<p>Some thoughts:</p>
<p>Suppose there are two players <span class="math-container">$i,j$</span> and two numbers <span class="math-container">$1,2$</span>. So type profiles are <span class="math-container">$\theta=(\theta_i,\theta_j) \in \{1,2\} \times \{1,2\}$</span>, and let them be independet draws from the same distribution with <span class="math-container">$P(\theta_i=2)=p$</span> and <span class="math-container">$P(\theta_i=1)=1-p$</span> . 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.</p>
<p>Let <span class="math-container">$b$</span> be the current standing bid and solve backwards as player <span class="math-container">$i$</span>:</p>
<p>First, <span class="math-container">$i$</span> overbids any <span class="math-container">$b\leq 1$</span> as challenging would mean losing for sure as for any <span class="math-container">$\theta, \theta_i+\theta_j\geq 2$</span>.
Similarly, <span class="math-container">$i$</span> challenges any <span class="math-container">$b\geq 5$</span> as it means winning for sure as for any <span class="math-container">$\theta, \theta_i+\theta_j&lt;5$</span> because 4 is the maximal sum.</p>
<p><span class="math-container">$\theta_i=1$</span> challenges <span class="math-container">$b=4$</span> as <span class="math-container">$1+\theta_j\leq3&lt;4$</span> to win for sure. <span class="math-container">$\theta_i=2$</span> challenges <span class="math-container">$b=4$</span> as overbidding to <span class="math-container">$b\geq5$</span> means player <span class="math-container">$j$</span> challenges the next round and wins. Challennging <span class="math-container">$b=4$</span> as <span class="math-container">$\theta_j=2$</span> means losing against type <span class="math-container">$\theta_j=2$</span>, but winning against type <span class="math-container">$\theta_j=1$</span>.</p>
<p><span class="math-container">$\theta_i=1$</span> challenges <span class="math-container">$b= 3$</span> as overbidding means losing for sure as any type would challenge bid <span class="math-container">$b\geq 4$</span> for sure in the next round and the sum is at most 3.
<span class="math-container">$\theta_i=2$</span> overbids <span class="math-container">$b=3$</span> to 4 as challenging means losing for sure, given that <span class="math-container">$2+\theta_i\geq3=b$</span> for sure.</p>
<p><span class="math-container">$\theta_i=2$</span> overbids <span class="math-container">$b=2$</span> as challenging means losing for sure, given that <span class="math-container">$2+\theta_i\geq3&gt;b$</span> for sure.
<span class="math-container">$\theta_i=1$</span> overbids <span class="math-container">$b=2$</span> as challenging means losing for sure, given that <span class="math-container">$2+\theta_i\geq2=b$</span> 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.</p>
<p>Consider the first round.</p>
<p>Suppose <span class="math-container">$p&gt;1-p&gt;\frac{1}{2}$</span>.</p>
<p>As <span class="math-container">$\theta_i=1$</span>, you could bid 4 in the first round. Both types must challenge, which means you lose against <span class="math-container">$\theta_j=2$</span> (sum is <span class="math-container">$3&lt;4$</span>) and lose against <span class="math-container">$\theta_j=1$</span> (sum is <span class="math-container">$2&lt;4$</span>). That is, you can only lose when bidding 4 as type 1 and you will never do this.</p>
<p>Alternatively, as <span class="math-container">$\theta_i=1$</span>, you can bid <span class="math-container">$b=3$</span> in the first round, which means you lose against <span class="math-container">$\theta_j=1$</span> who challenges for sure (sum is <span class="math-container">$2&lt;3$</span>) and win against <span class="math-container">$\theta_j=2$</span> 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 <span class="math-container">$p&gt;1-p$</span>.</p>
<p>Alternatively, you can bid 2 in the first round, which means both types will overbid.
Type <span class="math-container">$\theta_j=1$</span> 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 <span class="math-container">$1-p$</span>.
What does type 2 do upon seeing <span class="math-container">$b=2$</span>?
Given we assume no inference, they do the same as they would do if they started.</p>
<p>As <span class="math-container">$\theta_i=2$</span>, you can bid 4 in the first round, which means you lose against <span class="math-container">$\theta_j=1$</span> (sum is <span class="math-container">$3&lt;4$</span>) and win against <span class="math-container">$\theta_j=2$</span> (sum is <span class="math-container">$4$</span>). 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, <span class="math-container">$p&gt;1-p$</span>.</p>
<p>Alternatively as <span class="math-container">$\theta_i=2$</span>, you can bid 3 in the first round, which means you win against <span class="math-container">$\theta_j=1$</span> who challenges (sum is <span class="math-container">$3\geq3$</span>, and them overbidding means you challenge in the next round and win) and lose against <span class="math-container">$\theta_j=2$</span> who overbids to 4 which you have to challenge and lose - <span class="math-container">$\theta_j=2$</span> would not challenge as this would mean sure loss to them. Thus losing is more likely, <span class="math-container">$p&gt;1-p$</span>.</p>
<p>Thus, back to type 1 bidding 2 in the first round. If <span class="math-container">$p&gt;1-p$</span> and there was no inference, type <span class="math-container">$\theta_j=2$</span> bids 4 and you as <span class="math-container">$\theta_i=1$</span> win by challenging 4. But that would mean you always win as type <span class="math-container">$\theta_i=1$</span> when bidding 2. Then it must be that <span class="math-container">$j$</span> 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 <span class="math-container">$\theta_i=1$</span> means, winning with probability <span class="math-container">$1-p$</span> and losing with probability <span class="math-container">$p$</span>. Because <span class="math-container">$p&gt;1-p$</span>, this is worse than betting 3.</p>
<p>So <span class="math-container">$\theta_i=1$</span> bids 3 in the first round when <span class="math-container">$p&gt;1-p$</span>.</p>
<p>Now to type <span class="math-container">$\theta_i=2$</span> in the first round. Given the above, bidding 4 is better than bidding 3 given prior <span class="math-container">$p&gt;1-p$</span>. 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 <span class="math-container">$\theta_i=2$</span>. 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.</p>
<p>So <span class="math-container">$\theta_i=2$</span> bids 4 in the first round when <span class="math-container">$p&gt;1-p$</span>.</p>
<p>With <span class="math-container">$p&lt;1-p$</span>, 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.</p>

