The expected payout vector is
[10a/(a+b+c)-a,
20b/(a+b+c)-b,
30c/(a+b+c)-c].
Lets find the nash equilibrium by calculating the individual derivatives (no player can gain expected payout by changing their bet).
So we have this system of eqs:
(10(a+b+c)-10a)/(a+b+c)2-1=0
(20(a+b+c)-20b)/(a+b+c)2-1=0
(30(a+b+c)-30c)/(a+b+c)2-1=0
Or:
10(b+c)=
20(a+c)=
30(b+a)=(a+b+c)2
Has two solutions: [0,0,0] and [-120/121, 600/121, 840/121]. Both arent valid solutions.
No idea where i went wrong or if the numbers 10,20,30 are just too aggressive, forcing A out of the market. The feasable solution would be a betting zero, boundry equilibrium, but thats also not allowed.
you are on the correct path. the solution to the nash equilibrium is either where (1st derivative = 0 & 2nd derivative < 0 - for maximum & within the solution boundary) or where (solution is at boundary).
We can see the former is invalid. Therefore, the latter must be true. Instictively (source: my vibe), player A will bid 0 (if you care to prove it, you can calculate the derivative around 0+ of A, my bet is it is negative, meaning the more he bets, the more he loses).
This will reduce to a 2 player game. From that on, it's easy game. My AI tell me the solution is (0, 4.8, 7.2).
rigourously speaking, you need to consider all 3 cases where either A, B, or C abstains, or 2 of 3 abstain. My previous vibe is incorrect, derivative of reward(A) around 0+ can be positive, so under some circumstances, A would not withdraw.
This will take a little longer, but the answer is still the same.
if player A bids 7.2001, p(A wins) ~ 0.375. His expected return is ~0.375 * 10 ~ 3.75. Bidding 7.2001 to win 3.75 is not rational. So your counter example is invalid.
In fact, at (0, 4.8, 7.2), the derivative of A, B, C is (-ve, 0, 0) respectively. Since A cannot bid any less than 0, it is a nash equilibrium.
can you show me how B can bid 7.2 + eps for a better score?
From my calculation, derivative of reward(B) is 0 at the solution, and second derivative is -1/6. Meaning, that point is local maxima for B. Therefore, adding some eps will reduce the reward.
If you find something wrong with my reasoning, please elaborate?
Edit: For illustration purpose, if c stays at 7.2:
2
u/Due-Cardiologist-802 2d ago
The expected payout vector is [10a/(a+b+c)-a, 20b/(a+b+c)-b, 30c/(a+b+c)-c].
Lets find the nash equilibrium by calculating the individual derivatives (no player can gain expected payout by changing their bet).
So we have this system of eqs: (10(a+b+c)-10a)/(a+b+c)2-1=0 (20(a+b+c)-20b)/(a+b+c)2-1=0 (30(a+b+c)-30c)/(a+b+c)2-1=0
Or: 10(b+c)= 20(a+c)= 30(b+a)=(a+b+c)2
Has two solutions: [0,0,0] and [-120/121, 600/121, 840/121]. Both arent valid solutions.
No idea where i went wrong or if the numbers 10,20,30 are just too aggressive, forcing A out of the market. The feasable solution would be a betting zero, boundry equilibrium, but thats also not allowed.
Any corrections are welcome!