The document presents a method for determining Shapley values (φi) for elements in a set based on their marginal contributions. It defines the Shapley value formula and calculates the Shapley values for a set of 3 elements where different coalitions are assigned values. The Shapley values are calculated to be φ1=130/6, φ2=250/6, φ3=220/6. The sums of the Shapley values equals the total value of the grand coalition, demonstrating it is a valid allocation.

2. “It is not what we get. But who we become, what we contribute that
gives meaning to our lives.”
— Tony Robbins—

3.
(N, v)

4.
v({1,2,3}) = 100
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 40
v({1}) = 0
v({2}) = 0
v({3}) = 0
(N, v)

5.
(N, v)
v(S) − v(S − {i})
S S i
v({2,3}) = 80 v({1,2,3}) = 100
S i

6.
(N, v)
ϕi(v) =
∑
S:i∈S⊂N
(s − 1)!(n − s)!
n!
{v(S) − v(S − {i})}
i
Ss
v(ϕ) = 0, ϕ

7.
(s − 1)!(n − s)!
n!
S i
S
S
(2 − 1)!(3 − 2)!
3!
=
1
6
(3 − 1)!(3 − 3)!
3!
=
2
6
=
1
3

8.
(N, v)
ϕi(v) =
∑
S:i∈S⊂N
(s − 1)!(n − s)!
n!
{v(S) − v(S − {i})}
i
Ss
v(ϕ) = 0, ϕ

9.
(N, v)
ϕi(v) =
1
n! ∑
π∈Π
{v(Sπ,i ∪ {i}) − v(Sπ,i)}
i
Ππ
Sπ,i π i
πΠ
S(1→3→2), 3 = {1}
S(3→1→2), 2 = {1, 3}
π

10.
v({1,2,3}) = 100
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 40
v({1}) = 0
v({2}) = 0
v({3}) = 0
ϕi(v) =
1
n! ∑
π∈Π
{v(Sπ,i ∪ {i}) − v(Sπ,i)}
S(1→2→3), 1 = ϕ
v(ϕ) = 0
S(1→2→3), 1 ∪ {1} = {1}
v(S(1→2→3), 1 ∪ {1}) − v(S(1→2→3), 1)
= v({1}) − v(ϕ) = 0

11.
v({1,2,3}) = 100
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 40
v({1}) = 0
v({2}) = 0
v({3}) = 0
ϕi(v) =
1
n! ∑
π∈Π
{v(Sπ,i ∪ {i}) − v(Sπ,i)}
S(2→1→3), 1 = {2}
v(ϕ) = 0
S(1→2→3), 1 ∪ {1} = {1, 2}
v(S(1→2→3), 1 ∪ {1}) − v(S(1→2→3), 1)
= v({1, 2}) − v({2}) = 50

12.
v({1,2,3}) = 100
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 40
v({1}) = 0
v({2}) = 0
v({3}) = 0
ϕi(v) =
1
n! ∑
π∈Π
{v(Sπ,i ∪ {i}) − v(Sπ,i)}
S(2→3→1), 1 = {2, 3}
v(ϕ) = 0
S(2→3→1), 1 ∪ {1} = {1, 2, 3}
v(S(2→3→1), 1 ∪ {1}) − v(S(2→3→1), 1)
= v({1, 2, 3}) − v({2, 3}) = 20

13.
v({1,2,3}) = 100
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 40
v({1}) = 0
v({2}) = 0
v({3}) = 0
ϕi(v) =
1
n! ∑
π∈Π
{v(Sπ,i ∪ {i}) − v(Sπ,i)}
v(ϕ) = 0
ϕ1(v) =
1
3! ∑
π∈Π
{v(Sπ,i ∪ {1}) − v(Sπ,1)}
=
1
6
(0 + 0 + 50 + 20 + 40 + 20)
=
130
6

