Find the number where one choice stops being better than the other. Then work out what that number is actually good for.
Say a penalty round deals you this grid: shooting left pays 2 when the keeper goes left and 7 when they go right; shooting right pays 9 and 3. Shooting right looks great until you remember you are doing this two hundred times. Always shoot right and the keeper starts always going right, and you are stuck on 3 out of 10 forever. Every fixed habit ends in the worst cell of its own row.
Here is the part that trips people up. To find your mix you use the keeper's numbers, not your own. You are not squeezing extra goals out of p*. You are looking for the one share of left-footed shots that makes going left and going right worth the same to them. On this grid that is 6/11. Once both their options pay the same, there is nothing for them to read you for.
Which is also why you end up indifferent at q*. If shooting left paid more than shooting right against their mix, you would shoot left more often, and shooting left more often is the exact pattern they are watching for. Two options worth the same is not the answer failing. It is the answer.
The two numbers then do completely different jobs:
Nash only asks whether anyone wants to switch right now. ESS asks whether an arrangement survives being poked. Picture 95% of the market doing one thing and 5% trying something else. If the 5% earn less, they give it up and the 95% was safe. If they earn more, they grow, and whatever was there before is finished.
Every ESS is a Nash equilibrium. The reverse is false, and in a 2×2 game the whole thing comes down to one comparison you can read straight off the grid.
Take a night market that deals 5, 9, 7, 4. A stall in the square next to another square stall takes 5. A stall that went to the pier instead, while the other stayed in the square, takes 7. So being in the crowded spot is worse. Check the other side and it holds: 4 against 9. Crowding hurts both ways round. Put too many stalls in the square and the square stops paying, so they leave, and the share slides back to 5/7 on its own. You cannot invade that, because anything that becomes common punishes itself. That is an ESS.
Now a market that deals 6, 4, 1, 6. Square next to square takes 6; going to the pier alone takes 1. A busy spot now pulls the crowd, so being common is an advantage. The share where both spots pay the same still exists, at 2/7, but nothing holds it there. One extra stall in the square makes the square better, which brings the next one. Start a hair above 2/7 and you finish with every stall in the square. A hair below and you finish with every stall on the pier. It is a genuine Nash equilibrium that no real market would ever be sitting at.
Nobody in the night market is solving anything. The rule is only this: a stall that beat the average last night gets copied, one that did worse gets dropped, and the bigger the gap the faster it happens. Sixty stall-holders, no arithmetic between them, and the share still arrives at 5/7.
That is why the idea is worth having. Ordinary game theory needs players clever enough to find the equilibrium themselves. This needs nothing but imitation, so it covers bacteria, firms copying a rival's pricing, and a school where one haircut spreads through a year group in a fortnight.
It also splits apart two things that sound like one question. Where does the motion stop is answered by setting two payoffs equal, which is the algebra you have been doing. Where does the motion go is a separate question, and the two answers can disagree.
With one population there is a single number moving, and it can come to rest. In the night market it does.
With two populations there are two numbers, and each is driven entirely by the other. Sellers cheat more, so checking starts to pay and buyers check more. Buyers check more, so cheating stops paying and sellers go straight. Sellers are straight, so checking is a waste of time and buyers stop. Nobody is checking, so cheating pays again. Every link in that chain is correct and the chain has no end. The two shares go round a loop and never reach (p*, q*), even though (p*, q*) is a perfectly good equilibrium. Nothing in the system is pushing towards it.