CC BY-SA 4.0
2021-05-21T18:27:30.353Z
Thanks for the thoughts -- I'll take a close look at soon as I have time!
afreelunch
CC BY-SA 4.0
2021-05-20T20:22:02.170Z
2021-05-20T20:22:02.170Z
Bayesian
<p>Here is a game I like to play:</p>
<ul>
<li>Every player gets a random number independently drawn from a fixed distribution. To fix ideas, suppose everyone draws a number from <span class="math-container">$\{2, ..., 10\}$</span> where all numbers are equally likely (that way you can do this with playing cards...)</li>
<li>The first player says: 'I think that the sum of the numbers is <span class="math-container">$k$</span> or higher' where the 'bet' <span class="math-container">$k$</span> is an integer that they need to choose.</li>
<li>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 <span class="math-container">$k'$</span> or higher' for some integer <span class="math-container">$k' &gt; k$</span>.</li>
<li>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.</li>
<li>The above defines the rules in a particular round. And the players keep playing rounds until all but one is knocked out.</li>
</ul>
<p>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:</p>
<ul>
<li>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 <span class="math-container">$(2 + 10)/2 = 6$</span>. Suppose then that my number is <span class="math-container">$x$</span> and that I face <span class="math-container">$n$</span> opponents. Then I might predict the sum to be <span class="math-container">$x + 6n$</span>; and increase the betting until doing so would force me to bet higher than this (at which point I challenge).</li>
<li>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 <span class="math-container">$k$</span>, you may infer that <span class="math-container">$x + 6n \geq k$</span> where <span class="math-container">$x$</span> is their number, or equivalently <span class="math-container">$x \geq k - 6n$</span>. You might then compute the expected value of their number <span class="math-container">$x$</span> conditional on its being larger than <span class="math-container">$k - 6n$</span> (which is pretty easy in the uniform example).</li>
<li>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.</li>
</ul>

