Trilateral escalation in the dollar auction

Abstract

We find a new set of subgame perfect equilibria in a dollar auction that involves three active bidders. The player who falls to the third place continues making efforts to catch up until his lag from the frontrunner widens to a critical distance beyond which the catchup efforts become unprofitable. At that juncture the second-place player pauses bidding thereby bettering the chance for the third-place one to leapfrog to the front so as to perpetuate the trilateral rivalry. Once two players have emerged as the top two rivals, any such trilateral rivalry equilibrium produces larger total surplus for the three players than its bilateral rivalry counterpart does, where anyone who falls to the third place immediately drops out.

Appendices

A formal definition of the model

1.1 A.1 The dollar auction game

Let $I:=\{1, 2, 3\}$ denote the set of players; and $w_0:=b_{i,0}:=0$ for all $i \in I$. Let

$$\begin{aligned} n_0:=\left( w_0, (b_{i,0})_{i \in I}\right) \quad (=(0, 0, 0, 0)) \end{aligned}$$

be the initial node of the game. Given $n_0$, each player i’s set of feasible actions is $A_i(n_0):=\{0, 1\}$, with 1 signifying “to bid” and 0 signifying “to stay put.”

Pick any $t=1, 2, 3, \ldots$. Suppose that a node $n_{t-1}:=(w_{t-1}, (b_{i, t-1})_{i \in I})\in (I \cup \{0\}) \times \mathbb {R}_+^3$ and each player’s set of feasible actions at $n_{t-1}$ have been defined. For any $a_{t-1}:=(a_{i, t-1})_{i \in I}$ such that $a_{i, t-1}$ is a feasible action for player i at node $n_{t-1}$ ($\forall i \in I$), define the set of immediate successors of $n_{t-1}$ to be

$$\begin{aligned} N(n_{t-1}, a_{t-1}) \ := \left\{ (w_t, (b_{i, t})_{i \in I}) \in (I \cup \{0\} \times \mathbb {R}_+^3 \ \left| \ \begin{array}{l} w_t \ne w_{t-1} \Rightarrow \left[ \begin{array}{l} a_{w_t, t-1}=1, \\ b_{w_t, t}=\delta +\max _{i \in I}b_{i, t-1} \end{array}\right] \\ \left[ i \ne w_t \text{ or } i=w_t=w_{t-1} \right] \Rightarrow b_{i, t}=b_{i, t-1} \end{array} \right. \right\} . \end{aligned}$$

At any node $n_t:=(w_t, (b_{i, t})_{i \in I}) \in N(n_{t-1}, a_{t-1})$, which descends immediately from $(n_{t-1}, a_{t-1})$, define the set $A_i(a_{t-1}, n_t)$ of feasible actions for player i by:

$$\begin{aligned} A_i(a_{t-1}, n_t):=\left\{ \begin{array}{ll} \{0, 1\} &{} \hbox { if } w_t \ne 0 \hbox { and } i \ne w_t \; and \; \exists j \in I: a_{j, t-1}=1\\ \{0\} &{} \hbox { if } w_t=0 \hbox { or } i=w_t \; or\; a_{1,t-1}=a_{2, t-1}=a_{3, t-1}=0. \end{array}\right. \end{aligned}$$

The transition probability from $(n_{t-1}, a_{t-1})$ to any element $n_t:=(w_t, (b_{i, t})_{i \in I})$ of $N(n_{t-1}, a_{t-1})$ is defined by:

$$\begin{aligned}\{i \in I \mid a_{i, t-1}=1\} = \varnothing \Rightarrow & \Pr \{w_t=w_{t-1} \mid n_{t-1}, a_{t-1}\}=1; \\ \{i \in I \mid a_{i, t-1}=1\} \ne \varnothing \Rightarrow & \Pr \{w_t=i \mid n_{t-1}, a_{t-1}\}=\frac{a_{i, t-1}}{\left| \{i \in I \mid a_{i, t-1}=1\}\right| } \quad (\forall i \in I). \end{aligned}$$

The initial node $n_0$ is the 0-history. For any $t=1, 2, \ldots$, a t-history is a finite sequence $((n_k, a_k)_{k=0}^{t-1}, n_t)$ that satisfies all the following conditions: (i) $n_0$ is the 0-history; (ii) $((n_k, a_k)_{k=0}^{t'-1}, n_{t'})$ is a $t'$-history for any $t' \in \{1, \ldots , t-1\}$; (iii) $t \ge 2 \Rightarrow a_{t-1} \in \prod _{i \in I}A_i(a_{t-2}, n_{t-1})$, and $t=1 \Rightarrow a_{t-1} \in \prod _{i \in I}A_i(n_0)$; (iv) $n_t \in N(n_{t-1}, a_{t-1})$; and (v) the transition probability from $(n_{t-1}, a_{t-1})$ to $n_t$ is positive.

