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A mixed strategy is a probability distribution over a player's pure strategies. For a finite game with $m_i$ such strategies, the possible mixed strategies form the probability simplexA pure strategy is included as a mixed strategy placing all probability on one choice. The support of a mixed strategy consists of the choices with positive probability. In a sequential game, sampling a complete pure strategy once is the definition; it is not automatically equivalent to fresh randomization at every move.

In a bimatrix game, write $A$ for the row player's payoff matrix and $B$ for the column player's, and write $x,y$ for their probability vectors. Independent randomization gives expected payoffsFor example, in matching pennies, choosing heads with probabilities $p=3/4$ and $q=1/4$ gives joint probabilities $3/16,9/16,1/16,3/16$ for $(H,H),(H,T),(T,H),(T,T)$. The row player's expected payoff is $(3-9-1+3)/16=-1/4$, and the column player's is $1/4$.

Randomization can prevent an opponent exploiting a predictable choice. It does not mean every choice is equally good against every opponent: only best responses belong in the support of a mixed strategy at a Nash equilibrium. A shared random signal can instead produce a lottery over joint action profiles whose joint probabilities do not factor into independent mixed strategies.

A Nash equilibrium is a strategy profile in which every player's strategy is a best response to the others. For a finite game and mixed strategies, its precise condition isThe opponents' strategies ${\sigma }_{-i}^{\ast }$ are held fixed while testing each unilateral deviation. Equality is allowed: a player may have other equally good best responses. Because expected payoff is a linear function of the player's own mixture, it suffices to test every pure strategy. Equivalently, every choice in the support of a mixed strategy must attain the largest pure-choice expected payoff, and no unused choice may exceed it.

The two worked examples illustrate different uses of this condition. Matching pennies has no pure strategy Nash equilibrium, but its unique mixed strategy Nash equilibrium removes each player's vulnerability to predictable play. The asymmetric coordination example has three Nash equilibria, and its mixed one fails Pareto efficiency. A Nash equilibrium tests individual incentives, whereas Pareto efficiency compares feasible outcomes that may require several players to change together. In the prisoner's dilemma, for example, mutual defection is a Nash equilibrium even though mutual cooperation gives both players a larger payoff because $R>P$.

Existence in a finite game is guaranteed by Nash's theorem. The theorem does not select one Nash equilibrium, promise Pareto efficiency, or assert convergence of a learning process. Nash's ^[ref] used the Kakutani fixed-point theorem; his ^[ref] gave a proof using the Brouwer fixed-point theorem. The connection to a fixed point captures the idea of choices that remain mutually optimal when considered together.

The players are the two chains, each with the set of pure strategies $\{1,2,3,4,5\}$. A strategy profile $(x,y)$ gives the sites of $S$ and $C$. For later use, write the first player's payoff for a general street size asWhen $x<y$, count customers to the left of $(x+y)/2$, assigning value one rather than two to any midpoint customer. This gives $x+y-1$ whether the midpoint is an occupied building or lies between buildings. The other player receives $2k-u(x,y)$. At co-location, every customer's value is split, giving $k$ each.

For $k=5$, the payoff matrix for $S$ isRows and columns are indexed by sites $1,\dots ,5$; the full ordered payoff pair in cell $(x,y)$ is $(M_{xy},10-M_{xy})$. This discrete location game is a constant-sum game.

Neither player has a weakly dominant strategy among its pure strategies. Against an opponent at site one, the unique best response is site two, yielding $2k-2$. Co-location yields only $k$, and moving further right reduces one's payoff. Against an opponent at site $k$, the unique best response is $k-1$, by reflection. For $k\ge 4$, these two sites differ. No single pure strategy can therefore be a best response to both opposing actions, as a weakly dominant strategy would require.

Propagate an entire utility pair from a child chosen to maximize the component of the player whose turn it is. This is the two-player case of max-n search. At a terminal node let $V(s)=(U_A(s),U_B(s))$. At an A-node chooseand at a B-node replace $V_A$ by $V_B$. Do not maximize the two coordinates independently: their separate maxima may belong to incompatible outcomes. Fix a tie rule, such as the first child in the specified order. The rule recursively gives a pure strategy subgame perfect equilibrium in a finite perfect-information game: backward induction makes the moving player's selected continuation a best response at every node.^[ref]

In the original Figure 2, each of the three children of the root belongs to B. The left B-node prefers its first leaf because $1>0$, the middle B-node prefers its first leaf because $2>-2$, and the right B-node prefers its second leaf because $3>1$. Their propagated utility pairs, from left to right, areAt the root A compares $1,-1,-1$ and chooses the left child. Thus the root pair is $(1,1)$, with play following the root's left branch and then that B-node's first branch. No tie affects this outcome.

