Given 5000 people, chance of a single person being infectious 1:10,000, how big is the chance the virus spreads? This was an actual statement in my newspaper in an opinion piece, after the government said that at this moment, in The Netherlands, about 1 in 10,000 people are likely to be actively infectious of the coronavirus. The context was a demonstration where about 5,000 people gathered on a small square and the debate centered around whether it should've been forbidden. 
His assumption was that there's 50% chance the coronavirus spread a little. I think this is false, given the birthday paradox, but I couldn't substantiate my point, lacking the proper math skills wrt to this type of problem.
This may have been asked before, but I couldn't find it, or a way how to adopt other answers. If we assume that a single person would infect 2 other persons in 10 minutes (but this is not relayed, it takes time to spread further), and that during the event each person, on average, gets close to 30 other people (this is conservative), how would one calculate that? 
Photo of the actual gathering on the Dam Square in Amsterdam:

 A: It is important to understand that if we model the proportion of infectious individuals in a population by some parameter, say $\theta = 10^{-4}$, that this does not mean that for every $10000$ individuals, one is guaranteed to be infectious.  This is just an average rate among the population, akin to saying that if we randomly choose one person from the population, the probability they are infectious is $\theta$.  As such, in a cohort of $n = 5000$ people, there is a probability that none are infectious, but also a nonzero probability that more than one might be infectious, and so forth.
With this in mind, we can develop a crude model for estimating the probability of a transmission event in the cohort.  Because $\theta$ is so small and $n$ is large, but not so large relative to the entire population, we can use a Poisson distribution to model the random number of infectious individuals $X$ within the cohort, namely $$X \sim \operatorname{Poisson}(\lambda = n\theta = 1/2), \\ \Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!} = \frac{1}{e^{1/2} 2^x x!}, \quad x \in \{0, 1, 2, \ldots \}.$$  This gives us the following table:
$$\begin{array}{c|c}
x & \Pr[X = x] \\
\hline
0 & 0.606531 \\
1 & 0.303265 \\
2 & 0.0758163 \\
3 & 0.0126361 \\
4 & 0.00157951 \\
5 & 0.000157951 \\
\vdots & \vdots
\end{array}$$
As you can see, the probability of no infectious individuals is just over $60\%$, but the chance of more than one infectious individual is $$1 - \Pr[X = 0] - \Pr[X = 1] \approx 0.090204,$$ just a bit over $9\%$.  If we further assume that the reproduction number in this cohort is also a Poisson variable, that is to say, the number of transmission events $P$ per infected individual is Poisson with intensity $\rho = 1.5$, where I have chosen $\rho$ to be higher than $1$ due to the nature of such gatherings but not so high due to the current awareness of social distancing measures, then the average or expected number of total transmission events $T$ will be approximately $$\operatorname{E}[T] = \operatorname{E}[\operatorname{E}[T \mid X]] = \operatorname{E}[1.5 X] = 1.5 \operatorname{E}[X] = 0.75.$$  But this does not represent a probability of at least one transmission occurring.  To calculate this, we want 
$$\begin{align*}
\Pr[T \ge 1] &= 1 - \Pr[T = 0] \\
&\approx 1 - \sum_{x=0}^\infty \Pr[T = 0 \mid X = x]\Pr[X = x] \\
&= 1 - \Pr[X = 0] - \sum_{x=1}^\infty (\Pr[P = 0])^x \Pr[X = x] \\
&= 1 - e^{-1/2} - \sum_{x=1}^\infty e^{-3x/2} e^{-1/2} \frac{(1/2)^x}{x!} \\
&= 1 - \exp\left((e^{-3/2} - 1)/2\right) \\
&\approx 0.321883.
\end{align*}$$
The first approximation is due to the use of the Poisson model, when in fact we cannot have more than $5000$ transmissions.  I leave it as an exercise to compute this probability as a function of the mean reproduction number assumption $\rho$.

I would like to mention at this point that the reason why we used this Poisson/Poisson model is because it has convenient computational properties, yet sacrifices little in the way of accuracy.  As you can compute for yourself, the exact probability for no infectious individuals is not far off from that modeled by the Poisson distribution:  $$\Pr[X = 0] = (1 - \theta)^n = 0.606515\ldots.$$  Moreover, we can develop estimates of the unconditional variance and perform other computations with this hierarchical parametric model with minimal effort.
