Question about the famous 'statistics in medicine' puzzle Context of the question:

Given a test that predicts a disease with 99% accuracy. Furthermore it is know that one in 10.000 people suffer from this disease. If we assume that I test positive, then the question of the riddle is: "what are the odds that I am actually sick".

One can make a table like this:

$$
\begin{array}{c|lcr}
&amp; \text{Infected} &amp; \text{Healthy} \\
\hline
\text{Positive test} &amp; \frac{1}{10000}\times\frac{99}{100} &amp; \frac{9999}{10000}\times\frac{1}{100} \\
\text{Negative test} &amp; \frac{1}{10000}\times\frac{1}{100} &amp; \frac{9999}{10000}\times\frac{99}{100}\\
\end{array}
$$

The answer of the riddle is: $$\frac{\frac{1}{10000}\times\frac{99}{100}}{\frac{1}{10000}\times\frac{99}{100}+\frac{9999}{10000}\times\frac{1}{100}}=\frac{1}{102}$$

Now here is my question:

If I test positive, then I am either infected, or I am healthy. There are no other options. Therefore I would think that their combined odds add up to one. Then how come that the odds of postive test/infected + positive test/healthy (the values of the first row) do not add up to 1?

Thanks!
You are making 2 mistakes:

First, while you are right that if I test positive, then there is a 100% chance that I am either healthy or infected, but what that means is that $P(healthy|positive)+P(infected|positive)=1$, and you are describing this as $P(positive| healthy)$ + P(positive |infected)$

Second, you point to the top row as the values of $P(healthy|positive)$ and $P(infected|positive)$, but those are the values of $P(healthy \land positive)$ and $P(infected \land positive)$, and those will only add up to $P(positive)$, which is indeed not equal to $1$

Finally, note that what you calculated as $\frac{1}{102}$ is $P(infected|positive)$. So, use a similar method to compute $P(healthy|positive)$ (change the numerator to $\frac{9999}{10000} \times \frac{1}{100}$ to do this ... The denominator (which is $P(positive)$ stays the same), and you wil find that $P(healthy|positive)=\frac{101}{102}$, and now you see that they do add up to $1$.

In short: $P(positive| healthy)$, $P(healty|positive)$, and $P(positive \land healthy)$ are $3$ different things! (Though we do have that $P(positive \land healthy)=P(healthy \land positive)$)
The statement "If I test positive, then I am either infected, or I am healthy" corresponds to the probabilities of infected / positive + healthy / positive = 1; the event that you're conditioning on is "positive" not infected and healthy.