Four robots each get 1 unit of fuel and swim in 3 races. Before the trials, each robot secretly picks one of two strategies:
- Discrete (D): pick one of the 3 races uniformly at random (independently of the other robots) and put all its fuel (1) on it, with 0 on the other two races.
- Spread (S): put exactly 1/3 of its fuel on each of the 3 races.
Choosing the 3 winners. Look at every (robot, race) pair and the fuel that robot bid on that race. Repeat the following until every race has a winner:
- Among robots that have not won yet and races that have no winner yet, find the largest bid.
- If several (robot, race) pairs share that largest bid, pick one of them uniformly at random. (This includes bids of 0, if all remaining bids are 0.)
- That robot wins that race. Remove both the robot and the race.
A robot can win at most one race, so exactly 3 of the 4 robots win. Example: with bids D on race 1, D on race 1, S, S, the pairs with bid 1 are processed first, and one of the two D robots wins race 1 at random. Then the remaining bids of 1/3 on races 2 and 3 all tie, and the two S robots end up with races 2 and 3.
Every robot wants to maximise its probability of winning a race.
Each robot independently plays D with probability p and S with probability 1 − p. Find the value of p with 0 < p < 1 that makes this a symmetric Nash equilibrium, meaning that when the other three robots play this mix, a robot does equally well with D and with S (so it has no reason to change).
Answer format: give p in exact closed form (for example (sqrt(5)-1)/2), or as a decimal to at least 7 significant figures.