Use a simultaneous two-player normal-form game, with X choosing the row and Y the column. Each player's pure strategies are $b$ and $nb$; a payoff pair lists X first. Measure every dataset size in millions and use the stated normalization $M=1$. The source's reference to a shared half as “points” is interpreted in the same million-point units as the corpus, so the shared addition is $n/2$ to each initial dataset.

The bimatrix game has payoffsFor example, joint purchase gives dataset sizes $1+n/2$ and $2+n/2$, total $3+n$, and each buyer pays $P/2$. A sole buyer gets the full corpus and pays $P$. The total payoff is $1-P$ whenever a purchase occurs and $1$ when neither buys: the game is not a zero-sum game, even though gross market shares always sum to one.

For the following calculations, setThe payoff differences “buy minus do not buy,” with the other player's action held fixed, areThese four differences give all the best response comparisons without mixing up row and column payoffs.

A weakly dominant strategy must be a best response to either opponent action. For X this requiresThe first is stronger because $t>0$. ThusFor Y the comparisons are $t/3-P\ge 0$ and $(t-P)/2\ge 0$, givingBelow each upper endpoint the corresponding strategy is strictly better against both opponent actions; at the endpoint one comparison ties and the other is strict. Thus the answers also satisfy the convention that weak dominance requires a strict advantage somewhere. Since expected payoffs are linear in the opponent's mixing probability, checking both opponent pure strategies suffices for every opponent mixed strategy.

Neither asymmetric purchase outcome is a Nash equilibrium for any permitted $P>0,n>0$. At $(b,nb)$, X's chosen purchase requires $P\le 2t/3$, but Y's chosen nonpurchase, against X buying, requires $P\ge t$. These requirements are incompatible because $2t/3<t$.

At $(nb,b)$, Y's chosen purchase requires $P\le t/3$, whereas X's nonpurchase against a buying Y requires $P\ge t$. Again the inequalities are incompatible. Thus every Nash equilibrium in pure strategies has both players taking the same action; the conclusion is stronger than merely observing symmetry in their available actions, since their original market shares differ.

Use the two-by-two payoff entries visible in the original PDF. If $q$ is the probability that II uses its first action, I's expected payoffs from its two pure strategies are $4-q$ and $2q$. Their difference $4-3q$ is strictly positive for every $q\in [0,1]$. Thus I's first action strict dominance its second, and every Nash equilibrium has I use the first action with probability one. Against that action, II receives 8 from its first action and 4 from its second, so its first action is uniquely optimal. ThereforeThis excludes nontrivial mixed equilibria as well as the other pure pairs.

For the maximin bargaining solution, first compute the two security level payoffs. If I uses its first action with probability $p$, its payoffs against II's two pure strategies are $2+p$ and $4p$. Both increase with $p$, so its best worst-case payoff is attained at $p=1$ and is $d_1=3$. If II uses its first action with probability $q$, its payoffs against I's two actions are $4+4q$ and $6-6q$. The lower of these is maximized at their intersection,These are separate security calculations, not the expected payoffs from a pair of security strategies.

Cooperation permits jointly chosen lotteries, giving the convex hull of the four payoff. The vector $(3,8)$ dominates $(2,0)$ and $(0,6)$, so the Pareto frontier is the segment from $(3,8)$ to $(4,4)$, with equation $v=20-4u$. Individual rationality relative to $d=(3,24/5)$ restricts this toOn this negotiation set, the Nash product isIts unique maximizer is interior, givingOne implementing binding agreement always uses I's first action and uses II's first action with probability $3/5$, its second with probability $2/5$. This agreement raises both payoffs above their security levels, but II receives $32/5<8$, less than at the noncooperative equilibrium. Thus player II prefers the noncooperative game under the specified maximin bargaining rule. The cooperative feasible set contains the equilibrium payoff; it is the particular arbitration rule, not an inability to cooperate at that payoff, that produces II's lower allocation.

For a bimatrix game with row and column payoff matrices $A,B$, an equilibrium pair is a pair of mixed strategies $(p,q)$ for which neither player gains by a unilateral change:This is a Nash equilibrium, or mutual best response pair. Every finite game with nonempty finite pure-strategy sets has at least one mixed-strategy equilibrium by Nash's theorem. Equivalently, the strategy domains are compact convex probability simplexes and the expected payoffs are continuous and linear in each player's own strategy. The existence assertion need not hold if only pure strategies are allowed.

