3 players are playing a game where they get to pick independently without knowing the other players picks one of 2 prizes (A,B) and the payout is (a,b) for the two prizes, divided by how many people picked the specific prize.


For example, if the prizes are (3,1) and 2 people picks A and 1 person picks B, the 2 people get 3/2 = 1.5 each and the third person gets 1 for himself.


In a zero-sum setting, if the prizes are (3,1), it would always be best to pick the first location, since you are guaranteed a prize of at least 1. If prizes are (2,1), this is also true, since no other player can beat that strategy. But, if prizes are (3,2), and two players use the strategy, the third player can do better by always picking the 2nd prize.


Is it possible to find an equilibrium for this particular problem, and does it generalize to further players and prizes? Does it fit into some standard theory?



Simple(?) game theory
This looks like a homework exercise on mixed strategy Nash equilibria.


My ASCII art skills are a little rusty, but let's write up a "nice" litte 2x2x2 cube:



```

              +--------------+-----------------+
C plays 3    /              /                / |
            +--------------+----------------+  | 
C plays 2  /              /                /|  / 
          +---------------+---------------+ | /| 
B plays 3 |  (3/2,3/2,2)  |    (1,3,1)    | |/ /
          +---------------+---------------+-/ /
B plays 2 |    (3,1,1)    | (2/3,2/3,2/3) | |/
          +---------------+---------------+-/
               A plays 3     A plays 2

```

Yes I know it's ugly as sin. Just sketch it on a piece of paper instead and fill it out with the payoffs. And leave a spot somewhere for the (A:3,B:2,C:3) corner. 


Now let $p\_X$ denote the propability that player $X$ plays $3$ and let $U\_{X,y}$ denote the expeted payoff for player $X$ playing $y$


Because of the symmetries in this game, in a mixed strategy equilibrium $A$, $B$ and $C$ will all play the same mixed strategy, i.e. $p\_A = p\_B = p\_C$.


$$
U\_{C,3} = p\_Ap\_B\cdot 1 + p\_A(1-p\_B)\cdot\tfrac{3}{2} + (1-p\_A)p\_B\cdot\tfrac{3}{2} + 
(1-p\_A)(1-p\_B)\cdot 3 $$
$$ = p\_A^2 + \tfrac{3}{2}p\_A - \tfrac{3}{2}p\_A^2 + \tfrac{3}{2}p\_A - \tfrac{3}{2}p\_A^2 + 3
-6p\_A + 3p\_A^2 = p\_A^2 -3p\_A + 3$$


$$
U\_{C,2} = p\_Ap\_B \cdot 2 + p\_A(1-p\_B)\cdot 1 + (1-p\_A)p\_B\cdot 1 + (1-p\_A)(1-p\_B)\tfrac{2}{3} $$
$$ = \tfrac{2}{3}p\_A^2 + \tfrac{2}{3}p\_A + \tfrac{2}{3}$$


Then we just solve $U\_{C,3} = U\_{C,2}$ for $p\_A$ and we get $\tfrac{1}{3}p\_A^2 - \tfrac{11}{3}p\_A + \tfrac{7}{3} = 0$ which has solutions $p\_A = \frac{11 \pm \sqrt{93}}{2}$. Only the root corresponding to minus gives $p\_A \in [0,1]$, so we conclude that the players should play $3$ with a probability of approximately $0.678$.


PS: I've given up on `align`. It just doesn't work. Ever.


PPS: There's almost certainly a mistake. This was a pretty quick and dirty computation.


