rock-paper-scissors
a. Let xi = the probability that player R chooses action i, for i ∈ {Rock, Paper, Scissors}. Then Player R’s
maximin strategy can be found with the following optimization model:
min {x2 − x3 , −x1 + x3 , x1 − x2 }
maximize
subject to x1 + x2 + x3 = 1
x1 ≥ 0, x2 ≥ 0, x3 ≥ 0
This model is equivalent to the following linear program:
maximize
z
subject to z ≤ x2 − x3
z ≤ −x1 + x3
z ≤ x1 − x2
x1 + x2 + x3 = 1
x1 ≥ 0, x2 ≥ 0, x3 ≥ 0
b. Let vi = the probability the player C chooses action i, for i ∈ {Rock, Paper, Scissors}. Then Player C’s
minimax strategy can be found with the following optimization model:
max {−v2 + v3 , v1 − v3 , −v1 + v2 }
minimize
subject to v1 + v2 + v3 = 1
v1 ≥ 0, v2 ≥ 0, v3 ≥ 0
This model is equivalent to the following linear program:
minimize
w
subject to w ≥ −v2 + v3
w ≥ v1 − v3
w ≥ −v1 + v2
v1 + v2 + v3 = 1
v1 ≥ 0, v2 ≥ 0, v3 ≥ 0
c. Let’s rewrite player R’s linear program so that all the decision variables are on the left hand side of
the constraints, and all the constants are on the right hand side:
maximize
z
subject to z − x2 + x3 ≤ 0
(v1 )
z + x1 − x3 ≤ 0
(v2 )
z − x1 + x2 ≤ 0
(v3 )
x1 + x2 + x3 = 1
(w)
x1 ≥ 0, x2 ≥ 0, x3 ≥ 0
The dual of this linear program is
minimize
w
subject to v1 + v2 + v3 = 1
v2 − v3 + w ≥ 0
− v1 + v3 + w ≥ 0
v1 − v2 + w ≥ 0
v1 ≥ 0, v2 ≥ 0, v3 ≥ 0, w free,
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which is just player C’s linear program, rewritten so that all the decision variables are on the left hand
side of the constraints, and all the constants are on the right hand side.
d. We can represent the “play each alternative with probability 1/3” strategy for player R with the
is feasible for player R’s
solution x1 = x2 = x3 = 13 . Note that this solution linear program. The
objective function value of this feasible solution is min 31 − 13 , − 13 + 31 , 13 − 31 = 0.
e. Similarly, we can represent the “play each alternative with probability 1/3” strategy for player C with
the solution v1 = v2 = v3 = 31 , whichis feasible for player C’s linear program. The objective function
value of this feasible solution is max − 31 + 13 , 13 − 13 , − 31 + 31 = 0.
f. Note that we have feasible solutions to player R’s and player C’s linear programs with equal objective
function values. Since these linear programs form a primal-dual pair, by weak duality, these solutions
must be optimal. Therefore, we can conclude that the “play each alternative with probability 1/3”
strategy is optimal for both player R and player C.
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