Incomplete information bidding game with cutoff strategies

Question

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 $u_1(B|x_1)>-1$ $\implies P[x_2>\alpha_2]P[x_2>x_1](-2)+P[x_2>\alpha_2]P[x_2<x_1](2)+P[x_2<\alpha_2](1)>-1$

$\implies (1-\alpha_2)(1-x_1)(-2)+(1-\alpha_2)(x_1)(2)+\alpha_2>-1$

$\implies 4x_1-4\alpha_2x_1+3\alpha_2-1>0$

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 $x_1>(1+\alpha_2)/4$

Answer 1

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.