How do I calculate the odds of this? (Statistics and probability question)

It seems to me that the odds would be 0.001X0.81=0.00081 (0,001X0,81=0,00081: in the US we use commas to denote our decimal points) or i.e., 0,081%

How about looking at it a different way? Since they each tell the truth 90% of the time that means they lie 10% of the time. The odds of both not telling the truth are 0.1×0.1 or 1% .
That means you can be 99% sure it is true. It seems to me the rare event is not a factor here.

Take a look at section 48 of this article. It seems that the original likelihood of the event occurring is a factor and it can be brought into the calculation as if it were a third witness. I then calculated the odds of the event being true as only 0.9%.

@Brian1946 Indeed, if the events were entirely independent. But telling the truth is not by any stretch an independent action. The point would be to evaluate the likelyhood of the event from the fact that two reliable witnesses both testify to it, since they testify to it their testimony is, partly, dependent on the event, and that’s the problem, how do we calculate that?

@worriedguy You might be on to something there, afterall if two independent witnesses confirm eachother that should increase the likelyhood not decrease it as my calculations somehow suggested.

@flutherother uhm, that’s an interesting article, and it’s likely to be the explaiantion i need, even though it’ll take a while for me to wrap my head around it, expecially to connect it to the information i was supposed to complete the exercise with. The pp’/pp’+(1-p)(1-p’) part seems precisely the kind of thing one would expect, the two witnesses make eachother more believable, not the opposite.

@Thammuz

I think worriedguy is right.

Another way of looking at is that there’s a 90% chance that the first observer is right, which leaves a 10% chance that he didn’t see it.

When the other observer corroborates the first observer’s claim, then that reduces the margin of improbability by another 90% which leaves us with a 1% margin of improbability.

My reasoning goes as follows:

The question seems to assume that A or B lying is not dependent on what actually happened, and that they cannot be mistaken. Both of these are extremely unrealistic, but if we go with those assumptions, I think we can construct the full event space and the probabilities of each combination of events as follows:

The event happened, A tells the truth, and B tells the truth = 0.001*0.9*0.9
The event happened, A tells the truth, and B lies = 0.001*0.9*0.1
The event happend, A lies, and B tells the truth = 0.001*0.1*0.9
The event happened, A lies, and B lies = 0.001*0.1*0.1
The event did not happen, A tells the truth, and B tells the truth = 0.999*0.9*0.9
The event did not happen, A tells the truth, and B lies = 0.999*0.9*0.1
The event did not happen, A lies, and B tells the truth = 0.999*0.1*0.9
The event did not happen, A lies, and B lies = 0.999*0.1*0.1
(This sums up to one, so I think we have everything covered.)

Next, we observe that A and B both say the event happened. This matches two of the cases above:
P(T) = The event happened, A tells the truth, and B tells the truth = 0.001*0.9*0.9 = 0.0081
P(F) = The event did not happen, A lies, and B lies = 0.999*0.1*0.1 = 0.00999

Now, the probability that the event happened given that we observe A and B both telling it did is the proportion of those events where A and B both say so and the event happened, from all events where A and B both say so, or

P(T)/(P(T)+P(F)) = 0.000810/(0.000810+0.00999) = 0.075, or 7.5 percent.

@Janka Indeed, we looked at the exercise already yesterday and that’s exactly the result we found. Another formula but, still, that result.