If we can ignore the utility value of the negotiating pie (e.g. Alice may not want a 5th slice anyway), it is an interesting perspective. If only there were more examples and less explanation.
In general: Suppose the fallback positions are x slices for Alice, and y slices for Bob, and they get n slices total if they reach a deal. Then we can figure out the split by defining the gain, g, as the additional pieces they get if they reach a deal. So g=n-(x+y). Then the fair split should be x+g/2 for Alice and y+g/2 for Bob.
Another kind of situation is if there are 3 people, Alice, Bob, and Charlie, with fallbacks x, y, z respectively. Suppose they all need to reach an agreement to get n slices. Then g=n-(x+y+z) and Alice, Bob, and Charlie should get x+g/3, y+g/3, z+g/3 respectively.
Another possible situation is if there are 3 people, but they have unequal roles. Alice only needs to make a deal with at least one of Bob or Charlie, to get n slices. If Bob and Charlie both agree to the deal, then the total is still n slices. Intuitively, this would make Bob and Charlie less important, so they should get less of the gain. The split here should be x+2g/3 for Alice, y+g/6 for Bob and z+g/6 for Charlie, if they're all in on the deal. The reasoning there is that if we're adding games together, then we should also add the payoffs together. We can make a Alice&(Bob|Charlie) game by adding together Alice&Bob + Alice&Charlie - Alice&Bob&Charlie. Similarly, the payoffs should add like this: (g/2, g/2, 0) + (g/2, 0, g/2) - (g/3, g/3, g/3) = (2g/3, g/6, g/6)
Wow thanks for the very detailed explanation, mate, that is much appreciated. However, I still wish the author had delved into more real world examples, to let us have more sense of how this principle can be applied into more practical use cases. "Algorithms To Live By" https://www.goodreads.com/book/show/25666050-algorithms-to-l... did a really good job at that. Of course an article cannot be at book length but a man can hope.
Another kind of situation is if there are 3 people, Alice, Bob, and Charlie, with fallbacks x, y, z respectively. Suppose they all need to reach an agreement to get n slices. Then g=n-(x+y+z) and Alice, Bob, and Charlie should get x+g/3, y+g/3, z+g/3 respectively.
Another possible situation is if there are 3 people, but they have unequal roles. Alice only needs to make a deal with at least one of Bob or Charlie, to get n slices. If Bob and Charlie both agree to the deal, then the total is still n slices. Intuitively, this would make Bob and Charlie less important, so they should get less of the gain. The split here should be x+2g/3 for Alice, y+g/6 for Bob and z+g/6 for Charlie, if they're all in on the deal. The reasoning there is that if we're adding games together, then we should also add the payoffs together. We can make a Alice&(Bob|Charlie) game by adding together Alice&Bob + Alice&Charlie - Alice&Bob&Charlie. Similarly, the payoffs should add like this: (g/2, g/2, 0) + (g/2, 0, g/2) - (g/3, g/3, g/3) = (2g/3, g/6, g/6)