Sorry didn't explain my logic very well.
The classic example of Bayes' rule is when looking at the results of some sort of diagnostic (or test for a disease).
Suppose .001 of the population has a particular disease. A test is given to someone and it comes back positive. The test has a true positive rate of .99 and a false positive rate of .01 (i.e. if the person has the disease it will come back positive .99 probability and if they don't have the disease will come back positive with .01 probability). What is the probability that the person has the disease given that they get a positive test? If we us D to represent the event that the person has the disease, ~D to represent the event they don't have the disease, and T to represent the positive test we want to calculate the ratio of P(D|T) to P(D|~T)
By Bayes'
P(D|T)/P(~D|T) = (P(T|D)P(D)/P(T))/(P(T|~D)P(~D)/P(T)) = P(T|D)P(D)/(P(T|~D)P(~D)) (we cancel out P(T) from both numerator and denominator)
P(T|D)P(D)/(P(T|~D)P(~D)) = (.99)(.001)/((.01)(.999)) = .0991
That is it is more than 10 times more likely that someone doesn't have the disease vs. does have the disease given they tested positive. This is despite the fact that the test seems very accurate.
Here is how you might use Bayes' rule to figure out the original question.
Suppose we have two hypotheses:
(1) Girlfriend never cheated (call this H1)
(2) Girlfriend was cheating (call this H2)
To use Bayes' rule we want to calculate probability of H1 given the OP contracted the disease after 10 weeks as well as the probability of H2 given the OP contracted the disease after 10 weeks. We use P(H1|D) (D = got the disease) as a shorthand for probability of girlfriend never cheated given OP contracted disease ( P(H2|D) is defined similarly).
Bayes' rule states:
P(H1|D) = P(D|H1) P(H1)/P(D)
P(H2|D) = P(D|H2) P(H2)/P(D)
The term P(H1) and P(H2) are called priors. These represent how likely we believe each of these two scenarios to be before we observed the event D (OP getting the disease). Let's just assume they are equal.
Now we want to know which is greater P(H1|D) or P(H2|D). We can just take the ratio and test if it is greater than or less than 1.
P(H1|D)/P(H2|D) = P(D|H1)/P(D|H2) (where you cancel out P(D) since it is common to both terms).
The term P(D|H1) is called the likelihood. The idea is that if we assume that the hypothesis is true how likely would we be to observe the event D.
P(D|H1) is the easier to calculate. As you said for this to happen the girlfriend would have had to be a carrier and the probability of the OP catching the disease after 10 "sessions" would be the probability of not contracting it the first 9 sessions (.8^9) times the probability of catching on the 10th (.2). So P(D|H1) = .8^9 * .2 = .0268.
To calculate P(D|H2) we would have to make some additional assumptions. We would have to consider a few possibilities: (1) she was already a carrier *and* was cheating, (2) she was cheating then contracted the disease and then gave it to the OP.
Next, we would have to assign a prior probability to each of these scenarios and then calculate the likelihood of the data under each. (1) would be pretty easy to calculate. (2) we would have to consider the following factors: when did the girlfriend start cheating, how frequently was the girlfriend having sex with the person she was cheating with (since this impacts how likely she was to contract the disease), when did the gf contract the disease. As you state considering these factors seems weird, aren't we assuming something is true before testing how likely it is? The basic philosophy of Bayes' is that we have a prior belief over possibilities and then we imagine how likely our observations would be if each were true. This is what let's us then estimate how likely each of the possibilities is once we observe some data.
The logic of my original post was that in order for (2) to happen a lot of individual events have to take place. Each of these events may be more likely than what would have to happen for H1 to be true, but when taken in conjunction (multiplying the probabilities) they may multiply out to less than .0268
I am teaching a college course on some of this material this summer. Maybe I will use this as an example in my class

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Also, I don't want to come across as trivializing the OP's situation. I hope that he will understand that this is not my intention. I am a long time lurker on bluelight (as evidence by my join date and post count) and have a lot of respect for the folks that post on SLR. I will hopefully get BLer status within the next 10 years!
I don't want to turn this thread into a math thread, but ... I will (for the benefit of the OP, so it's ok %) )
No, it was the probability that he did NOT get the disease for 10 weeks. This must be the case if her story is true, assuming the medical information posted in this thread is correct. And this has around 4% of chance of being true, which may still be considered enough for giving her the benefit of the doubt, imho.
Since we are trying to determine if she was faithful or not, I am not sure that starting by assuming she wasn't is really relevant
With a 20% probability to contract it with each intercourse, I would not call that "a rare event" given that it seems not disputed that she is indeed the source of the infection. I am not familiar with Bayes though, only with basic probability, so please educate me if I am mistaken or if I misinterpreted you.