Mixed supports and saddle certificates

What you will learn

Construct two-sided pure-deviation bounds for a finite zero-sum game, repair a false support and characterize unused best-response ties and a continuum of equilibrium mixtures.

Prerequisites: Indifference · Mixed strategies · Stability, efficiency and solution assumptions · From outcomes to random variables

Specify a simultaneous game and a candidate support

Assume a single simultaneous move by two players, Row and Column. Row maximizes an expected net transfer of fictional chips and Column minimizes the same number; Column receives its negative. Players know the payoff matrix, randomize independently and do not observe the other current draw before acting. There are no fees, hidden types, repeated-play updates or wealth effects. These assumptions make the game zero sum with linear expected payoffs. They do not describe a complete poker hand. A mixed strategy is a distribution over available actions. Its support contains exactly the actions assigned positive probability, rather than every action that happens to be available.

Our original matrix has rows A, B, C and columns L, R. Row transfers are A=(4,−2), B=(−2,4), C=(c,c). The workshop changes c among zero, one and two while holding the other four cells fixed. Initially propose that Row uses only A and B. This proposal is a hypothesis about support, not permission to delete C. Column still has both columns. A successful support calculation must explain why neither player can improve using any available action, including a move assigned zero mass. Writing down fewer equalities can simplify a candidate calculation, but it cannot shrink the original deviation set.

Solve the restricted equalities and expose the omitted deviation

Let q be Column’s probability of L. Row’s three pure-action values are 6q−2, 4−6q and c. Indifference of A and B gives 6q−2=4−6q, hence q=1/2 and their common value is one. Within the proposed A,B support, let p be Row’s probability of A. Column L then delivers Row 6p−2 and Column R delivers 4−6p. Equating these gives p=1/2. The two probability constraints are satisfied, and the restricted two-by-two calculation is internally correct.

At c=0, C earns zero against this column mixture, so it cannot improve on one. At c=2, C earns two and defeats the candidate: switching all Row mass to C improves expected payoff by one chip. Nothing is wrong with either indifference equation. The failure is the unsupported inference from a solution in a subgame to a solution in the full game. C is available before and after the change; only its incentive changes. At c=1 it ties the supported actions. That tie is allowed even when C has zero probability. Equilibrium requires every used action to be best, not every best action to be used.

Build a two-sided certificate from every pure deviation

For any Row distribution x and Column distribution y, define the guarantee G as the minimum over columns of x’s expected row payoff. Define the cap H as the maximum over rows of the row payoff against y. The actual profile payoff u lies between G and H. To see the first inequality, u is a weighted average of the column values and cannot lie below their minimum. For the second, u is a weighted average of the row values and cannot exceed their maximum. These arguments also cover deviations to mixtures: a weighted average cannot improve beyond its best pure component.

If G=H=v, the bounds squeeze u to v. Row already earns its largest achievable value against y, and Column already holds Row to its smallest achievable value against x. Thus neither can improve unilaterally. This is a full-game saddle certificate. If the gap H−G is positive, the candidate is not an exact equilibrium: Row’s gain from a best unilateral response is H−u and Column’s gain, measured in Column’s own payoff, is u−G. Their sum equals H−G. Do not label the whole gap as one player’s gain. These are expected chips per play, not a probability of losing or an observed result from one random draw.

Repair the support and derive an equilibrium interval

At c=2, put all Row mass on C. Both column values then equal two, so G=2 regardless of the proposed column mixture. To obtain H=2 as well, require every row value against y to be at most two: 6q−2≤2 and 4−6q≤2. These inequalities give q≤2/3 and q≥1/3. Therefore every q in the closed interval [1/3,2/3], paired with pure C, is an equilibrium. This interval is continuous; the six-token workshop displays only its grid points 2/6, 3/6 and 4/6. A grid experiment is a visualization of the certificate, not proof that other fractions are excluded.

At q=1/3 the values of A,B,C are zero,two,two. B is an unused best response. That does not force Row to mix B into an equilibrium. For example, shift half the mass from C to B while keeping q=1/3. Row remains a best response to that fixed y, but the two column values become zero and three. Column can switch to L and lower Row’s payoff, so mutual optimality fails. This example separates a player’s best-response set from the set of equilibrium profiles. Against pure C both columns tie, yet an arbitrary column mixture is not necessarily an equilibrium: q=0 lets B earn four.

Check the arithmetic independently and construct contrasting witnesses

For the rejected c=2 candidate x=(1/2,1/2,0), y=(1/2,1/2), enumerate the four positive-probability cells. Each has probability 1/4, giving u=(4−2−2+4)/4=1. The column values are both one, while the largest row value is two. Hence G=1, H=2, and the gap is one. Independent direct deviation to C gives two instead of one, matching the Row improvement. This second calculation does not rely on solving equalities or trusting a software equilibrium label.

For the repaired x=(0,0,1), q=2/3, only cells C,L and C,R occur, with probabilities 2/3 and 1/3. Their payoffs are both two, so u=2. The pure row values are two,zero,two, and the pure column values are two,two. Both bounds equal two, even though A is tied and unused. At c=1, more degeneracy appears: any x=(t,t,1−2t) for 0≤t≤1/2 paired with q=1/2 has both column values one and all row values one. The formula proves the whole family, while token transfers can inspect selected members.

In the workshop, begin with c=0, move tokens to build equal A,B mass and equal L,R mass, enter G and H, and save the certificate. Raise c to two without changing the distributions and save the rejected candidate as evidence of the omitted deviation. Repair it by moving Row’s mass to C. Move Column to each interval endpoint and then outside the interval. Finally use c=1 to construct a tied support with some mass on C. Each edit clears the current certification; saved visit-local witnesses retain their assumptions. These experiments audit a finite matrix. They establish neither learning-algorithm convergence nor how a human opponent will behave.

Hands-on: Construct a full-game saddle certificate

Transfer conserved probability tokens, perturb an omitted row, build both payoff bounds and save rejected and repaired supports.

Common mistakes

• Deleting an available zero-probability action from the deviation audit.
• Requiring every tied best response to receive positive mass.
• Checking only Row after changing Row’s mixture.
• Treating the sum of both unilateral improvements as one player’s improvement.
• Reporting grid solutions as the entire continuous equilibrium set.

Keep these ideas

• Define support by positive mass and retain every available deviation.
• Compute the minimum column payoff G and maximum row payoff H.
• Equality of the bounds is a full-game equilibrium certificate.
• Unused ties and continuous families are compatible with equilibrium.

Further reading

• MIT OpenCourseWare: Mixed Strategy Nash Equilibrium, slides 31–32 and 36–38 ↗

  Read definitions of mixed strategies and mutual best responses, and the linearity argument on slides 36–38. Our three-by-two matrix, omitted-action perturbation, two-sided bounds and equilibrium families are original derivations. No full-course or video review claimed. Checked 2026-10-25.