exponentially distributed, and that everyone takes on average the same time to start a new topic (probably not true, but let's keep it simple for now). This means: $p(S_i = x) = \lambda e^{-\lambda x}$
The parameter $\lambda$ is the inverse of the average time a single person takes to come up with a new topic. In order to simplify things let's think about a version where there are only 2 people speaking, you want $p(|S_1 - S_2|) < 1$, which is $p(S_2 \in [S_1, S_1+1]) + p(S_1 \in [S_2, S_2+1])$ This is equivalent to:
$\int_0^{\infty} \int_{s_1}^{s_1+1} p(S_1=s_1) \cdot p(S_2=s_2) ds_2 ds_1 + \int_0^{\infty} \int_{s_2}^{s_2+1} p(S_1=s_1) \cdot p(S_2=s_2) ds_1 ds_2 $
Since the probability distributions for $S_1$ and $S_2$ are the same, we have:
$2 \cdot \int_0^{\infty} \int_{s_1}^{s_1+1} p(S_1=s_1) \cdot p(S_2=s_2) ds_2 ds_1$
Replacing the probability density function:
$ 2 \cdot \int_0^{\infty} \lambda e^{-\lambda s_1} ds_1 \int_{s_1}^{s_1+1} \lambda e^{-\lambda s_2} ds_2$
It is easy to integrate the exponential pdf (try it!) and the result is $1 - e^{-\lambda}$. Without surprise, the probability that two people start a topic at almost the same time grows with $\lambda$. A large $\lambda$ means that people take little time to start a new topic on average, so the probability of collision increases.
If you want to model this for two people that come up with a new topic at different rates, you can use two different distributions with $\lambda_1$ and $\lambda_2$. This time you have to compute the two original integrals. Computing this yields: $\frac{(1-e^{-\lambda_2})\lambda_1 + (1-e^{-\lambda_1})\lambda_2}{\lambda_1 + \lambda_2}$
Again, as the $\lambda$ increase the probability of getting two people talking at the same time increases. You can check the heatmap below to see its shape. The x-axis represents the average time the first person takes to initiate a new topic ($\lambda_1^{-1}$) and the y-axis the same for the second person ($\lambda_2^{-2}$).

So, now that we have solved this for the case of 2 people how can we solve it for the case of $n$? I think it's probably easier if we try to compute the probability of the opposite event: that a single person comes up with the topic alone, which means everyone else would have come up with a new topic more than 1 second later.
The probability that the first person comes up with a new topic alone is: $\prod_{i=2}^n p(S_i > S_1 + 1)$
The probability that any person comes up with the topic alone is the sum of the above for each person: $\sum_{i=1}^n \prod_{j \neq i} p(S_j > S_i + 1)$
Let's assume again that $S_i$ has an exponential distribution with parameter $\lambda_i$. Let us first determine what is the integral for $p(S_j > S_i + 1)$:
$\int_0^{\infty} \int_{s_1+1}^{\infty} p(S_i=s_i) \cdot p(S_j=s_j) ds_j ds_i$
Solving this integral yields:
$\frac{\lambda_i e^{-\lambda_j}}{\lambda_i + \lambda_j}$
Therefore the probability that two people initiate a new topic at "the same time" is:
$1 - \sum_{i=1}^n \prod_{j \neq i} \frac{\lambda_i e^{-\lambda_j}}{\lambda_i + \lambda_j} $
If everyone takes the same time $\lambda^{-1}$ on average to start a new topic, the above becomes:
$ 1 - \sum_{i=1}^n \prod_{j \neq i} \frac{\lambda e^{-\lambda}}{2 \lambda} = 1 - \sum_{i=1}^n \frac{e^{-(n-1)\lambda}}{2^{n-1}} = 1 - n \frac{e^{-(n-1)\lambda}}{2^{n-1}}$
Just as a sanity check, if you take $n$ to be 2, the above simplifies to our original formula $1 - e^{-\lambda}$ =)