Using the information that she is my grandmother, I speculate on the reason why she did not call on Thursday. Perhaps it is because she does not intend to come on Friday: P(Friday) is lowered. Perhaps it is because she does intend to come but judges the regularity of the event to make calling in advance unnecessary unless she had decided not to come: P(Friday) is raised. Grandmothers tend to be old and consequently may be forgetful: perhaps she intends to come but has forgotten to call: P(Friday) is raised. Grandmothers tend to be old, and consequently may be frail: perhaps she has been taken unwell; perhaps she is even now lying on the floor of her home, having taken a fall, and no-one is there to help: P(Friday) is lowered, and perhaps I should phone her. My answer to the problem is therefore: I phone her to see how she is and ask if she is coming tomorrow. I know -- this is not an answer within the terms of the question. However, it is my answer. The more abstract version you later posted is a different problem. We have two observations of A and B occurring together, and that is all. Unlike the case of Grandma's visits, we have no information about any causal connection between A and B. (The sequence of revealing A before B does not affect anything.) What is then the best estimate of P(B|~A)? We have no information about the relation between A and B, so I am guessing that a reasonable prior for that relation is that A and B are independent. Therefore A can be ignored and the Laplace rule of succession applied to the two observations of B, giving 3/4. ETA: I originally had a far more verbose analysis of the second problem based on modelling it as an urn problem, which I then deleted. But the urn problem may be useful for the intuition anyway. You have an urn full of balls, each of which is either rough or smooth (A or ~A), and either black or white (B or ~B). You pick two balls which turn out to be both rough and black. You pick a third and feel that it is sm

OK, I'll use the same model I use for text. The zeroth-order model is maxentropy, and the kth-order model is a k-gram model with a pseudocount of 2 (the alphabet size) allocated to the (k-1)th-order model. In this case, since there's never before been a Thursday in which she did not call, we default to the 1st-order model, which says the probability is 3/4 that she will come on Friday.