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.