Betting on the sum of scores
Some thoughts:
Suppose there are two players $i,j$ and two numbers $1,2$. So type profiles are $\theta=(\theta_i,\theta_j) \in \{1,2\} \times \{1,2\}$, and let them be independet draws from the same distribution with $P(\theta_i=2)=p$ and $P(\theta_i=1)=1-p$ . On equilibrium path, this is a continuation game of any larger game because in the end there are always only two players left and in equilibrium they should be the ones with the highest types. This is the continuation game where the standing bid is so high that only the two highest possible numbers remain as plausible candidates.
Let $b$ be the current standing bid and solve backwards as player $i$:
First, $i$ overbids any $b\leq 1$ as challenging would mean losing for sure as for any $\theta, \theta_i+\theta_j\geq 2$.
Similarly, $i$ challenges any $b\geq 5$ as it means winning for sure as for any $\theta, \theta_i+\theta_j&lt;5$ because 4 is the maximal sum.
$\theta_i=1$ challenges $b=4$ as $1+\theta_j\leq3&lt;4$ to win for sure. $\theta_i=2$ challenges $b=4$ as overbidding to $b\geq5$ means player $j$ challenges the next round and wins. Challennging $b=4$ as $\theta_j=2$ means losing against type $\theta_j=2$, but winning against type $\theta_j=1$.
$\theta_i=1$ challenges $b= 3$ as overbidding means losing for sure as any type would challenge bid $b\geq 4$ for sure in the next round and the sum is at most 3.
$\theta_i=2$ overbids $b=3$ to 4 as challenging means losing for sure, given that $2+\theta_i\geq3=b$ for sure.
$\theta_i=2$ overbids $b=2$ as challenging means losing for sure, given that $2+\theta_i\geq3&gt;b$ for sure.
$\theta_i=1$ overbids $b=2$ as challenging means losing for sure, given that $2+\theta_i\geq2=b$ for sure.
Hence, bidding 2 essentially means that the other one has the first move. You can construct multiple signaling equilibria here. Let's concentrate on the case in which nothing is inferred from such a move.
Consider the first round.
Suppose $p&gt;1-p&gt;\frac{1}{2}$.
As $\theta_i=1$, you could bid 4 in the first round. Both types must challenge, which means you lose against $\theta_j=2$ (sum is $3&lt;4$) and lose against $\theta_j=1$ (sum is $2&lt;4$). That is, you can only lose when bidding 4 as type 1 and you will never do this.
Alternatively, as $\theta_i=1$, you can bid $b=3$ in the first round, which means you lose against $\theta_j=1$ who challenges for sure (sum is $2&lt;3$) and win against $\theta_j=2$ who would lose a challenge and loses from overbidding to 4, which you would challenge. Winning is more likely as you beat type 2 and $p&gt;1-p$.
Alternatively, you can bid 2 in the first round, which means both types will overbid.
Type $\theta_j=1$ will bid 3 as bidding 4 or more means sure loss to them. You would challenge the 3 as type 1 and win (overbidding to 4 or more means sure loss to you).
That is you win at least with probability $1-p$.
What does type 2 do upon seeing $b=2$?
Given we assume no inference, they do the same as they would do if they started.
As $\theta_i=2$, you can bid 4 in the first round, which means you lose against $\theta_j=1$ (sum is $3&lt;4$) and win against $\theta_j=2$ (sum is $4$). Both will optimally challenge you even if they learn your type - the high type knows they lost upon seeing 4, the low types knows they won. Thus winning is more likely, $p&gt;1-p$.
Alternatively as $\theta_i=2$, you can bid 3 in the first round, which means you win against $\theta_j=1$ who challenges (sum is $3\geq3$, and them overbidding means you challenge in the next round and win) and lose against $\theta_j=2$ who overbids to 4 which you have to challenge and lose - $\theta_j=2$ would not challenge as this would mean sure loss to them. Thus losing is more likely, $p&gt;1-p$.
Thus, back to type 1 bidding 2 in the first round. If $p&gt;1-p$ and there was no inference, type $\theta_j=2$ bids 4 and you as $\theta_i=1$ win by challenging 4. But that would mean you always win as type $\theta_i=1$ when bidding 2. Then it must be that $j$ infers that you are type 1 when bidding 2. As a response, they would bid 3, which you can challenge or overbid and lose in any case. Hence, bidding 2 as type $\theta_i=1$ means, winning with probability $1-p$ and losing with probability $p$. Because $p&gt;1-p$, this is worse than betting 3.
So $\theta_i=1$ bids 3 in the first round when $p&gt;1-p$.
Now to type $\theta_i=2$ in the first round. Given the above, bidding 4 is better than bidding 3 given prior $p&gt;1-p$. What about bidding 2?
Under the belief that type 1 bids 2, both player types would overbid to 3. Here, challenging would mean sure loss for type $\theta_i=2$. Hence, we overbid to 4 with the same outcome as bidding 4 in the first round. Under a belief that type 2 bids 2, a type 2 would overbid to 4 and win, and a type 3 would overbid to 3 and win challenge against 3 and also after a challenge against 4.
So $\theta_i=2$ bids 4 in the first round when $p&gt;1-p$.
With $p&lt;1-p$, type 1 would still not bid 4 in the first round, but also type 2 would not want to bid 4, instead they bid 3 and type 1 bids 2.