Sequential Bidding With Private Dice Signals

Game Theory · Hard · Free problem

Two fair six-sided dice are rolled. You see die A (value $a$), and your opponent sees die B (value $b$) -- neither of you sees the other die. A "box" worth $a + b$ dollars is up for auction.

The two of you bid sequentially: Player 1 announces a bid, then Player 2 (having heard the bid) announces their bid. The higher bidder wins the box and pays their bid. (Assume ties go to Player 1.)

  1. Who benefits from being the second bidder, and why?
  1. What is Player 1's optimal bidding strategy as a function of $a$?
  1. What is Player 2's optimal response as a function of $b$ and the observed bid $x$?
  1. In equilibrium, what is the expected profit for each player?

Hints

  1. Think about what happens when you win -- does winning give you good or bad news about the box value? This is the winner's curse.
  2. The second bidder can infer the first player's die value from their bid. How does this information asymmetry affect each player's ability to profit?
  3. Try a linear bidding strategy $\beta_1(a) = a + c$ for Player 1 and compute the expected profit as a function of $c$. What value of $c$ avoids losses?

Worked Solution

How to Think About It: This is a sequential common-value auction with private signals. Each player sees half the box value ($a$ or $b$) and must infer the other half. Two forces dominate. First, the winner's curse: if you bid on your naive expectation and win, you tend to win precisely when the opponent's signal is *low* -- so winning is bad news about the box. Second, and decisively here, information leakage: Player 1 moves first, so the bid itself reveals information about $a$ to Player 2. Before any algebra, the gut read is: Player 2 is advantaged, and Player 1 cannot extract positive surplus. The flip side -- and the subtle part -- is that *how much* surplus Player 2 captures is not pinned down: it depends on which equilibrium is played.

Quick Estimate: Suppose Player 1 naively bids $a+3.5$ (their expected box value). Player 2 reads $x$, infers $a = x-3.5$, and outbids only when the box clears the price, i.e. when $b \ge 4$. So Player 1 wins only on $b \in \{1,2,3\}$ -- exactly the low-box states -- paying $a+3.5$ for a box worth $a+b$. Expected profit $= \tfrac{1}{6}[(1-3.5)+(2-3.5)+(3-3.5)] = -\tfrac{4.5}{6} = -\$0.75$. Player 1 loses money; the surplus flows to Player 2.

Approach: Model this as a signaling game and look for Perfect Bayesian Equilibria (PBE): Player 1's bid $\beta_1(a)$, Player 2's best response $\beta_2(b,x)$, and beliefs consistent with $\beta_1$. Two subtleties matter. (i) Incentive compatibility: Player 1's strategy must give *no type $a$* a profitable deviation, including mimicking another type's bid -- this kills the tempting "shade by 1" answer. (ii) Multiplicity: the game has many PBE -- separating *and* pooling -- and while they all hold Player 1 to $0$, they hand Player 2 different payoffs, so $\pi_2$ is not determined by PBE alone.

Formal Solution:

*Part 1 -- who is advantaged.* Player 2. Hearing the bid first lets Player 2 condition on the information leaked by $x$, then compete only when $a+b$ truly exceeds the price -- sidestepping the winner's curse. This is an informational edge, not a price discount: to win, Player 2 must still bid *above* Player 1.

*Part 3 (Player 2's response), since Player 1 best-responds to it.* Given beliefs, let $\hat a = E[a \mid x]$ be Player 2's posterior mean of $a$ after hearing bid $x$ (on a separating path $\hat a = \beta_1^{-1}(x)$; on a pooling path $\hat a$ is the pool average). Player 2 values the box at $\hat a + b$ and wins by bidding just above $x$ exactly when $\hat a + b > x$ (ties go to Player 1, and Player 2 never pays above its own value). So

$\beta_2(b,x) = \begin{cases} \text{bid just above } x & \text{if } \hat a + b > x,\\[2pt] \text{drop out (let P1 win)} & \text{otherwise.}\end{cases}$