CC BY-SA 4.0
2021-05-20T04:02:47.927Z
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.
Giskard
CC BY-SA 4.0
2021-05-20T04:05:15.037Z
I didn't think you can easily write down a BNE in the original game, as the space of pure strategies is huge.
Giskard
CC BY-SA 4.0
2021-05-20T09:50:30.653Z
@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
afreelunch
CC BY-SA 4.0
2021-05-20T09:51:34.933Z
And yes I do agree solving simplified versions is a good starting point (I have not tried this yet)
afreelunch
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2021-05-20T14:36:38.400Z
All players draw only once in the beginning? Do players know their own number or are numbers only revealed in the end?
Bayesian
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2021-05-20T17:02:49.753Z
@Bayesian I'm not sure I understand your question exactly. Players do indeed know their own number, and numbers are indeed revealed at the end of the round.
afreelunch
CC BY-SA 4.0
2021-05-20T17:04:24.933Z
I think perhaps you are asking if numbers are re-drawn at the start of a new round? The answer to this is 'yes' (although I suspect that focusing on multiple rounds is a bit of a distraction -- if we know how to play a given round, we probably have a good idea of how to play a sequence of rounds).
afreelunch
CC BY-SA 4.0
2021-05-20T17:20:56.610Z
I think I understand the game now. Each player gets a number/card and only the player observes their number, i.e., the own number is known when bidding. Once someone challenges all cards are revealed. That is, you keep your number and do not draw a new one in the next "bidding round" after someone overbids another player.
Bayesian
CC BY-SA 4.0
2021-05-20T17:25:08.887Z
I believe you will ruin your game by solving it ;) There should be a pure-strategy equilibrium in which the decision reveals the number such that the last player knows if he loses or wins and all other players know the probability of winning when they bid.
Bayesian
CC BY-SA 4.0
2021-05-21T09:38:41.733Z
@Bayesian Haha you may be right. Though fortunately learning the BNE usually doesn't tell you how to play a game :)
afreelunch
CC BY-SA 4.0
2021-05-21T10:36:38.847Z
True, and even then it would be hard to compute all cutoff strategies in your big example with 10 cards. However, note that $p=1/2$ is the knife-edge case in the t2o-player 2-numbers cont game. But $p\neq1/2$ in your example with cards! For instance, if you have 4 cards (two 1s and 2s each) and you receive the first card and it is a 1, you know that when the other player gets a card the probability to get a 1 is 1/3 and to get a 2 is 2/3. So in the card deck
Bayesian
<game-theory>
Betting on the sum of scores
afreelunch
# Betting on the sum of scores

(Asked by: username_0 on 2021-05-19)

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.

**Tags:** game-theory
> **Comments:**
> **username_1**: 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.  
> **username_1**: I didn't think you can easily write down a BNE in the original game, as the space of pure strategies is huge.  
> **username_0**: @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  
> **username_0**: And yes I do agree solving simplified versions is a good starting point (I have not tried this yet)  
> **username_2**: All players draw only once in the beginning? Do players know their own number or are numbers only revealed in the end?  
> **username_0**: @Bayesian I'm not sure I understand your question exactly. Players do indeed know their own number, and numbers are indeed revealed at the end of the round.  
> **username_0**: I think perhaps you are asking if numbers are re-drawn at the start of a new round? The answer to this is 'yes' (although I suspect that focusing on multiple rounds is a bit of a distraction -- if we know how to play a given round, we probably have a good idea of how to play a sequence of rounds).  
> **username_2**: I think I understand the game now. Each player gets a number/card and only the player observes their number, i.e., the own number is known when bidding. Once someone challenges all cards are revealed. That is, you keep your number and do not draw a new one in the next "bidding round" after someone overbids another player.  
> **username_2**: I believe you will ruin your game by solving it ;) There should be a pure-strategy equilibrium in which the decision reveals the number such that the last player knows if he loses or wins and all other players know the probability of winning when they bid.  
> **username_0**: @Bayesian Haha you may be right. Though fortunately learning the BNE usually doesn't tell you how to play a game :)  
> **username_2**: True, and even then it would be hard to compute all cutoff strategies in your big example with 10 cards. However, note that $p=1/2$ is the knife-edge case in the t2o-player 2-numbers cont game. But $p\neq1/2$ in your example with cards! For instance, if you have 4 cards (two 1s and 2s each) and you receive the first card and it is a 1, you know that when the other player gets a card the probability to get a 1 is 1/3 and to get a 2 is 2/3. So in the card deck  

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## 1 answers

### Answer by username_2 on 2021-05-20. Score: 2

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.

> **Comments:**
> **username_0**: Thanks for the thoughts -- I'll take a close look at soon as I have time!  

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