Write the probability vectors in the order Bach, Mozart, Schubert. Against $q=(q_B,q_M,q_S)$ the row player's three pure-action payoffs areAgainst $p=(p_B,p_M,p_S)$ the column player's payoffs areA supported action must attain the corresponding maximum, and an omitted action must give no larger payoff.

We first justify that both players have the same support of a mixed strategy, rather than assume it. Each player's maximum payoff is strictly positive for every opponent probability vector. Thus $p_M>0$ implies $q_M>0$, $q_B>0$ implies $p_B>0$, and $p_S>0$ is equivalent to $q_S>0$. If $p_B>0$ but $q_B=0$, then either $q_M>0$, making Mozart's payoff $2q_M$ strictly exceed Bach's $q_M$, or $q_S=1$, making Bach's payoff zero and Schubert's positive. Both contradict best response. Therefore $p_B>0$ also implies $q_B>0$. Similarly, if $q_M>0$ but $p_M=0$, then $p_B>0$ makes the column payoff for Bach $2p_B$ exceed Mozart's $p_B$, or $p_S=1$ makes Mozart's payoff zero and Schubert's positive. Thus $q_M>0$ implies $p_M>0$. The supports coincide, and there are only seven nonempty possibilities.

For singleton supports the three matching pure pairs are equilibria. For each two-element support, solve the two players' indifference equations and normalize their probabilities. For example, on Bach–Mozart, the equations are $4q_B=q_M$ and $p_B=4p_M$, giving $q=(1/5,4/5,0)$ and $p=(4/5,1/5,0)$. On Bach–Schubert they are $4q_B=3q_S$ and $2p_B=3p_S$. On Mozart–Schubert they are $2q_M=3q_S$ and $4p_M=3p_S$. In each case the omitted action gives strictly less than the common supported payoff.

For full support, the row indifference equations give $q_M=4q_B$ and $3q_S=2q_M$; the column equations give $p_B=4p_M$ and $3p_S=2p_B$. Normalization determines the remaining pair. The full list, with row strategy first, isAll displayed supported probabilities are positive. The supported payoff pairs in the four nonpure cases are respectively $(8/5,8/5)$, $(12/7,6/5)$, $(6/5,12/7)$ and $(24/23,24/23)$; these values also verify the omitted-action inequalities directly. The common-support argument excludes every other support pair, and each indifference system has a unique solution. Hence there are exactly seven equilibrium pairs, three pure and four genuinely mixed. This is a complete support enumeration for a bimatrix game, not just a list of selected equilibria.

Let $u_i$ denote player $i$'s expected payoff, so the definition also covers mixed strategies. If $p_i^{\ast }$ is a dominant strategy, thenfor every opponents' profile $p_{-i}$ and every alternative $q_i$. In particular this holds at $p_{-i}=p_{-i}^{\ast }$. Thus no player can improve its payoff by a unilateral deviation from $(p_1^{\ast },\dots ,p_n^{\ast })$, which is exactly the definition of a Nash equilibrium. A profile of dominant strategies is a Nash equilibrium.

For the second-price sealed-bid auction, use the usual private-value auction assumptions: a bidder with value $v$ has quasilinear utility $v$ minus its payment if it wins, and zero if it loses. Fix the highest opposing bid $h$. If $h<v$, winning gives $v-h>0$, and bidding $v$ wins. If $h>v$, winning gives $v-h<0$, and bidding $v$ loses. If $h=v$, winning or losing gives zero, irrespective of the tie rule. Any other bid can only change whether the bidder wins, not the payment $h$ conditional on winning. It therefore cannot improve on the truthful bid in any of these cases. Hence bidding the true value is a weakly dominant strategy. Neither symmetry nor independence of the opposing values is needed for this argument.

For part (a), in a first-price sealed-bid auction with $n\ge 2$, continuous nonnegative bids and $v>0$, no fixed bid is a dominant strategy. A positive bid $b$ can be improved against opponents all bidding zero: replace it by $b/2>0$, still win, and increase the payoff by $b/2$. Bid zero can be improved against an opposing highest bid $h=v/2$: bidding $3v/4$ wins and earns $v/4>0$, whereas zero loses. This proves the no dominant positive-value bid in a first-price auction result.

Allowing mixed strategies does not change that conclusion. Against all-zero opponents, a random bid has expected payoff strictly below $v$ unless it is always zero and the tie rule awards this bidder the item with certainty. Any positive bid incurs a strictly positive payment, and any probability of losing at zero also reduces the payoff below $v$. In the strict case choose a sufficiently small positive fixed bid $\delta$ so that $v-\delta$ exceeds the random bid's expected payoff. In the exceptional case the preceding deviation against $h=v/2$ still improves it. The assumptions matter: for value zero, bid zero is weakly dominant; with only one bidder it is also optimal. Consequently there is no dominant strategy for a positive-value bidder in the standard first-price setting.

