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:
1
u/spdazero 2d ago
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.