A feasible path is an infinite sequence $(n_0, a_0, (n_k, a_k)_{k=1}^\infty )$ in which $((n_k, a_k)_{k=0}^{t-1}, n_t)$ is a t-history for any $t=0, 1, 2, \ldots$.

Given any feasible path $(n_0, a_0, (n_k, a_k)_{k=1}^\infty )$, which contains the sequence $(n_k)_{k=1}^\infty$, namely $(w_k, (b_{i, k})_{i \in I})_{k=0}^\infty$, the ex post payoff to any player $i \in I$ is defined to be equal to

$$\begin{aligned} v\liminf _{t \rightarrow \infty }\varvec{1}_{w_t=i} -\sup _{t=1, 2, \ldots }b_{i, t}, \end{aligned}$$

(18)

where $\varvec{1}_{w_t=i}:=1$ if $w_t=i$, and $\varvec{1}_{w_t=i}=0$ if $w_t \ne i$.

Example 1 of a feasible path:

$$\begin{aligned}&\left( (0; 0, 0, 0), (1, 1, 0); (2; 0, \delta , 0), (0, 0, 1); (3; 0, \delta , 2\delta ), (0, 0, 0); \right. \\&\left. (3; 0, \delta , 2\delta ), (0, 0, 0); \cdots \right) , \end{aligned}$$

where $w_t=3$ for all $t=2$, namely, player 3 wins the prize in the third round of the game.

Example 2 of a feasible path:

$$\begin{aligned}&\left( (0; 0, 0, 0), (1, 1, 0); (2; 0, \delta , 0), (1, 0, 0); (1; 2\delta , \delta , 0), (0, 1, 0); \right. \\&\left. (2; 2\delta , 3\delta , 0), (1, 0, 0); (1; 4\delta , 3\delta , 0), \cdots \right) , \end{aligned}$$

where players 1 and 2 outbid each other, and hence take turn to be the frontrunner, in alternate order. Thus $\liminf _{t \rightarrow \infty } \varvec{1}_{w_i=i}=0$ for both players (and trivially so for player 3 as he never bids). According to (18), the payoff is equal to $-\infty$ for both players 1 and 2, and zero for player 3.

1.2 A.2 Formalization of the equilibrium concept

A strategy $\sigma _i$ for player i is a mapping that assigns, for any $t=0, 1, 2, \ldots$, to each t-history $((n_k, a_k)_{k=0}^{t-1}, n_t)$ a lottery on the set $A_i(n_{t-1}, a_{t-1})$ (or on the set $A_i(n_0)$ if $t=0$) of feasible actions for i. A strategy profile is a profile $(\sigma _i)_{i \in I}$ of strategies across all players. A subgame perfect equilibrium (SPE) is a strategy profile that satisfies sequential rationality at each t-history for every $t=0, 1, 2, \ldots$.

An SPE $(\sigma _i)_{i \in I}$ is said to be history independent iff for any $i \in I$, any $t, t' \in \{0, 1, 2, \ldots \}$, and any t- and $t'$-histories $((n_k, a_k)_{k=0}^{t-1}, n_t)$ and $((n'_k, a'_k)_{k=0}^{t'-1}, n'_{t'})$,

$$\begin{aligned} n_t=n'_{t'} \Longrightarrow \sigma _i((n_k, a_k)_{k=0}^{t-1}, n_t)=\sigma _i((n'_k, a'_k)_{k=0}^{t'-1}, n'_{t'}). \end{aligned}$$

Denote

$$\begin{aligned} N:=\{n_0\} \cup \{n_t \mid t=1, 2, \ldots ;\ ((n_k, a_k)_{k=0}^{t-1}, n_t) \text{ is } \text{ a } t{\text {-}}history\}. \end{aligned}$$

In any history-independent SPE, each player i’s strategy is function of only the last node of any t-history. Thus we write

$$\begin{aligned} \sigma _i(n_t):=\sigma _i((n_k, a_k)_{k=0}^{t-1}, n_t) \end{aligned}$$

for any $n_t \in N$ such that $((n_k, a_k)_{k=0}^{t-1}, n_t)$ is a t-history for some t. It follows that we can suppress the subscript t to write any element of N as $n=(w, (b_i)_{i \in I})$.