Part (b)'s printed nonexistence request is false under the stated symmetric independent private values model. Lack of a dominant strategy does not imply lack of a Nash equilibrium. For incomplete private information the appropriate equilibrium is a Bayesian Nash equilibrium, a Nash equilibrium of the game whose strategies are bidding rules as functions of private values. Here is an explicit counterexample with a full deviation check.

Let the $n\ge 2$ risk-neutral bidders have independent valuations with the uniform distribution on $[0,1]$, and put $a=(n-1)/n$. Suppose every opponent bids $\beta (w)=aw$. A bidder with value $v$ who bids $b\in [0,a]$ wins with probability $(b/a{)}^{n-1}$, since the opposing values must all be below $b/a$. Ties have probability zero, including at either endpoint. Its expected quasilinear utility isFor $b>0$,Thus $U_v$ increases up to $b=av$ and decreases thereafter. For $v=0$, bid zero is optimal directly. A bid above $a$ wins with certainty and pays more than bidding $a$, so it cannot improve on the maximum already found in $[0,a]$. A randomized bid is a mixture of these payoffs and also cannot improve on their maximum. The best response for every value is therefore the proposed bidding rule itself. This proves the uniform private-value first-price bidding equilibrium:The value-$v$ expected equilibrium payoff is $v^n/n$. The bidding rule is a pure strategy in the Bayesian game, even though values are random. Therefore the requested general nonexistence claim in (b) has a counterexample, rather than a valid proof. The defect is present in the original PDF, not introduced by the converted text.

In a two-player symmetric finite game, represent mixed strategies by probability vectors and let $e(x,y)=x^{T}Ay$ be the payoff to a player using $x$ against one using $y$. A resident strategy $x^{\ast }$ is an evolutionarily stable strategy if every distinct mutant strategy $y$ has an invasion barrier ${\epsilon }_0(y)>0$ such that, for $0<\epsilon <{\epsilon }_0(y)$,Thus when mutants are sufficiently rare, the resident has strictly larger fitness against the population mixture. The barrier may depend on the mutant; a common barrier is not part of this definition.

For fixed $y\ne x^{\ast }$ putBy bilinearity the resident-minus-mutant fitness difference isIf this is positive for all sufficiently small positive $\epsilon$, taking $\epsilon \downarrow 0$ gives $a_y\ge 0$. If $a_y=0$, strict positivity requires $b_y>0$. Conversely, if $a_y>0$, choose $0<{\epsilon }_0\le a_y/[2(a_y+|b_y|)]$; then the difference is at least $a_y-\epsilon (a_y+|b_y|)>a_y/2>0$. If $a_y=0$ and $b_y>0$, it is $\epsilon b_y>0$ for every $0<\epsilon <1$. This proves the two-condition criterion for evolutionary stability: for every $y\ne x^{\ast }$, eitherorThe weak first comparison is the symmetric Nash equilibrium condition; the second excludes neutrally competitive mutants when that comparison is an equality.

For the Hawk-Dove game, let $x^{\ast }=(1/2,1/2)$ and $y=(p,1-p)$, $0\le p\le 1$. Against the resident, both pure strategies give $1/2$, soDirect multiplication of the payoff matrix givesConsequentlyThe equality case of the two-condition criterion for evolutionary stability therefore applies to every distinct mutant. More explicitly, the fitness difference in a mixed population is $2\epsilon (p-1/2{)}^2$, positive for every $0<\epsilon <1$. Hence $\boxed{(1/2,1/2)\text{ is an evolutionarily stable strategy}}$, with the common invasion barrier one.

For pure strategy sets $S_1,\dots ,S_n$ and utility functions $u_i$, a Nash equilibrium is a strategy profile $a$ such thatFor mixed strategies, replace payoffs by their expectations under independent randomization and allow each player to choose any probability distribution. Since expected utility is linear in a player's own distribution, checking all pure deviations suffices.

Pure equilibria need not exist. In matching pennies, player 1 receives $1$ for equal choices and $-1$ for different choices, while player 2 receives the negative. At an equal pair, player 2 improves by switching; at a different pair, player 1 improves by switching. Thus no pure profile is a Nash equilibrium.

In the ranked university application game, every student must be admitted at a pure equilibrium. If someone were rejected, fewer than the total 50000 seats would be filled, so another university would have a vacancy. Moving there gives positive benefit instead of zero. Consequently all 50 universities have exactly 1000 applicants and admit all of them.