14.
v({1,2,3}) = 100
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 40
v({1}) = 0
v({2}) = 0
v({3}) = 0
ϕi(v) =
1
n! ∑
π∈Π
{v(Sπ,i ∪ {i}) − v(Sπ,i)}
v(ϕ) = 0
ϕ1(v) =
130
6
ϕ2(v) =
250
6
ϕ3(v) =
220
6
∑
i∈{1,2,3}
ϕi(v) =
130
6
+
250
6
+
220
6
= 100 = v({1, 2, 3})

15.
∑
i∈N
ϕi(v) =
∑
i∈N
1
n! ∑
π∈Π
{v(Sπ,i ∪ {i}) − v(Sπ,i)}
=
1
n! ∑
i∈N
∑
π∈Π
{v(Sπ,i ∪ {i}) − v(Sπ,i)}
=
1
n! ∑
π∈Π
∑
i∈N
{v(Sπ,i ∪ {i}) − v(Sπ,i)}
=
1
n! ∑
π∈Π
(v(Sπ,1 ∪ {1}) − v(Sπ,1) + v(Sπ,2 ∪ {2}) − v(Sπ,2) + ⋯v(Sπ,n ∪ {n}) − v(Sπ,n))
π = (1, 2, 4, 5, 3)
v(Sπ,1 ∪ {1}) − v(Sπ,1) + v(Sπ,2 ∪ {2}) − v(Sπ,2) + v(Sπ,3 ∪ {3}) − v(Sπ,3)) + v(Sπ,4 ∪ {4}) − v(Sπ,4)) + v(Sπ,5 ∪ {5}) − v(Sπ,5)
= v({1}) − v(ϕ) + v({1,2}) − v({1}) + v({1,2,3,4,5}) − v({1,2,4,5}) + v({1,2,4}) − v({1,2})) + v({1,2,4,5}) − v({1,2,4})
= v({1,2,3,4,5})
=
1
n! ∑
π∈Π
v(N) = v(N)
n!
n!
= v(N)

16.
ϕi(v) =
1
n! ∑
π∈Π
{v(Sπ,i ∪ {i}) − v(Sπ,i)} ≥
1
n! ∑
π∈Π
v({i}) = v({i})
v(Sπ,i ∪ {i}) − v(Sπ,i) ≥ v{i}

17.
x3
x1 = 20
x2 = 60
x3 = 50
ϕ(v) = (ϕ1(v), ϕ2(v), ϕ3(v)) = (
130
6
,
250
6
,
220
6
) ≒ (22,42,36)

18.
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

19.
∑
i∈N
ψi(v) = v(N)
∑
i∈N
ψi(v) < v(N)
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

20.
v(S ∪ {i}) = v(S)
ψi(v) = 0
i
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

21.
v(S ∪ {i}) = v(S ∪ {j})
ψi(v) = ψj(v)
i
i j
i, j
j
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

22.
ψi(v + w) = ψi(v) + ψi(w), ∀i ∈ N
v, w ∈ V
V
v + w (v + w)(S) = v(S) + w(S)
(cv)(S) = cv(S)
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

23.
v({1,2,3}) = 100
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 40
v({1}) = 0
v({2}) = 0
v({3}) = 0
v123(S) =
{
1, S = {1,2,3}
0
v12(S) =
{
1, S = {1,2}, {1,2,3}
0
v1(S) =
{
1, S = {1}, {1,2}, {1,3}, {1,2,3}
0
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

24.
v123(S) =
{
1, S = {1,2,3}
0
(N, v123)
ψ(v123) = (1/3, 1/3, 1/3)
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

25.
(N, v12)
ψ(v12) = (1/2, 1/2, 0)
v12(S) =
{
1, S = {1,2}, {1,2,3}
0
v({123}) = v({12})
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

26.
(N, v1)
ψ(v1) = (1, 0, 0)
v1(S) =
{
1, S = {1}, {1,2}, {1,3}, {1,2,3}
0
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

