Classic layman error about conditional probability. The probability you’re trying to compute Cheezit should not be the probability of event A occurring. Rather it should be the probability of event A occurring given that event B has already occured.
To illustrate:
I have two cats - Fifi and Trixibelle*. No other cats are capable of entering my house.
Suppose that the probability of one cat stealing another cat’s food is 0.01%**
After sleepily dishing out food at 9am, I have been lazing in bed with Fifi all morning. I therefore know with 100% certainty that Fifi hasn’t touched her food bowl. Meanwhile Trixibelle has been prowling the house.
Upon finally dragging myself out of bed and downstairs I discover (horror!) that Fifi’s food is gone!
What is the probability that Trixibelle stole Fifi’s food?
The answer is not 0.01%. The answer is 100%. The only possible solution to the conundrum is that Trixibelle did it.
We can flesh this out with a little use of symbols.
Let P(A) be the probability of event A occurring.
Let P(A|B) be the probability that event A occurs given that event B has already occurred
Let T and S be the events “Trixibelle steals Fifi’s food” and “Fifi’s food has been stolen” respectively.
Then P(T) = 0.01%, because the probability that a cat steals another cat’s food is 0.01%.
However P(T|S) = 100% because we know that food has been stolen.
Worryingly this error is not limited to cats and foodbowls. Even professionals routinely make the above error. I’ve seen lawyers use the above trick with DNA evidence and the media not realise the error. I’ve seen the medical profession misinterpret illness-testing data. I’ve even seen Cecil screw it up ( http://www.straightdope.com/classics/a3_189.html ). The long and the short of it is that statistics are counterintuitive.
TO bring it back to your problem: The question is not “What is the probability of the sequence of events involving star and planet formation, evolution and everything else”. It is “What is the probability of this, given that we are here to ask the question.” The probability of this is of course: 100%.
regards,
pan
*not really folks. This is an illustration only
**ditto. No cites please.
PS to give a fuller formulaic approach:
P(A|B) = P(A n B)/ P(B) where P(A n B) is the event “A and B both occur”
In the illness-testing scenario for instance we may have the case where the probability that a member of the population has a disease is 1/50000 and the probability that a test is positive where a member does not have the disease is 1%. A test comes back positive. What is the probability that the patient has the illness? Left as an exercise to the reader, but let me tell you that it is not 99%.