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Which (pure strategies) in each game are dominated? For each dominated strategy specify the (mixed) strategy that dominates it.


The solution manual says that

in Game 1, R is dominated by $\sigma_2 = (1/2, 1/2, 0)$

and B is dominated by $\sigma_1 = (1/2, 1/2, 0)$

in Game 2, B is dominated by $\sigma_1 = (3/5, 2/5, 0)$

and M is weakly dominated by $\sigma_2 = (1/4, 0, 3/4)$


I don't understand this solution completely. And also, how these sigma values are written?

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  • $\begingroup$ In game 1, $\sigma_2=(1/2,1/2,0)$ mean a mixed strategy half L and half M, with payoffs for Player 2 of $3,2,2$ which dominates the payoffs for Player 2 from R of $1,1,1$ $\endgroup$
    – Henry
    Apr 10, 2021 at 20:43

1 Answer 1

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We will see these in order:

In Game 1, R is dominated by $\sigma_2 = (1/2, 1/2, 0)$

Notation: $\sigma_2$ is a mixed strategy for player-2 to play the choices $L,M$ with probability $\frac{1}{2}$ each.

Explanation for dominance: $\sigma_2$ dominates R as in R, player-2 wins 1 irrespective of what player-1 plays (see the right side of the column 3 values on table 1). Thus the expected earnings for player-2 when player-1 plays strategies $\{T,C,B\}$ is $\{1,1,1\}$. On the other hand, if Player-2 plays $\sigma_2$, then the expected earning is $\{\frac{4+2}{2},\frac{0+4}{2},\frac{3+1}{2}\} = \{3,2,2\}$ when player-1 plays $\{T,C,B\}$.

Note that $\{3,2,2\}$ dominates $\{1,1,1\}$. Therefore, irrespective of what player-1 plays, player-2 should play $\sigma_2$. Thus, $\sigma_2$ dominates R


In Game 1, B is dominated by $\sigma_1 = (1/2, 1/2, 0)$ The expected payoff to player-1 under B when player-2 plays $\{L,M,R\}$ is $\{1,1,1\}$ The expected payoff to player-1 under $\sigma_1$ when player-2 plays $\{L,M,R\}$ is $\{\frac{4+2}{2},\frac{0+4}{2},\frac{3+1}{2}\} = \{3,2,2\}$, which dominates $\{1,1,1\}$. Thus $\sigma_1$ dominates B.


In Game 2, B is dominated by $\sigma_1 = (3/5, 2/5, 0)$

The expected payoff to player-1 under B when player-2 plays $\{L,M,R\}$ is $\{2,2,2\}$ The expected payoff to player-1 under $\sigma_1= (3/5, 2/5, 0)$ when player-2 plays $\{L,M,R\}$ is $\{\frac{3}{5}\times 0 + \frac{2}{5}\times 6,\frac{3}{5}\times 3 + \frac{2}{5}\times 1,\frac{3}{5}\times 5 + \frac{2}{5}\times 6\} = \{\frac{12}{5},\frac{11}{5},\frac{27}{5}\}$, which dominates $\{2,2,2\}$. Thus $\sigma_1$ dominates B.


In Game 2, M is weakly dominated by $\sigma_2 = (1/4, 0, 3/4)$

The expected payoff to player-2 under M when player-1 plays $\{T,C,B\}$ is $\{3,3,6\}$ The expected payoff to player-2 under $\sigma_2= (\frac{1}{4}, 0, \frac{3}{4})$ when player-1 plays $\{T,C,B\}$ is $\{\frac{1}{4}\times 2 + \frac{3}{4}\times 6,\frac{1}{4}\times 6 + \frac{3}{4}\times 2,\frac{1}{4}\times 0 + \frac{3}{4}\times 8\} = \{5,3,6\}$, which weakly dominates $\{3,3,6\}$.

We say $\sigma_2$ weakly dominates M because (i) Payoff($\sigma_2$,Player-1 move) $\geq$ Payoff(M,Player-1 move) $\forall$ Player-1 moves, but (ii) $\exists$ Player-1 move such that Payoff($\sigma_2$,Player-1 move) $>$ Payoff(M,Player-1 move)

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  • $\begingroup$ Thank you for your great explanation. Then, for example, in the game 2 and the case of B, $\sigma_1=(3/5, 2/5,0)$ is chosen randomly? That’s, according to your explanation, when I calculate $\sigma_1=(p,1-p,0)$, I have found that $p\in (1/2, 2/3)$. That’s, can we choose any p value randomly in this range such that B is dominated? $\endgroup$
    – 1190
    Apr 11, 2021 at 12:12
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    $\begingroup$ yes, and then you may even have dominance between such strategies. ie given some $p_1, p_2$, find if $\sigma_1^1=(p_1,1-p_1,0)$ dominates $\sigma_1^2=(p_2,1-p_2,0)$. Optimizing over such simplices, one can find a single optimal (or an interval of equally optimal) strategies $\endgroup$ Apr 11, 2021 at 12:19
  • $\begingroup$ Okay thank you. I clearly understand the answer thanks to you:) $\endgroup$
    – 1190
    Apr 11, 2021 at 12:20
  • $\begingroup$ Dear Madhavan, can you please look at my this question as well? I will be glad if you will make a clear explanation as it is. Thank you:) math.stackexchange.com/questions/4101269/… $\endgroup$
    – 1190
    Apr 14, 2021 at 8:47

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