Choose a university with largest arithmetic mean admitted rank and let $r_{\min }$ be its lowest rank. If an outside student of rank $r>r_{\min }$ moved there, that student would be admitted in place of $r_{\min }$; the new arithmetic mean would exceed the university's previous arithmetic mean, which was at least the student's current payoff. This is a profitable deviation. Therefore no higher rank is outside: that university consists of the highest 1000 ranks. Remove that block and repeat with the remaining university of largest arithmetic mean. At each step the same argument excludes omitted higher ranks from the remaining set. The admitted cohorts are necessarily the consecutive blocks

Conversely, assigning these blocks to labelled universities is an equilibrium. A student moving to a higher block is rejected. A student moving to a lower block displaces its lowest rank; all the other admitted ranks there are below every rank in the student's original block. Comparing the two arithmetic means with that student's own rank held fixed shows a strict loss. Thus every block assignment is an equilibrium, and the preceding argument excludes all other pure ones:

There is more than one further mixed equilibrium. We give two genuinely mixed profiles, rather than count the pure equilibria as degenerate mixed ones. First, let every student apply independently and uniformly to all 50 universities. For each fixed student the other players' distribution is invariant under every permutation of university labels. All 50 pure applications therefore give the same expected utility. Uniform mixing is a best response, so this is a Nash equilibrium.

For a distinct mixed equilibrium, assign each of the 48 highest consecutive rank blocks deterministically to a different university. Let each of the lowest 2000 students independently choose either of the two remaining universities with probability $1/2$. For a low-ranked student these two universities have identical expected utility by label symmetry, and that payoff is positive: there is a positive-probability event on which the student is admitted. Applying to one of the 48 full higher-ranked blocks would guarantee rejection. Hence the specified mixture is a best response.

For a student in a deterministic higher block, moving to another deterministic block loses by the pure-equilibrium comparison above. It remains to rule out a move to a mixed university. Let the student's own block start at $L\ge 2001$, and let their rank be $r\le L+999$. Their present payoff is $L+499.5$. The number $M$ of low-ranked applicants at a specified mixed university has a binomial distribution with parameters $(2000,1/2)$, and the higher-ranked deviator is always admitted. If $M\ge 1$, all other admitted ranks are at most 2000, so the new arithmetic mean is at most $(r+2000)/2$, and thus at most $(L+2999)/2$. If $M=0$, it is $r\le 50000$. Therefore the expected deviation payoff is at mostbecause the gap without the last term is $(L-2000)/2\ge 1/2$. No such deviation improves the payoff. This proves a second mixed equilibrium, with only the lowest 2000 students randomizing. It is visibly different from the fully uniform profile. In particular,

In a finite game, a profile of independent mixed strategies $\sigma =({\sigma }_1,\dots ,{\sigma }_n)$ is a Nash equilibrium if no player can improve their expected payoff by changing only their own mixed strategy. In symbols,Because expectation is linear in a player's own mixed strategy, checking this inequality for all their pure strategies suffices. Nothing requires the players' payoffs to sum to zero.

For the symmetric finite game, write $e_i(p)=e(i,p)$ and $\bar{e}(p)={\sum }_i{p}_i{e}_i(p)$. Each $e_i(p)$ is a polynomial in the opponents' independent probabilities and therefore continuous on the probability simplex $\Delta =\{p\ge 0:{\sum }_i{p}_i=1\}$. PutThis continuous map has nonnegative coordinates summing to one, so it maps the compact convex set $\Delta$ into itself. The Brouwer fixed-point theorem gives $p=F(p)$. At that fixed point $r_i=p_{i}S$. If $S>0$, every $i$ in the strategy support has $r_i>0$ and therefore $e_i(p)>\bar{e}(p)$. Butwhereas every supported term would be strictly positive, a contradiction. Thus $S=0$, all pure payoffs are at most $\bar{e}(p)$, and their weighted average equals $\bar{e}(p)$. Every supported pure strategy is consequently a best response to the common opposing mixed strategy. The profile in which every player uses $p$ is a symmetric Nash equilibrium. This proves the gain-map proof of symmetric equilibrium for any finite number of players, including boundary fixed points of the probability simplex.

