Probability of a household being covid-19 contact traced The United Kingdom is moving to a system of contact tracing where people who test positive for the virus will be asked who they've been in contact with and the chain of contacts traced. 
Assuming each household has a fixed $1/1000$ weekly chance of catching the virus (regardless of how much or little contact they have), contact tracers trace contacts back 1 week, and each household has contact with $X$ other households, in terms of $X$, what is the weekly probability that a given household will be contact traced?
I think the number of households who is they test positive the given household will be contact traced is something like $1 + X + (X^2)/2 + (X^3)/2^2 + (X^4)/2^3 + \dots$ But that goes on to infinity. How do I simplify it?
In the above attempt, the $1$ is the given household themselves, $X$ is first order contacts, $(X^2)/2$ is second order contacts etc. 
 A: I'm not sure what's the best thing to tell you as there are two aspects here. One is: does my mathematical model accurately reflect reality? The other is: how to evaluate the infinite sum showing up in the model?
As for the first: if the model predict that contact tracers will track down and interview infinitely many people in one week time, then it stands to reason that somewhere during the week the reality and model start to look quite different. We could try and build a better model but this is tricky, so instead I focus on the other question: how to evaluate the sum
$$S = 1 + X + X^2/2 + X^3/2^2 + X^4/2^3 + X^5/2^4 + \ldots $$
First step: it is a bit annoying that the powers of $X$ are different from the powers of $2$. What to do about that? Well $X^2/2$ is $X^2/2^2$ multiplied by 2. $X^3/2^2$ is $X^3/2^3$ multiplied with 2. So we can rewrite in a more pleasant form:
$$S = 1 + 2(X/2 + X^2/2^2 + X^3/2^3 + \ldots)$$
To make things even more pleasant we can introduce a new letter. Say $Y = X/2$. Then we get
$$S = 1 + 2*(Y + Y^2 + Y^3 + Y^4 + Y^5 + \ldots)$$
Yaaay! No fractions anymore. Now things are getting easy. Let's make it even easier for ourselves and first try to understand the thing inside the brackets, the sum:
$$T = Y + Y^2 + Y^3 + \ldots$$
Now the first thing to notice is: if $Y \geq 1$ then $Y^2 \geq 1$ AND $Y^3 \geq 1$ etc etc so we add up infinitely many things that are all at least 1 so the answer is infinity. Done. We have our answer to the mathematical question. How to make the model more realistic is a different question.
But what now if $Y < 1?$ (Or equivalently $X < 2$, remembering that $Y = X/2$.)
That is when things get interesting. 
Look at the sum
$$U = 1 + Y + Y^2 + Y^3 + \ldots$$
How does this relate to the sum $T$ above? Well on one hand you can say:
$$U = 1 + T$$
On the other hand you can say:
$$U = T/Y$$
Both seem equally defensible. So who is right?
The answer is: both!
We get $T/Y = 1 + T$ and this gives us a way of solving $T$ using high school algebra:
$$T = Y + YT$$
$$T - YT = Y$$
$$(1 - Y)T = Y$$
$$T = \frac{Y}{1-Y}$$
Here we see that this can only make sense if $Y < 1$: we don't want a negative number out of a sum of positive numbers. But this fits nicely with our earlier observation that the computation can only makes sense when $Y < 1$ since otherwise $T$ is infinite. So nothing new here, really. 
Now
$$S = 1 + 2T = \frac{1 - Y}{1 - Y} + \frac{2Y}{1 - Y} = \frac{1 + Y}{1 - Y} = \frac{1 + 2X}{1 - 2X}$$
Very nice, but remember this equation can only make sense if $X < 2$ because otherwise we already had that $S$ is infinite.
The great thing is that in fact for every $0 \leq X < 2$ we do have that this equation holds. Just try it out for $X = 1$:
$$1 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + \ldots$$
gets arbitrarily close to 3. Just compute a few more terms if you don't believe it, eventually you will be convinced that the limit is 3.
And plugging $X = 1$ in in our formula we find
$$\frac{1 + X/2}{1 - X/2} = \frac{3/2}{1/2} = 3$$
Other example: when $X = 1.2$  our formula predicts that infinite ends up at 
$$\frac{1 + X/2}{1 - X/2} = \frac{1.6}{0.4} = 4$$
And feeding the first, say 30, terms of the infinite sum into wolfram alfa will convince you that this is indeed correct.
I just taught you this trick because it can be helpful in many situation, but perhaps not so much in this.
Here is a related riddle. I have two parents. They each have two parents and so on. It follows that by the time the dinosaurs died out there were many many more people alive than are today, and that is just counting my own predecessors, not those of everyone else alive now. How is this possible? 
