A variant of odds or evens

Problem

A variant of odds or evens: if the sum of the numbers shown is even, G1 wins $2$ EUR; if it is odd, G2 wins $1$ EUR. Is it a zero-sum game? What is G1’s optimal mixed strategy?

Solution

The game is not zero-sum: the two outcomes have payoffs of different value ($+2$ EUR for G1 against $+1$ EUR for G2), so one player’s winnings are not the exact opposite of the other’s loss. Let $p$ denote the probability that G1 plays «even» and $q$ the probability that G2 plays «even». The sum is even when both play even or both play odd. G1’s expected winnings are $\bar V_1(p,q) = 2\bigl[pq + (1-p)(1-q)\bigr] - \bigl[p(1-q) + (1-p)q\bigr].$ Seeking the strategy of G1 that makes $\bar V_1$ independent of the opponent’s choice $q$ (min-max setting), one obtains $\boxed{p = \tfrac{1}{3}}$ while the corresponding strategy of G2 turns out to be $q = \tfrac{2}{3}$. It is worth checking the calculations by substituting these values.

Links

Topics: Probabilita
Concepts: Gioco a somma zero · Strategia mista
Skills: Calcolo probabilita · Modellizzare
Exercise type: Problema modellizzazione