Two-finger Morra

Idea. Neither player has a safe pure strategy, so both mix. Each mixes so as to make the other indifferent between showing 1 and 2 fingers.

Step 1: A's payoffs. With A's choice first and B's second: $(1,1)\to +2$, $(2,2)\to +4$, $(1,2)$ and $(2,1)\to -3$.

Step 2: A's mix. If A shows 1 with probability $p$:

• against B showing 1: $2p-3(1-p)=5p-3$;
• against B showing 2: $-3p+4(1-p)=4-7p$.

These are equal when $12p=7$, so $p=\frac{7}{12}$, and then A gets $5\cdot \frac{7}{12}-3=-\frac{1}{12}$ whatever B does.

Step 3: B's mix. By the same symmetric payoffs, B showing 1 with probability $\frac{7}{12}$ holds A to $-\frac{1}{12}$ whatever A does. So the value is $-\frac{1}{12}$.

The trap. The matches pay A a total of $2+4=6$ and the mismatches cost $3+3=6$, so the game looks fair. It is not: B can steer away from the expensive $(2,2)$ outcome, and wins about 8 cents per round.