Fahad Khalil
Econ 485
Answer Key 3
For 2 person games with 3 pure strategies, we will prove that a player will never use a
strictly dominated strategy with a positive probability in a NE.
This proof
can be
applied for strategy sets with two pure strategies also.
(c
1
)
(c
2
)
(1c
1
c
2
)
L
C
R
(r
1
)
T
x
1
, 
x
2
, 
x
3
, 
(r
2
)
M
y
1
, 
y
2
, 
y
3
, 
(1  r
1
 r
2
)
B
, 
, 
, 
We know that strictly dominated strategies are not used in NE if only pure strategy NE
are considered.
Here we show that the same is true even when mixed strategies are
considered.
Only the Row player's payoffs for T and M are needed for this proof.
The
other payoffs are not really relevant, but you can put in any if you want.
Suppose T strictly dominates M for Row.
Then, x
1
> y
1
, x
2
> y
2
, and x
3
> y
3
.
If Row plays T with probability 1 while Column plays with probability c = (c
1
, c
2
, 1  c
1

c
2
), then Row's expected payoff is:
U
R
(r
1
=1, c) = c
1
·x
1
+ c
2
·x
2
+ (1  c
1
 c
2
)·x
3
.
And Row's payoff if he plays M with probability one is:
U
R
(r
2
=1, c) = c
1
·y
1
+ c
2
·y
2
+ (1  c
1
 c
2
)·y
3
.
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 Spring '11
 eric
 Game Theory, Row, pure strategies, strictly dominated strategy, positive probability

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