*Why the "shade by 1" answer is wrong (the incentive-compatibility trap).* Consider the candidate $\beta_1(a) = a+1$. On its own path, Player 2 infers $\hat a = x-1$ and outbids when $(x-1)+b > x$, i.e. $b \ge 2$; Player 1 then wins only at $b=1$, paying $a+1$ for a box worth $a+1$ -- zero profit. So *on-path* Player 1 earns $0$. But this is not an equilibrium, because a high type can deviate. Let true type $a$ instead bid $x' = 2$ (mimicking the lowest type $a'=1$). Player 2 now infers $\hat a = 1$ and outbids only when

+b > 2$, i.e. $b \ge 2$, so Player 1 wins at $b=1$ -- but now pays only
$ for a box worth $a+1$, netting $a-1 > 0$ whenever $a > 1$ (e.g. $a=6$ earns $\tfrac{1}{6}(6-1)\approx\$0.83$). Since a profitable deviation exists, $\beta_1(a)=a+1$ fails incentive compatibility.

*Part 2 -- the best Player 1 can secure (always $0$).* Take any PBE and any bid $x$ that Player 1 actually plays. If Player 1 *wins* at $x$, it is because Player 2 declined to outbid, i.e. $\hat a + b \le x$, where $\hat a = E[a\mid x]$ is the average type bidding $x$. Player 1's realized profit when winning is $(a+b) - x \le (a+b) - (\hat a + b) = a - \hat a$. Averaging $a - \hat a$ over the types that pool at $x$ gives $0$ (by definition of the conditional mean), and the loss states only subtract, so no bid can yield positive expected profit summed across the types that use it; the best Player 1 can guarantee is $0$. Concretely, two clean PBE attain it:

*Part 4 -- expected profits, and why $\pi_2$ is not unique.* In every PBE, Player 1 is held to

$\boxed{\pi_1 = 0,}$

but Player 2's payoff is not determined by the PBE concept. The truthful equilibrium gives $\pi_2 = 3.5$; the pooling equilibria give $\pi_2 = 7 - x_0$ for any $x_0 < 4.5$, which ranges over $(2.5,\,7)$ as $x_0$ ranges over $(0,4.5)$ -- for example pooling at $x_0 = 2$ gives Player 2 a payoff of $5$, well outside the separating family's range. (The earlier claim that the payoff set is just the separating family $a+c$ with $\pi_2 \in [2.5,3.5]$ is incomplete: it omits these pooling/no-win PBE.) So the only robust prediction is the split $(\pi_1,\pi_2)=(0,\ 7-\pi_1^{\text{price}})$ with Player 1 always at $0$; pinning down Player 2's exact dollar take requires an equilibrium refinement.

Answer: 1. Player 2 is advantaged -- the first bid leaks information about $a$, so Player 2 competes only when it pays, avoiding the winner's curse. 2. Player 1 cannot do better than zero expected profit; a focal optimal strategy is the truthful bid $\beta_1(a)=a$ (pooling at any $x_0<4.5$ also nets Player 1 zero). 3. Player 2 bids just above $x$ iff $\hat a + b > x$, where $\hat a = E[a\mid x]$ (the inferred type/pool mean), and drops out otherwise. 4. $\pi_1 = 0$ in every PBE. Player 2's payoff is not unique: $\pi_2 = 3.5$ at the truthful separating equilibrium, and $\pi_2 = 7 - x_0$ for any pooling equilibrium at $x_0 < 4.5$ (e.g. $\pi_2 = 5$ when $x_0 = 2$). Player 1 is always held to $0$, but Player 2's take depends on the equilibrium selected.

Intuition

This problem is a miniature version of the winner's curse that shows up constantly in trading. Whenever you win an auction or get filled on a trade, you should ask: "What does the fact that I won tell me about the true value?" If you win, it is because nobody else wanted it at that price -- which is bad news. The second bidder here is like a market maker who sees order flow before quoting: they only trade when the information favors them. The first bidder is like someone posting a limit order blind, knowing they will get picked off by informed flow.

The deeper lesson is about information asymmetry in sequential games. The first mover reveals their private signal through their bid, and the second mover exploits this. In equilibrium, Player 1 is forced to shade their bid so aggressively that they earn zero profit -- all the surplus goes to the informed side. This is exactly the adverse selection problem in market microstructure: liquidity providers (first movers) earn zero or negative edge against informed traders, which is why bid-ask spreads exist in real markets.

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