# Am I correct in thinking this game has neither Nash Equilibria nor dominant strategies?

I've taken this example from some lecture slides. The slides state there is no Nash equilibrium. I suspect there is also no dominant strategy for either player. Is this true?

Two players $i$ and $j$ simultaneously call either $heads$ or $tails$. If they call the same, player $i$ wins. Otherwise player $j$ wins.

The pay-off matrix is:

$$\begin{matrix} i/j & H & T\\ H & 1,-1 & -1,1 \\ T & -1,1 & 1-1 \end{matrix}$$

I think there are no dominant strategies because for $i$ the utility of (H,H) $\geq$ (T,T) $\gt$ (H,T) $\geq$ (T,T) where (A,B) denotes $i$ played A and $j$ played B. (1 $\geq$ 1 $\gt$ -1 $\geq$ -1). A similar situation occurs for $j$.

Does that reasoning seem correct?

Thanks!

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There is, indeed, no pure-strategy Nash equilibrium, and no dominant strategy; your arguments are correct. On the other hand, there is a mixed-strategy Nash equilibrium: choose either move with probability 0.5.

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There are on the contrary mixed strategy equilibria. This equilibrium can be found by noting that a strategy pair $(\sigma_1,\sigma_2)$ is a mixed strategy NE, iff every player is indifferent between the pure strategies played with positive probability in an equilibrium and each player weakly prefers the strategies played with positive probability to those played with zero probability.
Thus assuming player $j$ plays heads with probability $p$, player $i$ would obtain: $u(H,p)=p-(1-p)=2p-1$ playing $H$, and obtain $u(T,p)=-p+(1-p)=1-2p$ playing $T$. These two utilities are equal only if $p=1/2$. Assuming $i$ plays heads with probability $q$, by similar reasoning we obtain $q=1/2$.
In this case, this is the reason the equilibrium is at $p = q = 0.5$. For other games, the equilibrium point may lie elsewhere. –  Johannes Kloos Apr 2 '12 at 19:03