I think $n=3$ will just involve a lot of equations. Here's my sketch.
It must be that a player gets an equal payoff from all actions specified in their strategy. Note that the lowest bid, 0, wins only if the other two bid the same thing. Let $x=(x_0,x_1,\dots,x_{99})$ be the weight each player places on each action (naturally $x_0$ is the weight on 0, etc). So the expected payoff is
$$100\sum_{i=1}^{99}x_i^2$$
Let's skip the bid of 1 for now.
For the bid 2, a player only wins if the others bid the same thing or below 2.
$$(100-2)( \sum_{i\neq 2}x_i^2 + 2x_0x_1)$$
Now for a general $j$. A player wins if the others bid the same thing or all below $j$.
$$(100-j)[ \sum_{i=j+1}^{99} x_i^2 + (\sum_{i=0}^{j-1} x_i)^2 ]$$
Finally, this vector lives in the simplex. $\sum_{i=0}^{99}x_i=1$.
This gives $100$ equations for $100$ unknowns, so you should be able to solve. Here, the player is indifferent between all strategies, so it must be a best response, and we have an equilibrium.
This is more of a discussion point than an answer. Let us assume $c_e < \sum_{i=1}^Nc_i$, otherwise this is a game nobody would play. It would also be illogical to terminate before the player has made the money $c_e$ back during the game (otherwise the player would just not play). Let $R_j=\sum_{i=1}^jc_i$, and let $1 \leq r \leq N$ be the minimum value for which $R_r=\sum_{i=1}^qc_i > c_e$. Let us consider what happens for the strategy "terminate after $k\geq r$ wins".
Notice that for this strategy there are only two outcomes, either the player wins $R_k-c_e$ with probability $p_k=\frac{N}{N+1}\times \cdots \frac{N-k+1}{N-k}=\frac{N-k+1}{N+1}$, or 'wins' $-c_e$ with probability $q_k=1-p_k=\frac{k}{N+1}$. So for each $k$ considered, we find ourselves a bernoulli distribution (on $\{C_k-c_e, -c_e\}$) with probability of success $p_k=\frac{N-k+1}{N+1}$.
At this point, we can look at the expected value $W_k$ for each of these strategies to get $$W_k = R_k\frac{N-k+1}{N+1} - c_e$$
if for any $k \geq r$ we get $W_k>0$, then our strategy is to find the largest $W_k$ and then terminate at $k$ and repeat. That is, if the player is allowed to play this game infinitely many times, then this would be a profitable strategy. This is kind of like playing the lottery, suppose the lottery gave a positive expected return (which it doesn't in reality), then playing the lottery infinitely often should result in a net profit over time, though playing it once is highly likely to result in a net loss.
So now suppose the player is only allowed to play the game once, then now it's a matter of a single bernoulli trial. Notice that for our strategy to terminate at $k$, $p_k$ is monotone decreasing in $k$, so we want to use the smallest posible value $k$. However, we also do not want to make any losses, so the smallest possible value of $k$ for which we still can make a profit is $r$, i.e we should terminate as soon as we win more money than we paid to play the game. Having said this, if $p_r$ is quite small (say $10\%$) then we win with only $10\%$ chance and it may not be worthwile for the player to play the game in the first place. One could argue that a large $R_r$ might make the game worthwile, but now this is a matter of personal opinion. Would you like to bet $£100$ say to win $£1001$ with $10\%$ chance? How about to win $£10001$ with $1\%$ chance?. Of course if we can play the game infinitely often, then these numbers are worthwile, but if we can only play once, most people would not play.
If say we are a casino trying to implement this game, we should ensure
- $W_k$ is never positive for any $k\geq r$
- $r$ is fairly large. i.e for anyone using the strategy to terminate after $r$ wins, the probability of them making money is $p_r=\frac{N-r+1}{N+1}$, where if $r$ is large then $p_r$ is small
Best Answer
This is effectively an iterated prisoners' dilemma. If it was a single round then you would do better to play Blue rather than Red, no matter which colour your opponent selects in the same round, but if there is a series of repeated rounds then trust between the two sides can lead to both choosing Red and ending up with positive scores so long as the trust is maintained.
There is no provably optimal strategy which leads to positive outcomes that does not involve communication with the other team. The assumption is that such commmunication only takes place within the game and any punishments can only be delivered in the game.
The problem is that there is no reason to be trusting in the final round if it is known to be the final round, and this break-down of trust feeds back through the earlier round if both players are game-theoretic rational. Despite this, some trusting strategies with retaliation can evolve in a population with other similar strategies present.