<ul>
<li>Players <span class="math-container">$A,B$</span> are bidding for the outcome of 200 coinflips <span class="math-container">$X\sim Bin(200,1/2)$</span>.
The winning player gains <span class="math-container">$X$</span> minus their bid.</li>
<li>Player <span class="math-container">$B$</span> has advantage of knowing the outcome of 20 randomly selected coins. (Let's call this number <span class="math-container">$K\sim Bin(20,1/2)$</span>)</li>
<li>Upon equal bids, the winner is picked randomly.</li>
<li>This is a first price auction and the players want to maximize expected payoffs.</li>
</ul>
<ol>
<li>What are the optimal strategies/expectations under simultaneous bidding?</li>
</ol>
<p>My attempt:
restrict ourselves to pure strategies.</p>
<ul>
<li><p><span class="math-container">$A$</span>'s strategy profile is just the their bid <span class="math-container">$a\in[0,99]$</span></p>
</li>
<li><p><span class="math-container">$B$</span>'s strategy profile is the set of (likely nondecreasing) assignments <span class="math-container">$b(k)\colon k\to \left[0,\dfrac{180}{2}+k-1\right]$</span></p>
</li>
<li><p>A's best response to <span class="math-container">$b(k)$</span>:
<span class="math-container">$$\begin{align*}BR_A(b(k)) &amp;=\text{argmax}_a\,\,  E_{k\sim K} \,\,  \begin{cases}\left(k +\dfrac{180}{2}-a\right)  &amp; \text{if }a&gt;b(k) \\ \dfrac{1}{2} \left(k +\dfrac{180}{2}-a\right) &amp;  \text{if }a=b(k)\\ 0 &amp;\text{else}\end{cases}\\
&amp;=\text{argmax}_a \left[
\mathbf 1_{a=b(T)} P(K=T) \dfrac{1}{2}\left(k +\dfrac{180}{2}-a\right) + \sum_{k&lt;T}P(K=k) \left(k+\dfrac{180}{2}-a\right)\right]\end{align*}$$</span>
where <span class="math-container">$T(b(k),a)$</span> (for &quot;threshold&quot;) is defined as <span class="math-container">$T:=\min_k \colon b(k)\geq a$</span>. In other words, given <span class="math-container">$B$</span>'s bid curve <span class="math-container">$b(k)$</span>, <span class="math-container">$A$</span> must decide the upper bound <span class="math-container">$k^*$</span> of possible <span class="math-container">$K$</span>-values for which they want to outbid <span class="math-container">$B$</span>. Once <span class="math-container">$k^*$</span> is decided upon, the actual bid <span class="math-container">$a^*$</span> is just <span class="math-container">$b(k^*-1)+1$</span></p>
</li>
<li><p><span class="math-container">$B$</span>'s best response to <span class="math-container">$a$</span> is <span class="math-container">$a+1$</span> (or <span class="math-container">$a$</span> to split) as long as <span class="math-container">$$\dfrac{180}{2}+k - (a+1)&gt;0\iff a &lt; \dfrac{180}{2}+k-1 \iff k&gt;a+1 -\dfrac{180}{2}$$</span> and staying out otherwise. This corresponds to <span class="math-container">$T(a)=a+1 -(180/2)$</span></p>
</li>
<li><p>I'm not sure that this is the right thing to do, but I'm compelled to plug in this <span class="math-container">$T(a)$</span> into <span class="math-container">$A$</span>'s best response and find the maximizing <span class="math-container">$a$</span>. It feels like the range of <span class="math-container">$k$</span> values for which <span class="math-container">$B$</span> wants to outbid <span class="math-container">$A$</span> should meet the range of <span class="math-container">$k$</span>s for which <span class="math-container">$A$</span> wants to outbid <span class="math-container">$B$</span>, or else there is arbitrage.</p>
</li>
</ul>
<ol start="2">
<li>I'd appreciate any ideas on what would happen if <span class="math-container">$B$</span> makes a bid that is visible to <span class="math-container">$A$</span>. It feels like this would be a much harder problem, as it opens up mixed bluffing strategies for <span class="math-container">$B$</span>.</li>
</ol>

