For the bargaining outcome, I’ll assume we’re looking for a Nash Bargaining Solution (as suggested in another comment thread).
The defection point has expected utility 3p/2 for Theory I and expected utility 3q/2 for Theory II (using the same notation as I did in this comment).
I don’t see immediately how to calculate the NBS from this.
Then Theory I has expected utility 1, and Theory 2 has expected utility 1⁄2.
Assume (x,y) is the solution point, where x represents probability of voting for A (over B), and y represents probability of voting for C (over D). I claim without proof that the NBS has x=1 … seems hard for this not to be the case, but would be good to check it carefully.
Then the utility of Theory 1 for the point (1, y) = 1 + y/2, and utility of Theory 2 = 1 - y. To maximise the product of the benefits over the defection point we want to maximise y/2*(1/2 - y). This corresponds to maximising y/2 - y^2. Taking the derivative this happens when y = 1⁄4.
Note that the normalisation procedure leads to being on the fence between C and D at p = 2⁄3.
If I’m correct in my ad-hoc approach to calculating the NBS when p = 2⁄3, then this is firmly in the territory which prefers D to C. Therefore the parliamentary model gives different solutions to any normalisation procedure.
For the bargaining outcome, I’ll assume we’re looking for a Nash Bargaining Solution (as suggested in another comment thread).
The defection point has expected utility 3p/2 for Theory I and expected utility 3q/2 for Theory II (using the same notation as I did in this comment).
I don’t see immediately how to calculate the NBS from this.
Assume p = 2⁄3.
Then Theory I has expected utility 1, and Theory 2 has expected utility 1⁄2.
Assume (x,y) is the solution point, where x represents probability of voting for A (over B), and y represents probability of voting for C (over D). I claim without proof that the NBS has x=1 … seems hard for this not to be the case, but would be good to check it carefully.
Then the utility of Theory 1 for the point (1, y) = 1 + y/2, and utility of Theory 2 = 1 - y. To maximise the product of the benefits over the defection point we want to maximise y/2*(1/2 - y). This corresponds to maximising y/2 - y^2. Taking the derivative this happens when y = 1⁄4.
Note that the normalisation procedure leads to being on the fence between C and D at p = 2⁄3.
If I’m correct in my ad-hoc approach to calculating the NBS when p = 2⁄3, then this is firmly in the territory which prefers D to C. Therefore the parliamentary model gives different solutions to any normalisation procedure.