Under the following assumptions + simplifications:
- The goal of each player is to maximize their expected profit.
- The only valid bids are $0 and $99
- For the optimal strategy, no one person can deviate and perform better.
- The players are greedy - if there exist multiple strategies that satisfy the deviation policy, then each player chooses the strategy that maximizes the probability of a winner.
Then for n = 3 the optimal strategy ends up being bidding $0 10/11 of the time and $99 1/11 of the time.
Suppose two players bet $0 with probability p and $99 with probability 1-p. The third bets with probabilities q, 1-q respectively. Then, the expected profit is
100 * q * (1-p)^2 + (1-q) * p^2
Now, we want to maximize this profit with respect to q. Note that this is linear in q - it equals
(99p^2 - 200p + 100) q + (constant with respect to q)
For p < 10/11, this coefficient is positive, and q = 1 outperforms p
For 1 >= p > 10/11, this coefficent is negative, and q = 0 outperforms p
So, the only value of p for which no deviation can improve expected profit is p = 10/11.
Now, in principle you can generalize this to the game where you can make any bet from $0 to $99, if you add more variables, but at that point this approach quickly becomes infeasible. I think even with a computer it's not solvable exactly, since I think you'd end up with non-linear constraints, but I could be wrong.