For any $n, n' \in N$ such that $n=(w, (b_i)_{i \in I})$ and $n'=(w', (b'_i)_{i \in I})$, we say that n is isomorphic to $n'$, or $n \equiv n'$, iff there exists a permutation $\psi : I \rightarrow I$ such that $b_i-b_j=b'_{\psi (i)}-b'_{\psi (j)}$ for all $i, j \in I$. This $\psi$ is called a permutation associated with $n \equiv n'$.

A history-independent SPE $(\sigma _i)_{i \in I}$ is said to be symmetric iff $\sigma _i(n)=\sigma _{\psi (i)}(n')$ for any $n \equiv n'$ with any associated permutation $\psi$.

For any $n \in N$ such that $n=(w, (b_i)_{i \in I})$, define $\phi _n: I \rightarrow \{\alpha , \beta , \gamma \}$ by

$$\begin{aligned} \phi _n(i):=\left\{ \begin{array}{ll} \alpha &{} \hbox { if}\ i=w\\ \gamma &{} \hbox { if}\ i \in \arg \min _{j \in I}b_j\\ \beta &{} \hbox { if}\ i \in I {\setminus } \left( \{w\} \cup \arg \min _{j \in I}b_j\right) . \end{array}\right. \end{aligned}$$

Examples: At the initial node $n_0$, since $w=0$, $\phi _{n_0}(i)=\gamma$ for all players i. At a period-1 node say $n_1:=(2; 0, \delta , 0)$, $\phi _{n_1}(2)=\alpha$ and $\phi _{n_1}(1)=\phi _{n_1}(3)=\gamma$. At any node $n_t$ with $t \ge 2$ such that $a_{t-1} \ne (0, 0, 0)$, $\phi _{n_t}$ is one-to-one.

For any $n, n' \in N$ such that $n \equiv n'$, there exists a permutation $\psi : I \rightarrow I$ such that

$$\begin{aligned} \phi _n(i)=\phi _{n'}\left( \psi (i)\right) \end{aligned}$$

for any $i \in I$. Since $n \equiv n'$, if $n=(w, (b_i)_{i \in I})$ and $n'=(w', (b'_i)_{i \in I})$ then (by the definitions of $\phi _n$ and $\equiv$) we have $b_i-b_j=b'_k-b'_l$ whenever $\phi _n(i)=\phi _{n'}(k)$ and $\phi _n(j)=\phi _{n'}(l)$. Thus $\psi$ is a permutation associated with $n \equiv n'$. Consequently, by the definition of symmetry,

$$\begin{aligned} \phi _n(i)=\phi _{n'}(j) \Rightarrow \sigma _i(n)=\sigma _j(n'). \end{aligned}$$

(19)

For any $n \in N$ such that $n=(w, (b_i)_{i \in I})$, denote

$$\begin{aligned} \tilde{s}(n):=\frac{1}{\delta }\left( \max _{i \in I} b_i-\min _{i \in I} b_i\right) . \end{aligned}$$

Claim: For any $n, n' \in N$, $n \equiv n' \iff \tilde{s}(n)=\tilde{s}(n')$. This follows from two observations, each proved inductively. First, for any $n \in N$ such that $n=(w, (b_i)_{i \in I})$ and any $i, j \in I$, $b_i-b_j=m\delta$ for some integer m. Second, for any $n \in N$ such that $n=(w, (b_i)_{i \in I})$,

$$\begin{aligned} \tilde{s}(n) \ne 0 \Longrightarrow \left[ w \in I \text{ and } b_w-\max _{j \in I {\setminus } \{w\}}b_j=\delta \right] . \end{aligned}$$

That is, the gap of committed payments between the frontrunner and any second-place bidder is always $\delta$. Thus, the only payment-gap that may be different between any two elements of N is the gap between the frontrunner and the lowest bidder.