In the least unique bid auction, let $p$ be the common probability of bid 1. Bidding 1 wins only when the other two bids are both 2, and its winning payoff is $3-1=2$. Bidding 2 wins only when the other two bids are both 1, and its winning payoff is $3-2=1$. Hence the two expected payoffs areNeither symmetric pure strategy profile is an equilibrium: against two bids of 1 a deviation to 2 earns 1 instead of 0, and against two bids of 2 a deviation to 1 earns 2 instead of 0. An interior symmetric equilibrium must make the two expected payoffs equal. Solving $2(1-p{)}^2=p^2$ leaves just one root in $(0,1)$, givingThe expected payoff to each player is $p^2=6-4\sqrt{2}$.

To count all nonsymmetric Nash equilibria, let $p_i$ be player $i$'s probability of bidding 1. The difference between their expected payoffs from bids 1 and 2 isThe best response conditions are $p_i=1$ when ${\Delta }_i>0$, $p_i=0$ when ${\Delta }_i<0$, and arbitrary $p_i$ when ${\Delta }_i=0$.

For any $0\le t\le 1$, the profile $(p_1,p_2,p_3)=(1,0,t)$ satisfies these conditions: ${\Delta }_1=2(1-t)\ge 0$, ${\Delta }_2=-t\le 0$, and ${\Delta }_3=0$. Thus every permutation of this profile is a Nash equilibrium. At $t=0,1$ these give the six pure profiles with at least one bid of each kind; for $0<t<1$, one player mixes freely and the other two use opposite pure strategies.

There are no other nonsymmetric equilibria. If all three probabilities are interior, every ${\Delta }_i$ must vanish. Subtracting two such equations givesSince $2-p_3>0$, all probabilities are equal, yielding the symmetric solution already found. If exactly two players mix and the third uses bid 2, either mixer strictly prefers bid 1 because $\Delta =2(1-p_j)>0$. If the third uses bid 1, either mixer strictly prefers bid 2 because $\Delta =-p_j<0$. Both cases are impossible. If exactly one player mixes, their indifference requires the other players to use opposite bids. Finally, with no mixing, the all-equal pure profiles fail and the remaining six satisfy the best response conditions.

Consequently the complete Nash equilibrium set consists of the single symmetric mixed profile together with the six closed one-parameter families given by permutations of $(1,0,t)$. Their endpoints overlap at the six pure equilibria. In particular,Counting only the six pure ones would miss the interior points of these families in the two-bid three-player least unique bid auction.

A Nash equilibrium is a profile in which each player's strategy is a best response to all the others: no unilateral change increases that player's payoff. In this routing game the payoff can be taken as minus delay. A pure strategy is a single route; a mixed strategy is a probability distribution over routes, with different players randomizing independently. A non-zero-sum game does not require the players' payoffs to add to a fixed total. Nash's theorem guarantees a mixed Nash equilibrium for every finite game with finitely many pure strategies. Pure equilibria need not exist in general; the congestion game here does have them, since a unilateral delay change equals the change in its finite congestion potential.

Put $x=n_1+n_3$ and $y=n_2+n_3$. The three current delays, in the order $ABD,ACD,ABCD$, areFor a player using the first route, switching to the second makes the new second-route load $y+1$, while switching to the third retains that player's use of $AB$ and adds one user to $CD$. The two comparisons are thereforeThey reduce to $x\le y+1$ and $y\ge 74$. Reversing the two outer routes gives $y\le x+1$ and $x\ge 74$ for each second-route user. A third-route user who switches to the first retains their own use of $AB$, and one who switches to the second retains their own use of $CD$. The corresponding comparisons give $y\le 75$ and $x\le 75$. Thus the necessary and sufficient conditions, together with nonnegative integer counts summing to $n$, areThey are sufficient because each used route has been compared with both alternatives, and necessary because each comparison is an available unilateral deviation. The unit changes in loads matter in a finite-player congestion game.

For $n=100$, the counts $(25,25,50)$ give $x=y=75$ and every player's delay is $15/4$. They satisfy all the conditions. To find every pure Nash equilibrium, first note that $n_3=0$ is impossible: if either outer route is empty, its other route users fail the lower-load condition; if both are used, $x,y\ge 74$ contradict $x+y=100$. Thus $n_3>0$, so $x,y\le 75$. Since $x=100-n_2$ and $y=100-n_1$, this forces $n_1,n_2\ge 25$. Both are positive, so the remaining conditions force $x,y\ge 74$ and hence $n_1,n_2\le 26$. Their differences are at most one, so every such pair works. The complete equilibrium count vectors areEach vector also represents many pure profiles obtained by choosing which labeled players use each route. Therefore the displayed equilibrium is not unique, even as an aggregate count vector.

