@sorin.cristescu Don’t you think this must be like;
A=1
B=2
C=3
A+B=1+2/2=1.5
B+C=2+3/2=2.5
A+C=1+3/2=2
A+B+C=1+2+3/3=2
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@sorin.cristescu Don’t you think this must be like;
A=1
B=2
C=3
A+B=1+2/2=1.5
B+C=2+3/2=2.5
A+C=1+3/2=2
A+B+C=1+2+3/3=2
Posted using Partiko iOS
If I understand well you'd split the added value evenly. A+B gives indeed an added value of 3 (total is 6). But B+C has an added value of 0: B produces 2 alone and C produces 3 alone and B+C produces exactly 5.
Anyway, splitting the added value half-half seems fair when there are only two participants but becomes problematic at A+B+C. The added value is 4 (1 of A + 2 of B + 3 of C + 4 of added value from cooperation= 10). A receives 4/3 = 1.33 on top of his 1, for a total of 2.33. But if A only works with B then it gets 1+1.5 = 2.5. So for A, it makes more sense to work only with B than with B and C, although the A+B+C combination produces more (10) than the (A+B) combination and C taken separately (6+3 = 9)
The Shapley calculation (which you find on Wikipedia) takes this into account in order to compensate A such as it's in his own interest to enter into the A+B+C combination
yes A+B+C produce bigger value at the total but C always gets lower value because of A in return
"Lower" with respect to ?... Each of them starts by receiving what they would have gotten working alone, so C will anyway always get at least 3. Whether he receives more than that and how much more depends on how successful he is at cooperating. For instance, with B, the team is not great as it produces 0 added value. With A, the total added value is 2.5 so C receives 1.25. For A+B+C, as long as C gets at least 1.26 (instead of 4/3=1.33) it will be in his own interest to work in the trio. That 0.07 allow A and B to go from 1.33 to 1.36-1.37 for instance. Still not 1.5 though, but a bit better
Thaks for the witness vote!