The Marble Game

Players A and B each hold a red and a blue marble and secretly present one to the other. If both present red, A wins $3. If both present blue, A wins $1. If the colours differ, B wins $2. The winnings come from an outside bank, not from the other player. Is it better to be A, better to be B, or does it not matter?

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#Set up the payoffs

Let A show red with probability $p$ and B show red with probability $q$ . A is paid only on a match and B only on a mismatch, so

\E[A] = 3pq + 1(1-p)(1-q), \qquad \E[B] = 2\big[p(1-q) + (1-p)q\big]. \tag{1}

A collects on the matching diagonal, three dollars on red-red and one on blue-blue, while B collects two dollars on either mismatch. At the mixed equilibrium B nets one dollar a round against A's seventy-five cents.

#Best responses

A gains from raising $p$ exactly when $\dfrac{\partial \E[A]}{\partial p} = 4q - 1 > 0$ , so A goes all red above $q = \tfrac14$ and all blue below it. B gains from raising $q$ when $\dfrac{\partial \E[B]}{\partial q} = 2(1 - 2p) > 0$ , so B goes all red below $p = \tfrac12$ and all blue above. Neither best response is constant, so the stable play is mixed.

#The equilibrium

The responses balance at $p = \tfrac12$ and $q = \tfrac14$ , where

\E[A] = 3 \cdot \tfrac12 \cdot \tfrac14 + 1 \cdot \tfrac12 \cdot \tfrac34 = \tfrac38 + \tfrac38 = \tfrac34, \tag{2}

\E[B] = 2\big[\tfrac12 \cdot \tfrac34 + \tfrac12 \cdot \tfrac14\big] = 2 \cdot \tfrac12 = 1. \tag{3}

#Result

At equilibrium B expects $1.00 and A expects $0.75, so take the role of B. A's headline $3 prize is rare and easy to dodge, while B banks a steady $2 every time the colours disagree.