{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# a3 - Fahad Khalil Answer Key 3 Econ 485 For 2 person games...

This preview shows pages 1–2. Sign up to view the full content.

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 ) (1-c 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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

a3 - Fahad Khalil Answer Key 3 Econ 485 For 2 person games...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online