The way I see it is this: We consider A likely, but B unlikely. Say P(A)=1-a and P(B)=b, where a and b are small. A and B are currently independent. Then we observe that A implies B (i.e. “¬A v B”). We get probabilities
P(A|A=>B) = P(A=>B|A)P(A)/P(A=>B) = P(B)P(A)/[P(B)P(A)+P(B)P(¬A)+P(¬B)P(¬A)] = [b-ab]/[b+a-ab] or approximately P(A|A=>B) = b/[a+b]
P(B|A=>B) = P(A=>B|B)P(B)/P(A=>B) = 1.P(B)/[P(B)P(A)+P(B)P(¬A)+P(¬B)P(¬A)] = b/[b+a-ab] or approximately P(B|A=>B) = b/[a+b].
So (to first order) we have that observing “A implies B” gives a probability of b/[a+b] for both A and B. So if we’re more sure that A is true than that B is false we have a 0.5; both A and B are likely (Modus Ponens). But if we’re more sure that B is false than that A is true we have b<a so b/[a+b] < 0.5; both A and B are unlikely (Modus Tollens).
To summarise: “one man’s modus tollens is another man’s modus ponens” occurs when two beliefs that we strongly believe come into conflict. In this situation our final beliefs depend on our relative confidences in our two beliefs. We keep the one we are more confident in, and discard the one we are less confident in. This means that two people who both strongly held those two beliefs could suddenly find themselves in disagreement; this happens when one of them thought that a<b and the other one thought that b<a.
EDIT: Note that I used A likely and B unlikely because the standard phrasing of MP is A and A=>B together give B. If we had taken ¬A instead of A we would have the situation where we have two unlikely beliefs, and we suddenly learn that one or the other has to be true. Similarly, if we had had ¬B instead of B we would have had the situation where we have two stong beliefs, and we suddenly learn that they conflict with each other. All these situations are equivalent.
The way I see it is this: We consider A likely, but B unlikely. Say P(A)=1-a and P(B)=b, where a and b are small. A and B are currently independent. Then we observe that A implies B (i.e. “¬A v B”). We get probabilities
P(A|A=>B) = P(A=>B|A)P(A)/P(A=>B) = P(B)P(A)/[P(B)P(A)+P(B)P(¬A)+P(¬B)P(¬A)] = [b-ab]/[b+a-ab] or approximately P(A|A=>B) = b/[a+b]
P(B|A=>B) = P(A=>B|B)P(B)/P(A=>B) = 1.P(B)/[P(B)P(A)+P(B)P(¬A)+P(¬B)P(¬A)] = b/[b+a-ab] or approximately P(B|A=>B) = b/[a+b].
So (to first order) we have that observing “A implies B” gives a probability of b/[a+b] for both A and B. So if we’re more sure that A is true than that B is false we have a 0.5; both A and B are likely (Modus Ponens). But if we’re more sure that B is false than that A is true we have b<a so b/[a+b] < 0.5; both A and B are unlikely (Modus Tollens).
To summarise: “one man’s modus tollens is another man’s modus ponens” occurs when two beliefs that we strongly believe come into conflict. In this situation our final beliefs depend on our relative confidences in our two beliefs. We keep the one we are more confident in, and discard the one we are less confident in. This means that two people who both strongly held those two beliefs could suddenly find themselves in disagreement; this happens when one of them thought that a<b and the other one thought that b<a.
EDIT: Note that I used A likely and B unlikely because the standard phrasing of MP is A and A=>B together give B. If we had taken ¬A instead of A we would have the situation where we have two unlikely beliefs, and we suddenly learn that one or the other has to be true. Similarly, if we had had ¬B instead of B we would have had the situation where we have two stong beliefs, and we suddenly learn that they conflict with each other. All these situations are equivalent.