The claim and (19) together imply that, for any symmetric history-independent SPE $(\sigma _i)_{i \in I}$,

$$\begin{aligned} \left[ \phi _n(i)=\phi _{n'}(j) \text{ and } \tilde{s}(n)=\tilde{s}(n') \right] \Longrightarrow \sigma _i(n)=\sigma _j(n'). \end{aligned}$$

Let $\sigma _i(n)(1)$ denote the probability of playing the action 1—to bid—at the node n by player i according to strategy $\sigma _i$. By the above-displayed fact, there is no loss of generality to represent a symmetric history-independent SPE by a list

$$\begin{aligned} \left( \pi _0, \pi _1, (\pi _{\beta , s}, \pi _{\gamma , s})_{s=1}^\infty \right) \end{aligned}$$

such that $\pi _0=\sigma _i(n_0)(1)$, $\pi _1=\sigma _i(n)(1)$ such that $\tilde{s}(n)=1$ for any $i \in I$ with $\phi _n(i) \ne \alpha$, and, for any $s=2, 3, \ldots$ and any $r \in \{\beta , \gamma \}$, $\pi _{r, s}=\sigma _i(n)(1)$ such that $\tilde{s}(n)=s$ and $\phi _n(i)=r$.

A symmetric history-independent SPE is said to be independent of non-participants iff (1) holds in the $\pi$-notation introduced above.

Our equilibrium concept is therefore defined to be any symmetric history-independent SPE that is independent of non-participants.

B Proofs

1.1 B.1 Lemma 1

Claim (i) has been proved in Ødegaard and Zheng (2020). Here we prove (ii), the uniqueness claim. Consider any equilibrium where the underdog stays put forever. Then only the follower’s strategy needs to be specified. By independence condition (1) of nonparticipants, the follower’s strategy is independent of the state variable, which measures merely the distance from the forever inactive underdog. Thus, the follower’s strategy is a constant probability $\pi$ of bidding. Let $V_*$ denote the expected payoff for the frontrunner in any such equilibrium. In the current round, he either wins the prize v with probability $\tilde{\sigma }_2$ or becomes the follower next round. In the latter case, by the symmetry condition in our equilibrium concept, he bids with probability $\pi$ thereby paying $2\delta$ to become the frontrunner again. Thus, the Bellman equation is

$$\begin{aligned} V_*=\left( 1-\pi \right) v+\pi ^2 \left( -2\delta +V_*\right) . \end{aligned}$$

Note that $\pi >0$, otherwise ($\pi =0$), $V_*=v$; with $v>2\delta$ by assumption, the follower would bid for sure, contradicting the supposition $\pi =0$. Also note that $\pi <1$, otherwise the Bellman equation implies $V_*=-2\delta +V_*$, contradiction. Now that $0<\pi <1$, the follower is indifferent about bidding, hence $V_*=2\delta$. Plug this into the Bellman equation to obtain $\pi =1-2\delta /v$. That proves Claim (ii). $\square$

1.2 B.2 Lemma 2

Lemma 2

By definition of $L_s$, the equilibrium expected payoff for an underdog whose lag from the frontrunner is s, we know that $L_s=0$ for all $s \ge v/\delta$. Starting from any such s and use backward induction towards smaller s, together with the law of motion (8) and the fact $V_2-(s+1)\delta <0$ for all $s \ge s_*$ due to the definition of $s_*$, we observe that $L_s=0$ for all $s \ge s_*$. At any state $s \ge s_*$, by (8), an underdog gets zero expected payoff if he does not bid; if he bids then by Eq. (2) there is a positive probability with which he gets a negative payoff $V_2-(s+1)\delta$; hence his best response is uniquely to not bid at all. Hence

$$\begin{aligned} s \ge s_*&\Longrightarrow&L_s=0 \text{ and } \pi _{\gamma , s}=q_{\gamma , s}=0, \end{aligned}$$

