Here is a game I like to play:

  • Every player gets a random number independently drawn from a fixed distribution. To fix ideas, suppose everyone draws a number from $\{2, ..., 10\}$ where all numbers are equally likely (that way you can do this with playing cards...)
  • The first player says: 'I think that the sum of the numbers is $k$ or higher' where the 'bet' $k$ is an integer that they need to choose.
  • The second player can either challenge or make a higher bet. If they challenge, the numbers are revealed and the sum computed. One of the two players is then knocked out depending on who was right -- and we move to the next round. If the second player doesn't want to challenge, they must then state 'I think that the sum of the numbers is $k'$ or higher' for some integer $k' > k$.
  • In that case, we move to the third player who can either challenge or make a still higher bet. And so the round continues.. until one player does inevitably challenge when the betting gets too high.
  • The above defines the rules in a particular round. And the players keep playing rounds until all but one is knocked out.

Despite having played this a few times, I really don't know what a sensible strategy (or alternately a Bayes-Nash equilibrium strategy) might be. Does anyone have any ideas? [If the game above is too complicated, I invite simplifications -- for instance, you might suppose that there are just 2 players.] For what it is worth, here are a couple of speculations on my part:

  • A very naive strategy is ignore the informational content of bets and stick with one's prior. In this example, the expected value of each number is $(2 + 10)/2 = 6$. Suppose then that my number is $x$ and that I face $n$ opponents. Then I might predict the sum to be $x + 6n$; and increase the betting until doing so would force me to bet higher than this (at which point I challenge).
  • Obviously, one can try to best respond to players who follow this kind of strategy. For example, if you see them betting some large amount $k$, you may infer that $x + 6n \geq k$ where $x$ is their number, or equivalently $x \geq k - 6n$. You might then compute the expected value of their number $x$ conditional on its being larger than $k - 6n$ (which is pretty easy in the uniform example).
  • I suspect that in a (Bayes-Nash) equilibrium, players must use mixed strategies. If your bet is a (strictly increasing) function of your number, then I can perfectly infer your number from your bet (if I know your strategy). This gives me a big edge -- and suggests that your strategy can't form part of an equilibrium.
  • $\begingroup$ Have you tried solving the simplified game with numbers 1,2,3 and two players? Seems like a good first step that would tell you a bit about its complexity. $\endgroup$
    – Giskard
    May 20, 2021 at 4:02
  • $\begingroup$ I didn't think you can easily write down a BNE in the original game, as the space of pure strategies is huge. $\endgroup$
    – Giskard
    May 20, 2021 at 4:05
  • $\begingroup$ @Giskard yes I realised that a few mins after I wrote it! (But didn't have my laptop to hand to edit...) Will fix it now $\endgroup$
    – afreelunch
    May 20, 2021 at 9:50
  • 1
    $\begingroup$ And yes I do agree solving simplified versions is a good starting point (I have not tried this yet) $\endgroup$
    – afreelunch
    May 20, 2021 at 9:51
  • 1
    $\begingroup$ @Bayesian Haha you may be right. Though fortunately learning the BNE usually doesn't tell you how to play a game :) $\endgroup$
    – afreelunch
    May 21, 2021 at 9:38

1 Answer 1


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 $\theta, \theta_i+\theta_j<5$ because 4 is the maximal sum.

$\theta_i=1$ challenges $b=4$ as $1+\theta_j\leq3<4$ 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 $2+\theta_i\geq3>b$ 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 $p>1-p>\frac{1}{2}$.

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 $3<4$) and lose against $\theta_j=1$ (sum is $2<4$). 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 $2<3$) 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 $p>1-p$.

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 $3<4$) 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, $p>1-p$.

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, $p>1-p$.

Thus, back to type 1 bidding 2 in the first round. If $p>1-p$ 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 $p>1-p$, this is worse than betting 3.

So $\theta_i=1$ bids 3 in the first round when $p>1-p$.

Now to type $\theta_i=2$ in the first round. Given the above, bidding 4 is better than bidding 3 given prior $p>1-p$. 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 $p>1-p$.

With $p<1-p$, 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.

  • $\begingroup$ Thanks for the thoughts -- I'll take a close look at soon as I have time! $\endgroup$
    – afreelunch
    May 21, 2021 at 18:27

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