<p>Given that <span class="math-container">$X_1,X_2\sim\text{Unif}(0,1)$</span> and are independently distributed. Suppose the bidding strategy of player 2 is to bid if and only if <span class="math-container">$X_2$</span> is greater than <span class="math-container">$a$</span>.</p>
<p>Player 1 after observing <span class="math-container">$X_1=x_1$</span> will bid if</p>
<p><span class="math-container">$2\Pr(X_2\leq x_1, X_2&gt;a) -2\Pr(X_2&gt; x_1, X_2&gt;a)+ 1\Pr(X_2\leq a)\geq -1$</span></p>
<p>For <span class="math-container">$x_1\geq a$</span>, the above inequality is</p>
<p><span class="math-container">$2(x_1-a)-2(1-x_1)+a\geq -1$</span></p>
<p>For player 1's best response strategy to be bid if and only if <span class="math-container">$X_1$</span> is greater than <span class="math-container">$a$</span>, it must be the case that at <span class="math-container">$x_1=a$</span>,</p>
<p><span class="math-container">$2(x_1-a)-2(1-x_1)+a= -1$</span></p>
<p>Substituting <span class="math-container">$x_1=a$</span> in above, we get the value of <span class="math-container">$a$</span>,</p>
<p><span class="math-container">$-2(1-a)+a= -1$</span> gives</p>
<p><span class="math-container">$a = \frac{1}{3}$</span>.</p>
<p>So the Symmetric Bayesian Nash equilibrium strategy is to bid if and only if the observed number is greater than <span class="math-container">$\frac{1}{3}$</span>. Now we can quickly check that this is a Bayesian Nash equilibrium,
suppose player 2 is to bid if and only if <span class="math-container">$x_2$</span> is greater than <span class="math-container">$\frac{1}{3}$</span>,</p>
<p>Player <span class="math-container">$1$</span>'s expected payoff from bidding when <span class="math-container">$x_1\geq\frac{1}{3}$</span> exceeds than that from folding:</p>
<p><span class="math-container">$2(x_1-\frac{1}{3})-2(1-x_1)+\frac{1}{3} = 4x_1-\frac{7}{3}\geq -1$</span>.</p>
<p>Player <span class="math-container">$1$</span>'s expected payoff from folding when <span class="math-container">$x_1&lt;\frac{1}{3}$</span> exceeds than that from bidding:</p>
<p><span class="math-container">$-1\geq -2\times\frac{2}{3}+\frac{1}{3}$</span>.</p>
<p>By symmetry, this is a Bayesian Nash equilibrium.</p>

CC BY-SA 4.0
2024-11-11T17:37:55.430Z
oh! Thank you so much!
Pc1
CC BY-SA 4.0
2024-11-11T16:04:45.527Z
2024-11-11T16:49:17.550Z
2024-11-11T16:49:17.550Z
Amit
<p>I have tried solving a question from Watson's book &quot;Strategy&quot; and my answer doesn't match with the solutions provided online. I am unable to figure out why. Please help.</p>
<p>The question:
Consider a simple simultaneous-bid poker game. First, nature selects numbers x1 and x2 . Assume that these numbers are independently and uniformly
distributed between 0 and 1. Player 1 observes x1 and player 2 observes x2 ,
but neither player observes the number given to the other player. Simultaneously and independently, the players choose either to fold or to bid. If both
players fold, then they both get the payoff −1. If only one player folds,
then he obtains −1 while the other player gets 1. If both players elected
to bid, then each player receives 2 if his number is at least as large as the
other player’s number; otherwise, he gets −2. Compute the Bayesian Nash
equilibrium of this game. (Hint: Look for a symmetric equilibrium in which
a player bids if and only if his number is greater than some constant a. Your
analysis will reveal the equilibrium value of a.)</p>
<p>My solution:</p>
<p>I have assumed player 2 Bids only if <span class="math-container">$x_2&gt;\alpha_2$</span>. Now, player 1 bids only if her utility from bidding, given her type, is larger than her utility from folding.
That is,
Player 1 bids if and only if <span class="math-container">$u_1(B|x_1)&gt;-1$</span>
<span class="math-container">$\implies P[x_2&gt;\alpha_2]P[x_2&gt;x_1](-2)+P[x_2&gt;\alpha_2]P[x_2&lt;x_1](2)+P[x_2&lt;\alpha_2](1)&gt;-1$</span></p>
<p><span class="math-container">$\implies (1-\alpha_2)(1-x_1)(-2)+(1-\alpha_2)(x_1)(2)+\alpha_2&gt;-1$</span></p>
<p><span class="math-container">$\implies 4x_1-4\alpha_2x_1+3\alpha_2-1&gt;0$</span></p>
<p>P1 bids iff <span class="math-container">$4x_1(1-\alpha_2)+3\alpha_2 -1&gt;0$</span></p>
<p>since <span class="math-container">$\alpha_2\in[0,1]$</span> P1 will definitely bid if <span class="math-container">$\alpha_2&gt;\frac{1}{3}$</span>
and she will be indifferent between B or F if <span class="math-container">$\alpha_2=1/3$</span> and <span class="math-container">$x_1&gt;0$</span>.
and if <span class="math-container">$\alpha_2&lt;\frac{1}{3}$</span> then P1 bids iff <span class="math-container">$x_1&gt;\frac{1-3\alpha_2}{4(1-\alpha_2)}$</span>.</p>
<p>Can someone please tell me if my answer(up until this point is correct.) It differs from the solution available online.</p>
<p>Solution available online
If <span class="math-container">$\alpha_2&gt;1/3$</span> then player 1 bids. if <span class="math-container">$\alpha_2&lt;1/3$</span> P1 bids if <span class="math-container">$x_1&gt;(1+\alpha_2)/4$</span></p>