A coordinated allocation with fifty players on each outer route and nobody on $ABCD$ gives every player delay $7/2$, strictly better than $15/4$. It is not a Nash equilibrium: a first-route user can switch to $ABCD$ and obtain delayThis exhibits the conflict between coordinated welfare and unilateral incentives. In fact $7/2$ is below every user's delay in all four pure equilibrium count vectors above.

For a symmetric Nash equilibrium when $n=100$, let each player independently choose the two outer routes with probability $q$ each and the middle route with probability $1-2q$. Conditional on choosing the first route, that player certainly uses $AB$, while each of the other $99$ players uses it with probability $1-q$. The expected delays of the first and third routes are consequentlyThe second has the same expected delay as the first by symmetry. Equalizing these givesAll three probabilities are positive and all three pure routes have expected delay $15/4$, so every route is a best response to the other players' common mixture. This proves a symmetric mixed Nash equilibrium. Using probabilities $1/4,1/4,1/2$ would omit the player's own-load correction and would not make all three expected delays equal.

For completeness, the same calculation works for every positive number of players. If $n\le 75$, every player choosing $ABCD$ is a symmetric pure Nash equilibrium. For $75<n<149$, a symmetric mixed equilibrium hasFor $n\ge 149$, each player choosing either outer route with probability one half is a symmetric mixed equilibrium. In that last regime the unused middle route has expected cost at least that of the outer routes, since their difference is $-3/4+(n+1)/200\ge 0$. These formulas agree at the regime boundaries.

Use net monetary gain as payoff and order each player's pure strategies as $1,2,4$. The payoff matrix for the first player isThe second player's matrix is $-A$. Equal choices return both stakes, so the diagonal is zero; when the first player wins, their net gain is the other's stake, and when they lose it is minus their own stake. This is a matrix game, hence a zero-sum game and a normal-form game.

The game is degenerate in the nondegeneracy of a bimatrix game sense. Against the first player's pure strategy $4$, the second player's choices $1$ and $2$ are both best responses, with payoff four. A mixed strategy of support size one therefore has two pure best responses. The usual Lemke-Howson algorithm path needs additional tie handling or perturbation in such a game; the zero-sum game structure gives a simpler direct linear program.

Add five to every entry of the first player's matrix, obtainingThis does not change either player's best responses or Nash equilibria; it raises the game value by five. Since all entries are positive, its value $v_B$ is positive. If $p$ is a row mixed strategy guaranteeing $v_B$, put $x=p/v_B$. Then $B^{T}x\ge 1$ and $1^{T}x=1/v_B$.

Conversely any feasible $x$ has $s=1^{T}x>0$, and $p=x/s$ guarantees payoff $1/s$ in the shifted game. Maximizing that guaranteed payoff is therefore equivalent to minimizing $s$ under $B^{T}x\ge 1$, $x\ge 0$. These are precisely the displayed constraints. The dual program maximizes $1^{T}y$ subject to $By\le 1$, $y\ge 0$; normalizing an optimal $y$ gives the column strategy. This is positive-payoff linear programming for a matrix game.

Write $S=R+\delta$ and let $C_i=\delta +{\sum }_{j\ne i}{b}_j$. Player $i$'s payoff in the proportional contest with outside effort is $u_i(b)=v_{i}b/(b+C_i)-b$. When $C_i>0$, this is strictly concave andThe best response is zero when $v_i\le C_i$, and otherwise is $\sqrt{v_i{C}_i}-C_i$. Thus the nonnegative-effort Karush-Kuhn-Tucker conditions at a pure strategy Nash equilibrium giveFor an active player the first-order equation is $v_i(S-b_i)=S^2$. For an inactive player the derivative at zero is $v_i/S-1$. Strict concavity makes these conditions sufficient as well as necessary.

If $\delta =0$, an equilibrium cannot have just one active player: that player would win with certainty and could lower its positive effort. An all-zero profile also cannot be an equilibrium under the usual completion of proportional allocation at zero: at least one player can gain by investing an arbitrarily small amount. Consequently there are at least two active players, so every $C_i$ is positive. If $\delta >0$ this positivity is automatic. The allocation rule's otherwise undefined all-zero value at $\delta =0$ is therefore immaterial to the equilibrium calculation.

Let $k=\hat{n}>0$, $H_k={\sum }_{i=1}^{k}1/v_i$, and use the harmonic mean ${\bar{v}}_k=k/H_k$. Summing the active efforts givesThe positive root supplies the total-effort formula for a proportional contest with outside effort:At $\delta =0$ this reduces to $R=(k-1){\bar{v}}_k/k$. There is also a genuine zero-active case: if $\delta \ge v_1$, then $k=0$ and $R=0$; no harmonic mean of an empty family is needed.