(20)

which proves Claim (i) of the lemma. Apply backward induction to (8) starting from $s=s_*$ and we obtain

$$\begin{aligned} 2 \le s \le s_*-1 \Longrightarrow V_2-(s+1)\delta \ge L_s \ge L_{s+1} \ge 0, \end{aligned}$$

(21)

with the inequality $L_s \ge L_{s+1}$ being strict whenever $s<s_*-1$. Thus, for any $s<s_*-1$, $V_s-(s+1)\delta >L_{s+1} \ge 0$; hence Eqs. (2) and (8) together imply that an underdog’s best response is uniquely to bid for sure:

$$\begin{aligned} 2 \le s < s_*-1&\Longrightarrow&L_s>0 \text{ and } \pi _{\gamma , s}=1, \end{aligned}$$

(22)

which proves Claim (ii) of the lemma. $\square$

1.3 B.3 Claims (i) and (iv) of Theorem 1

First, we make several observations first. By (8) and (21), $L_2$ is a convex combination between $L_3$ and $V_2-3\delta$, with $V_2-3\delta \ge L_3$ when $s_* \ge 3$. Thus,

$$\begin{aligned} s_* \ge 3 \Longrightarrow L_2 \le V_2-3\delta . \end{aligned}$$

(23)

Equation (10), combined with (7) and (8), implies

$$\begin{aligned} M_{s_*-1}=q_{\gamma , s_*-1}L_2 {\mathop {\le }\limits ^{(23)}} \left( V_2-3\delta \right) ^+. \end{aligned}$$

(24)

Lemma 4

$s_* \ne 3$

Proof

Suppose, to the contrary, that $s_*=3$. Hence $0 \le V_2-3\delta <\delta$. Thus, by Eq. (24), $M_2<\delta$. Then (4) requires that $\pi _1<1$, otherwise $V_1=M_2<\delta$, implying a contradiction that no one would bear the sunk cost $\delta$ to become the initial $\alpha$ player. Now consider the decision of any non-$\alpha$ player at the state $s=1$, as depicted by (5). Since $V_2-2\delta >V_2-3\delta \ge L_2$, with the second inequality due to (23), each non-$\alpha$ player at $s=1$ would maximize the probability of becoming the $\alpha$ in the next round, i.e., $\pi _1=1$, contradiction. $\square$

Lemma 5

If $s_* \ge 4$ then $V_3-2\delta \ge M_2 \ge L_2>0$.

Proof

Suppose that $V_3-2\delta < L_2$. Then, by the fact $\pi _{\gamma , 2}=1$ (Lemma 2.ii and $s_* \ge 4$) and Eq. (2), the $\beta$ player at state $s=2$ would rather stay put than bid, hence $\pi _{\beta , 2}=0$. This, combined with (6) in the case $s=2$ and the fact $\pi _{\gamma , 2}=1$, implies that $V_2=M_2$. Since $V_3-2\delta <L_2$ coupled with (7) implies $M_2 \le L_2$, we have a contradiction $V_2 \le L_2 < V_2$, with the last inequality due to (8). Thus we have proved $V_3-2\delta \ge L_2$. Therefore, with $M_2$ a convex combination between $V_3-2\delta$ and $L_2$ (since $\pi _{\gamma , 2}=1$), $V_3-2\delta \ge M_2 \ge L_2$. Finally, to show $L_2>0$, note from the hypothesis $s_* \ge 4$ and definition of $s_*$ that $V_2-3\delta >0$. This positive payoff the underdog at state $s=2$ can secure with a positive probability through bidding. Hence $L_2>0$ follows from (8). $\square$

Lemma 6

If $s_* \ge 4$ then $\pi _{\gamma , s_*-1}>0$.

Proof

Section 4.1. $\square$

Proof of Theorem 1.i (impossibility of odd dropout states)