CC BY-SA 4.0
2024-11-11T16:54:09.747Z
$x_2>\alpha_2$ and $x_2>x_1$ are not independent events. You cannot multiply their probabilities.
Amit
<game-theory><bayesian-game>
Incomplete information bidding game with cutoff strategies
Pc1
# Incomplete information bidding game with cutoff strategies

(Asked by: username_0 on 2024-11-11)

I have tried solving a question from Watson's book "Strategy" and my answer doesn't match with the solutions provided online. I am unable to figure out why. Please help.

The question:
Consider a simple simultaneous-bid poker game. First, nature selects numbers x1 and x2 . Assume that these numbers are independently and uniformly
distributed between 0 and 1. Player 1 observes x1 and player 2 observes x2 ,
but neither player observes the number given to the other player. Simultaneously and independently, the players choose either to fold or to bid. If both
players fold, then they both get the payoff −1. If only one player folds,
then he obtains −1 while the other player gets 1. If both players elected
to bid, then each player receives 2 if his number is at least as large as the
other player’s number; otherwise, he gets −2. Compute the Bayesian Nash
equilibrium of this game. (Hint: Look for a symmetric equilibrium in which
a player bids if and only if his number is greater than some constant a. Your
analysis will reveal the equilibrium value of a.)

My solution:

I have assumed player 2 Bids only if $x\_2>\alpha\_2$. Now, player 1 bids only if her utility from bidding, given her type, is larger than her utility from folding.
That is,
Player 1 bids if and only if $u\_1(B|x\_1)>-1$
$\implies P[x\_2>\alpha\_2]P[x\_2>x\_1](-2)+P[x\_2>\alpha\_2]P[x\_2-1$

$\implies (1-\alpha\_2)(1-x\_1)(-2)+(1-\alpha\_2)(x\_1)(2)+\alpha\_2>-1$

$\implies 4x\_1-4\alpha\_2x\_1+3\alpha\_2-1>0$

P1 bids iff $4x\_1(1-\alpha\_2)+3\alpha\_2 -1>0$

since $\alpha\_2\in[0,1]$ P1 will definitely bid if $\alpha\_2>\frac{1}{3}$
and she will be indifferent between B or F if $\alpha\_2=1/3$ and $x\_1>0$.
and if $\alpha\_2<\frac{1}{3}$ then P1 bids iff $x\_1>\frac{1-3\alpha\_2}{4(1-\alpha\_2)}$.

Can someone please tell me if my answer(up until this point is correct.) It differs from the solution available online.

Solution available online
If $\alpha\_2>1/3$ then player 1 bids. if $\alpha\_2<1/3$ P1 bids if $x\_1>(1+\alpha\_2)/4$

**Tags:** game-theory, bayesian-game
> **Comments:**
> **username_1**: $x_2>\alpha_2$ and $x_2>x_1$ are not independent events. You cannot multiply their probabilities.  

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## 1 answers

### Answer by username_1 on 2024-11-11. Score: 2 | Accepted answer

Given that $X\_1,X\_2\sim\text{Unif}(0,1)$ and are independently distributed. Suppose the bidding strategy of player 2 is to bid if and only if $X\_2$ is greater than $a$.

Player 1 after observing $X\_1=x\_1$ will bid if

$2\Pr(X\_2\leq x\_1, X\_2>a) -2\Pr(X\_2> x\_1, X\_2>a)+ 1\Pr(X\_2\leq a)\geq -1$

For $x\_1\geq a$, the above inequality is

$2(x\_1-a)-2(1-x\_1)+a\geq -1$

For player 1's best response strategy to be bid if and only if $X\_1$ is greater than $a$, it must be the case that at $x\_1=a$,

$2(x\_1-a)-2(1-x\_1)+a= -1$

Substituting $x\_1=a$ in above, we get the value of $a$,

$-2(1-a)+a= -1$ gives

$a = \frac{1}{3}$.

So the Symmetric Bayesian Nash equilibrium strategy is to bid if and only if the observed number is greater than $\frac{1}{3}$. Now we can quickly check that this is a Bayesian Nash equilibrium,
suppose player 2 is to bid if and only if $x\_2$ is greater than $\frac{1}{3}$,

Player $1$'s expected payoff from bidding when $x\_1\geq\frac{1}{3}$ exceeds than that from folding:

$2(x\_1-\frac{1}{3})-2(1-x\_1)+\frac{1}{3} = 4x\_1-\frac{7}{3}\geq -1$.

Player $1$'s expected payoff from folding when $x\_1<\frac{1}{3}$ exceeds than that from bidding:

$-1\geq -2\times\frac{2}{3}+\frac{1}{3}$.

By symmetry, this is a Bayesian Nash equilibrium.

> **Comments:**
> **username_0**: oh! Thank you so much!  

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