CC BY-SA 4.0
2024-06-27T06:24:30.843Z
1. Perhaps include more details, such as this is a first price auction and players want to maximize expected payoffs.
Giskard
CC BY-SA 4.0
2024-06-27T06:25:12.010Z
2. Do you have reason to believe that an 'elegant' analytical solution exists, e.g. is this an exercise from a book or class? An alternative to an elegant solution would be something that a computer program can find using trial and error, which seems very possible here given the limited strategy space.
Giskard
CC BY-SA 4.0
2024-06-27T06:50:27.993Z
The question was allegedly asked in a trading interview. It wouldn't be surprising for it to not have a nice solution. My hope is to find a way to confirm the intuition of adverse selection: that A can only ever win bids that have negative EV.
enryuxbt
CC BY-SA 4.0
2024-06-27T16:47:35.797Z
"*A can only ever win bids that have negative EV.*" If, instead of asking for a complete solution, you post that conjecture as a question, users here should be able disprove it.
Giskard
<game-theory><nash-equilibrium>
Bidding game with asymmetric information
enryuxbt
# Bidding game with asymmetric information

(Asked by: username_0 on 2024-06-25)

* Players $A,B$ are bidding for the outcome of 200 coinflips $X\sim Bin(200,1/2)$.
  The winning player gains $X$ minus their bid.
* Player $B$ has advantage of knowing the outcome of 20 randomly selected coins. (Let's call this number $K\sim Bin(20,1/2)$)
* Upon equal bids, the winner is picked randomly.
* This is a first price auction and the players want to maximize expected payoffs.

1. What are the optimal strategies/expectations under simultaneous bidding?

My attempt:
restrict ourselves to pure strategies.

* $A$'s strategy profile is just the their bid $a\in[0,99]$
* $B$'s strategy profile is the set of (likely nondecreasing) assignments $b(k)\colon k\to \left[0,\dfrac{180}{2}+k-1\right]$
* A's best response to $b(k)$:
  $$\begin{align\*}BR\_A(b(k)) &=\text{argmax}\_a\,\, E\_{k\sim K} \,\, \begin{cases}\left(k +\dfrac{180}{2}-a\right) & \text{if }a>b(k) \\ \dfrac{1}{2} \left(k +\dfrac{180}{2}-a\right) & \text{if }a=b(k)\\ 0 &\text{else}\end{cases}\\
  &=\text{argmax}\_a \left[
  \mathbf 1\_{a=b(T)} P(K=T) \dfrac{1}{2}\left(k +\dfrac{180}{2}-a\right) + \sum\_{k
  where $T(b(k),a)$ (for "threshold") is defined as $T:=\min\_k \colon b(k)\geq a$. In other words, given $B$'s bid curve $b(k)$, $A$ must decide the upper bound $k^\*$ of possible $K$-values for which they want to outbid $B$. Once $k^\*$ is decided upon, the actual bid $a^\*$ is just $b(k^\*-1)+1$
* $B$'s best response to $a$ is $a+1$ (or $a$ to split) as long as $$\dfrac{180}{2}+k - (a+1)>0\iff a < \dfrac{180}{2}+k-1 \iff k>a+1 -\dfrac{180}{2}$$ and staying out otherwise. This corresponds to $T(a)=a+1 -(180/2)$
* I'm not sure that this is the right thing to do, but I'm compelled to plug in this $T(a)$ into $A$'s best response and find the maximizing $a$. It feels like the range of $k$ values for which $B$ wants to outbid $A$ should meet the range of $k$s for which $A$ wants to outbid $B$, or else there is arbitrage.

2. I'd appreciate any ideas on what would happen if $B$ makes a bid that is visible to $A$. It feels like this would be a much harder problem, as it opens up mixed bluffing strategies for $B$.

**Tags:** game-theory, nash-equilibrium

> **Comments:**
> **username_1**: 1. Perhaps include more details, such as this is a first price auction and players want to maximize expected payoffs.  
> **username_1**: 2. Do you have reason to believe that an 'elegant' analytical solution exists, e.g. is this an exercise from a book or class? An alternative to an elegant solution would be something that a computer program can find using trial and error, which seems very possible here given the limited strategy space.  
> **username_0**: The question was allegedly asked in a trading interview. It wouldn't be surprising for it to not have a nice solution. My hope is to find a way to confirm the intuition of adverse selection: that A can only ever win bids that have negative EV.  
> **username_1**: "*A can only ever win bids that have negative EV.*" If, instead of asking for a complete solution, you post that conjecture as a question, users here should be able disprove it.  

---

## 0 answers

