 Solution

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.