For a fully explicit active-set rule, setThe equilibrium equation is $G(S)=0$. For $\delta >0$, $G$ is strictly decreasing from $+\infty$ to $-1$. For $\delta =0$ its limit at zero is $n-1>0$, and it is strictly decreasing wherever a zero could occur. Hence the positive root is unique. Moreover $S\ge \delta$, since its equation gives $1-\delta /S\ge 0$, and $b_i=S(1-S/v_i{)}_{+}$ supplies an equilibrium with total effort $S-\delta$.

Player $j$ is active exactly when $v_j>S$, equivalently $G(v_j)<0$. Ordering the valuations, including ties, givesTherefore the active-set threshold for a proportional contest with outside effort isThe qualifying indices form a prefix because $G$ is decreasing. Equality excludes the marginal player, as required by strict positivity of effort. Equivalently, the positive root $S_k$ of the displayed quadratic must satisfy $v_k>S_k\ge v_{k+1}$, with $v_{n+1}=0$. These formulas include one active player when $\delta >0$ and exclude that case when $\delta =0$.

Let $u,v$ be the row and column mixed strategies, so their entries are nonnegative and sum to one. The payoffs are $u^{\mathsf{T}}Pv$ and $u^{\mathsf{T}}{P}^{\mathsf{T}}v$. A Nash equilibrium is a pair in which each strategy is a best response to the other: neither player can improve its expected payoff by a unilateral change of mixed strategy. Equivalently, every positive-probability pure strategy attains that player's maximum pure-strategy payoff against the opposing mixture.

For the complementarity construction of a symmetric Nash equilibrium put $s={\sum }_i{x}_i>0$ and $u=x/s$. The equation $Px+z=1$ gives $Px\le 1$. Nonnegativity and $x^{\mathsf{T}}z=0$ imply $x_i>0\Rightarrow z_i=0\Rightarrow (Px{)}_i=1$. Therefore every used pure strategy has payoff $1/s$ against $u$, while all other pure strategies have payoff at most $1/s$. The same vector of pure-strategy payoffs applies to the column player in this symmetric bimatrix game. Thus a symmetric Nash equilibrium is

For a genuine Lemke-Howson algorithm path, use two separate vectors, $\xi$ for the row player and $\eta$ for the column player, with slacksGive labels $i=1,2,3$ to ${\xi }_i=0$ or $r_i=0$, and labels $3+j$ to ${\eta }_j=0$ or $s_j=0$. A nonzero completely labeled pair has ${\xi }_i{r}_i=0$ and ${\eta }_j{s}_j=0$, and normalization yields mutual best responses. At the artificial pair $(\xi ,\eta )=(0,0)$ all six labels are present.

Drop label $3$ by increasing ${\xi }_3$. The two limiting rows of $P\xi \le 1$ are $3{\xi }_3\le 1$ and $2{\xi }_3\le 1$, so the minimum-ratio pivot stops at ${\xi }_3=1/3$, where $s_2=0$. Label $5$ is now duplicated, since ${\eta }_2=0$ also carries it. Increase ${\eta }_2$ to remove that duplicate. The limiting inequalities are $3{\eta }_2\le 1$ and $4{\eta }_2\le 1$, so the next pivot stops at ${\eta }_2=1/4$, where $r_3=0$ restores the dropped label $3$. The endpoint isIt has all six labels and satisfies both complementarity conditions. After normalization, the Lemke-Howson algorithm returnsAgainst column strategy $2$, row strategy $3$ pays $4$, exceeding $3$ and $0$. Against row strategy $3$, column strategy $2$ pays $3$, exceeding $0$ and $2$. This verifies the endpoint directly.

To find every other Nash equilibrium, observe that row strategy $3$ strictly dominates row strategy $1$: the payoff differences against the three columns are $(2,1,2)$, all positive. By symmetry, column strategy $1$ is also strictly dominated. Neither can appear in an equilibrium. The reduced strategies $2,3$ have row payoff matrixThe two pure Nash equilibria are $(e_2,e_3)$ and $(e_3,e_2)$. If the opponent uses strategy $2$ with probability $q$, the payoffs of strategies $2,3$ are $3(1-q)$ and $2+2q$. Indifference requires $q=1/5$. The same computation applies to the other player. A player mixing both strategies forces this exact opposing mixture; a player playing a pure strategy has a unique opposing best response, producing one of the two pure equilibria. Hence there are exactly three Nash equilibria:For the symmetric mixed equilibrium, the original complementarity construction can use $x=(0,1/12,1/3)$ and $z=(3/4,0,0)$; it gives payoff $12/5$ to each player after normalization.