27.
ψ(v1) = (1, 0, 0)
ψ(v2) = (0, 1, 0)
ψ(v3) = (0, 0, 1)
ψ(v12) = (1/2, 1/2, 0)
ψ(v13) = (1/2, 0, 1/2)
ψ(v23) = (0, 1/2, 1/2)
ψ(v123) = (1/3, 1/3, 1/3)
v = α1v1 + α2v2 + α3v3 + α12v12 + α13v13 + α23v23 + α123v123
ψ(v) = ψ(α1v1 + α2v2 + α3v3 + α12v12 + α13v13 + α23v23 + α123v123)
= ψ(α1v1) + ψ(α2v2) + ψ(α3v3) + ψ(α12v12) + ψ(α13v13) + ψ(α23v23) + ψ(α123v123)
= α1ψ(v1) + α2ψ(v2) + α3ψ(v3) + α12ψ(v12) + α13ψ(v13) + α23ψ(v23) + α123ψ(v123)
= (α1 +
1
2
α12 +
1
2
α13 +
1
3
α123, α2 +
1
2
α12 +
1
2
α23 +
1
3
α123, α3 +
1
2
α13 +
1
2
α23 +
1
3
α123)
vxx
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

28.
v = α1v1 + α2v2 + α3v3 + α12v12 + α13v13 + α23v23 + α123v123
v({1}) = α1v1({1}) + α2v2({1}) + α3v3({1}) + α12v12({1}) + α13v13({1}) + α23v23({1}) + α123v123({1})
= α1
v({1}) = 0 α1 = 0
α2 = 0, α3 = 0
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

29.
v = α1v1 + α2v2 + α3v3 + α12v12 + α13v13 + α23v23 + α123v123
= α1 + α2 + α12 = α12
v({1,2}) = 50 α12 = 50
α13 = 40, α23 = 80
v({1,2}) = α1v1({1,2}) + α2v2({1,2}) + α3v3({1,2}) + α12v12({1,2}) + α13v13({1,2}) + α23v23({1,2}) + α123v123({1,2})
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

30.
v = α1v1 + α2v2 + α3v3 + α12v12 + α13v13 + α23v23 + α123v123
= α1 + α2 + α3 + α12 + α13 + α23 + α123 = 170 + α123
v({1,2,3}) = 100 α123 = − 70
v({1,2,3}) = α1v1({1,2,3}) + α2v2({1,2,3}) + α3v3({1,2,3}) + α12v12({1,2,3}) + α13v13({1,2,3}) + α23v23({1,2,3}) + α123v123({1,2,3})
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

31.
v = α1v1 + α2v2 + α3v3 + α12v12 + α13v13 + α23v23 + α123v123
v = 50v12 + 40v13 + 80v23 − 70v123
ψ(v) = ψ(50v12 + 40v13 + 80v23 − 70v123)
= ψ(50v12) + ψ(40v13) + ψ(80v23) + ψ(−70v123)
= 50ψ(v12) + 40ψ(v13) + 80ψ(v23) − 70ψ(v123)
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))

32.
ψ(v) = (ψ1(v), ψ2(v), ⋯, ψn(v))
v = α1v1 + α2v2 + α3v3 + α12v12 + α13v13 + α23v23 + α123v123
v + 70v123 = 50v12 + 40v13 + 80v23
ψ(v + 70v123) = ψ(50v12 + 40v13 + 80v23)
ψ(v) + ψ(70v123) = ψ(50v12) + ψ(40v13) + ψ(80v23)
ψ(v) + 70ψ(v123) = 50ψ(v12) + 40ψ(v13) + 80ψ(v23)
ψ(v) = 50ψ(v12) + 40ψ(v13) + 80ψ(v23) − 70ψ(v123)
= 50(
1
2
,
1
2
,0) + 40(
1
2
,0,
1
2
) + 80(0,
1
2
,
1
2
) − 70(
1
3
,
1
3
,
1
3
) = (
65
3
,
125
3
,
110
3
) = (
130
6
,
250
6
,
220
6
)

33.
ϕi(v) =
∑
S:i∈S⊂N
(s − 1)!(n − s)!
n!
{v(S) − v(S − {i})}

36. See you next time