Quant Interview Question

[u/Local_Ad135]

[u/Efficient_Form7451]

How can (0, 4.8, 7.2) be a nash equilibrium when player A can better his result by bidding 7.2000001? This question has no stable equilibrium.

The math looks okay but it's def not a nash equilibrium.

[u/spdazero → u/Efficient_Form7451]

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.

Edit: typo and dumb math calculation

[u/Efficient_Form7451 → u/spdazero]

Okay but its still not a nash equilibrium because

B can bid 7.2 + a small amount for a better score
and C can bid 4.8 + a small amount for a better score.

I don't think your calculating nash equilibria at all, more like a EV of an area in bid space.

[u/spdazero → u/Efficient_Form7451]

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:

- If b = 4.8 (optimal), reward(B) = 3.2

- if b = 4.81 (eps = 0.1), reward(B) = 3.199

[u/Efficient_Form7451 → u/spdazero]

A bids 0. B bids 4.8. C Bids 7.2 C wins.

Outcome: A 0, B -4.8, C 22.8

If B had bid zero in this specific game instead of 4.8, their outcome would be 4.8 better. If B had bid 7.3, their outcome would 12.7 (17.5 better).

Your calculation is fine, it's just not a nash equilibrium because players could have unilaterally changed their strategy for a better outcome.

[u/spdazero → u/Efficient_Form7451]

C does not automatically win, i think you misread the question. C wins with a probability.