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## Hot answers tagged dominant-strategy

8

Fixing the strategy of the opponent, a mixed strategy never yields a strictly higher utility if you are expected utility-maximizing. The reason is that the expected utility from a mixed strategy is at most as high as the highest utility from the pure strategies which this mixed strategy plays with positive probability. That is not to say that a mixed ...

4

Isn't the mixed extension of matching pennies a good example of this? The strategies $p,q$ are both elements of $[0,1]$, and \begin{align*} U_1(p,q) & = pq + (1-p)(1-q) \\ \\ U_2(p,q) & = p(1-q) + (1-p)q. \end{align*} There is a unique Nash-equilibrium at $p = q = 1/2$ but all $p$ and all $q$ are rationalizable, because in the equilibrium all ...

3

As I said in the comment, Nash's theorem shows the existence of a Nash equilibrium (possibly but not necessarily in mixed strategies). If you are interested in Nash equilibria is proper mixed strategies, i.e., NE in which all players play at least two actions with positive probability, you can easily show the following impossibility result: No such NE can ...

3

Consider the following game between P1 (row player) and P2 (column player): \begin{array}{|c|c|c|}\hline & L & R \\\hline T& 1,1 & 2,0 \\\hline B& 0,0 & 1,1 \\\hline \end{array} $T$ is P1's dominant strategy $T$ is P1's best response to both of P2's strategies $L$ and $R$ $L$ is P2's best response to P1's strategy $T$ $R$ is P2's ...

3

To ensure that the game has a single, completely-mixed, Nash equilibrium there needs to be "cyclic unique best responses", i.e., either BR(L) = O, BR(O) = R, BR(R) = U, BR(U) = L, or BR(L) = U, BR(U) = R, BR(R) = O, BR(O) = L. The strict cyclic preferences both rule out pure equilibria and allow a (unique) completely mixed equilibrium. There are ...

3

A Nash equilibrium that consists of weakly dominant strategies is a stronger solution concept than a NE itself. Consider the following simple matrix game where best replies have been marked with * \begin{array}{c|cc} P1/P2&\text{left}&\text{right}\\ \hline \text{Up}&1^*,1^*&0^*,0\\ \text{Down}&0,0^*&0^*,0^* \end{array} Both Up and ...

2

I remember slaving over the notation in this book when I was a bad undergraduate. It brings up some interesting memories, some which may help you. $F(x)$ is the cumulative distribution of a single bidder's valuation. $G(x)$ is the cumulative distribution of the highest bidder's valuation, given $N$ bidders. For the example you are referring to, values ...

2

I think you have a typo -- the equilibrium bidding strategy in the first-price auction you specify should be $$\beta(x) = \frac{N-1}{N}x$$ Here's a hint that might help. The CDF of the order statistic, $Y_1$, is $$G(x) = x^{N-1}$$ To see why this is the case, notice that this is exactly the probability that some given quantity $x$ is greater than or ...

2

Do you have a formal notion of what 'stable' means? Nash equilibria are often thought of informally as the strategies that support stable outcomes. If that's all that the term 'stable' means, then of course the outcomes $(5,10)$ and $(10,5)$ are also stable. As Lee Sin notes above, the outcome $(5,5)$ is also a Nash equilibrium outcome, and so is stable in ...

2

You're on the right track here. You need to check every outcome for its potential to be a NE. You're correct in stating that outcomes (5,10) and (10,5) are NEs however you didn't identify that (5,5) is also an NE. (5,5) If player one deviates he receives a payout of 5. If player two deviates he receives a payout of 5. Therefore no player has an incentive ...

2

\begin{array}{|c|c|c|} \hline &L&R\\\hline T&1,1&0,0\\\hline B&0,0&0,0\\\hline \end{array} In the game above, there are two pure strategy Nash equilibria: $(T,L)$ is an equilibrium in weakly dominant strategies; $(B,R)$ is an equilibrium in weakly dominated strategies. Noting that "dominant" and "dominated" are two different words,...

2

Repeated games and nonlinear utility Let's assume a trivial two-player game where each player has two options A and B; and the payout is +1/-1 if players pick the same and -1/+1 if players pick differently. Let's assume that the game is repeated 100 times with the strategies chosen and committed to beforehand. This means that if your opponent picks a fixed ...

2

A game outcome that is Pareto optimal or Pareto efficient is one where no one player can be made better off without making at least one player worse off. So a Nash equilibrium can easily be Pareto sub-optimal (or Pareto inefficient), which means that it is possible to one player can be made better off without making at least one player worse off. However, "...

2

In 2-player games, the strategies that survive iterated elimination of strictly dominated strategies are called rationalizable. Note that even if no strategy is strictly dominant, there can be strictly dominated strategies. If you cannot eliminate any strategy, then all strategies are rationalizable. Only if correlation of players' randomization is allowed, ...

1

The set of rationalizable strategies is the set of strategies that survive the iterated elimination of strictly dominated strategies, i.e., strategies that are never a best response. It is a weaker concept than Nash equilibrium. For player 1, you can eliminate strategy M, which is strictly dominated by T. You cannot eliminate any strategy for player 2 as ...

1

Hint Let $h_i^*(h_{-i}$) be firm $i$'s best response to the other $N-1$ firms' strategy profile $h_{-i}$. If $h_i^*(\cdot)$ is a dominant strategy, then it must be independent of the other firms' strategies, i.e. $h_i^*(h_{-i})=h_i^*$ for all $h_{-i}$.

1

The definition of a monopoly is one company having the exclusive control of a good or service. Since there are 2+ companies it does not fit the definition of a monopoly. A duopoly is a situation where 2 companies control the majority of the market. It sounds like this describes your situation. A more correct term may be oligopoly which describes a market ...

1

Start with the following matrix. $$\begin{matrix}1 && 0 && 0\\ 1 && 1 && 0\\ 1 && 1 && 1 \end{matrix}$$ You will choose a $1$ in the matrix and change it to a $0$. The other player will choose a $0$ and change it to a $1$. Your goal is for the resulting matrix to have an even determinant. The other player's ...

1

Q1 Your table seems to be correct. Here is a quick Python implementation for generating the payoffs: def payoff_calculator(x, y): if x+y < 14: return (x+1,y+1) else: if x==y: return (x,y) else: return (y,20-x) if x < y else (20-y,x) primes = [2,3,5,7,11,13,17,19] payoffs = [[payoff_calculator(i,...

1

I agree with Herr, the payoff matrix looks right. Also, there are no strictly dominated strategies because a strictly dominated strategy cannot be a best response for any possible belief. However, If any player believes that the other player is choosing 19, then every strategy (both pure and mixed) is a best response.

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