I think your example eviscerates the first "Increase in Probability" definition, at least as presented here, and shows that it doesn't account for non-independent evidence.
There's a deep philosophical point at stake here. Is probability a 1) quantification of a person's uncertainty, or 2) a statement about the universe?
2) is a position that is not well-regarded here, and I would recommend Probability is in the Mind and then possibly Probability is Subjectively Objective as to why.
If 1), then their previous knowledge matters. It's one particular person's uncertainty. You start off very uncertain about the lottery result. Then you read the first newspaper report, which makes you much less uncertain about the lottery result. Then you read the second newspaper report, which makes you very slightly less uncertain about the lottery result. As you watch your uncertainty decrease, you see that the first report has a huge effect (and thus is strong evidence), and the second report has a small effect (and thus is weak evidence). With more background knowledge about correctness, the second report drops to 0 effect (and thus is not evidence to you).
The mathematical way to think about this is that the strength of evidence is a function of both e and b. This is necessary to ensure that there's a probability shift, and if you don't force a shift, you can have even more silly 'evidence', like sock color being evidence for logical tautologies.
One might object that they're not particularly interested in measuring their personal uncertainty, but in affecting the beliefs of others. If you wanted to convince someone else that Bill Clinton is probably going to win the lottery, it seems reasonable to be indifferent to whether they read the Times or the Post, so long as they read one. But your personal measure of evidence is wildly different between the two papers! How do we reconcile your personal measure of uncertainty, and your desire to communicate effectively?
The answer I would give is being more explicit about the background b. It's part of our function, and so let's acknowledge it. When b is "b & e1", then e2 is not significant evidence. When b is just b, e2 is significant evidence, and so if you want to convince someone else of h and their current knowledge is just b, you can be indifferent between e1 and e2 because P(h|b,e1)=P(h|b,e2).
Let's further demonstrate that with a modification to the counterexample like the one Manfred suggested. Suppose I learn e1, that the New York Times reports that Bill Clinton owns all but one of the tickets, by reading that day's copy of the Times, which I'll call r1.
Suppose my background, which I'll call t, is that the New York Times (and my perception of it) is truthful. So P(e1|t,r1)=1. I liked the prose so much, I give the newspaper a second read, which I'll call r2. Is r2 evidence for e1? Well, P(e1|t,r1,r2)=1, which is the same as P(e1|t,r1), which suggests by the "increase in probability" definition that reading the newspaper a second time is not evidence for what the newspaper says. (Note that if we relax the assumption that my reading comprehension is perfect, then reading something a second time is evidence for what I thought it said, assuming I think the same thing after the second read. If we only relax the assumption that the newspapers are perfectly correct, we don't get a change in evidence.)
Does it seem reasonable that, given perfect reading comprehension, that considering the same piece of evidence twice should only move your uncertainty once? If so, what is the difference between that and the counterexample where one newspaper says "Times" on the front, and the other says "Post"?
(If you relax the assumption that the newspapers are perfectly correct, then the second newspaper is evidence by the "increase in probability" definition, because of the proposition g discussed in the OP.)
Right, so, one think that is left open by both definitions is the kind of interpretation given to the function P. Is that suppose to be interpreted as a (rational) credence function? If so, the Positive Relevance account would say that e is evidence that h when one is rational in having a bigger credence in h when one has e as evidence than when one does not have e as evidence. For some, though, it would seem that in our case the agent that already knows b and e1 wouldn't be rational in having a bigger credence that Bill will win the lottery if she learns ...
I would like to share a doubt with you. Peter Achinstein, in his The Book of Evidence considers two probabilistic views about the conditions that must be satisfied in order for e to be evidence that h. The first one says that e is evidence that h when e increases the probability of h when added to some background information b:
The second one says that e is evidence that h when the probability of h conditional on e is higher than some threshold k:
A plausible way of interpreting the second definition is by saying that k = 1/2. When one takes k to have such fixed value, it turns out that P(h|e) > k has the same truth-conditions as P(h|e) > P(~h|e) - at least if we are assuming that P is a function obeying Kolmogorov's axioms of the probability calculus. Now, Achinstein takes P(h|e) > k to be a necessary but insufficient condition for e to be evidence that h - while he claims that P(h|e&b) > P(h|b) is neither necessary nor sufficient for e to be evidence that h. That may seem shocking for those that take the condition fleshed out in (Increase in Probability) at least as a necessary condition for evidential support (I take it that the claim that it is necessary and sufficient is far from accepted - presumably one also wants to qualify e as true, or as known, or as justifiably believed, etc). So I would like to check one of Achinstein's counter-examples to the claim that increase in probability is a necessary condition for evidential support.
The relevant example is as follows:
The point now is that, although it seems right to regard e2 as being evidence in favor of h, it fails to increase h's probability conditional on (b&e1) - at least so says Achinstein. According to his example, the following is true:
Well, I have my doubts about this counterexample. The problem with it seems to me to be this: that e1 and e2 are taken to be the same piece of evidence. Let me explain. If e1 and e2 increase the probability of h, that is because they increase the probability of a further proposition:
and, as it happens, g increases the probability of h. That The New York Times reports g, assuming that the New York Times is reliable, increases the probability of g - and the same can be said about The Washington Post reporting g. But the counterexample seems to assume that both e1 and e2 are equivalent with g, and they're not. Now, it is clear that P(h|b&g) = P(h|b&g&g), but this does not show that e2 fails to increase h's probability on (b&e1). So, if it is true that e2 increases the probability of g conditional on e1, that is, if P(g|e1&e2) > P(g|e1), and if it is true that g increases the probability of h, then it is also true that e2 increases the probability of h. I may be missing something, but this reasoning sounds right to me - the example wouldn't be a counterexample. What do you think?