Suppose, to the contrary, that the dropout state $s_*$ of an equilibrium is an odd number. Since $s_* \ge 3$ by hypothesis of the lemma, $s_* \ge 5$. By Lemma 5, $L_2 \le M_2 \le V_3-2\delta$. Apply (6) to the case $s=s_*-2$ and use the fact that $\pi _{\gamma , s_*-2}=1$ (thereby ruling out $V_{s_*-2} \rightarrow v$) due to Lemma 2.ii. Then we have $M_{s_*-1} \le L_2$, due to (24), and hence $M_{s_*-1} \le M_2 \le V_3-2\delta$. Reason backward along the transition chain (13), starting from $M_{s_*-1} \le V_3-2\delta$ and using the fact $L_2 \le M_2 \le V_3-2\delta$ (Lemma 5), to obtain

$$\begin{aligned}M_{s_*-1} \le V_3-2\delta \Rightarrow & V_{s_*-2} \le \max \{V_3-2\delta , M_2\} \le V_3-2\delta \\ \quad \Rightarrow & M_{s_*-3} \le \max \{-2\delta +V_{s_*-2}, L_2\} \le \max \{V_3-2\delta , L_2\} \le V_3-2\delta \\ \quad \Rightarrow & V_{s_*-4} \le \max \{M_{s_*-3}, M_2\} \le V_3-2\delta \\ \Rightarrow & \cdots \end{aligned}$$

Since $s_*$ is an odd number and $s_*\ge 5$, this chain eventually reaches $V_3$, i.e., $3=s_*-2m$ for some positive integer m. Hence we obtain the contradiction $V_3 \le V_3-2\delta$. $\square$

Proof of Theorem 1.iv

That the $\beta$ player stays put for sure at state $s_*-1$ has been proved in Sect. 4.2, where the hypothesis $s_* \ge 4$ is true due to Lemma 4. For the rest of the claim, we first prove $0<\pi _{\gamma , s_*-1} <1$. The first inequality follows from Lemma 6 since $s_* \ge 4$. To prove $\pi _{\gamma , s_*-1} <1$, suppose to the contrary that $\pi _{\gamma , s_*-1}=1$. Then by the fact $\pi _{\beta , s_*-1}=0$ and (6) applied to the case $s=s_*-1$, we have $V_{s_*-1}=M_2$ and $M_{s_*-1}=L_2$. Reason backward along the transition chains (12) and (13), starting from $V_{s_*-1}=M_2$ and $M_{s_*-1}=L_2$ and using the fact $L_2 \le M_2$ (Lemma 5). Then we have two chains of inequalities:

$$\begin{aligned} V_{s_*-1}=M_2\Rightarrow & {} M_{s_*-2} \le \max \{-2\delta +M_2, L_2\} \le M_2 \\\Rightarrow & {} V_{s_*-3} \le \max \{M_{s_*-2}, M_2\} \le M_2 \\\Rightarrow & {} M_{s_*-4} \le \max \{-2\delta +V_{s_*-3}, L_2\} \le \max \{M_2, L_2\} \le M_2 \ \Rightarrow \cdots \\ M_{s_*-1}=L_2\Rightarrow & {} V_{s_*-2} \le \max \{L_2, M_2\} \le M_2 \\\Rightarrow & {} M_{s_*-3} \le \max \{-2\delta +V_{s_*-2}, L_2\} \le \max \{M_2, L_2\} \le M_2 \\\Rightarrow & {} V_{s_*-4} \le \max \{M_{s_*-3}, M_2\} \le M_2 \ \Rightarrow \cdots \end{aligned}$$

The two chains combined lead to $V_s \le M_2$ for all $s \le s_*-1$. Hence $V_3 \le M_2$, contradicting Lemma 5. Thus we have proved that $\pi _{\gamma , s_*-1}<1$.

With $\pi _{\gamma , s_*-1}<1$, bidding is not the unique best response for the $\gamma$ player at state $s_*-1$, hence $V_2 \le s_*\delta$ (otherwise the bottom branch of (8) in the case $s=s_*-1$ is strictly positive and, by (20), is strictly larger than the middle branch, so the $\gamma$ player would strictly prefer to bid). By definition of $s_*$, $V_2 \ge s_*\delta$. Thus $V_2=s_*\delta$. $\square$

1.4 B.4 Claims (ii) and (iii) of Theorem 1 (action of $\beta$ when $s\le s_*-2$)

The part of Theorem 1.iii on the $\gamma$ player follows from Lemma 2.ii. We prove the part on the $\beta$ player here, through proving Lemmas 8 and 10. The former shows that bidding is a follower’s unique best response at even-number states, and the latter, odd-number states. Lemma 10 uses a Lemma 9, which also implies Claim (ii) of the theorem.

Lemma 7

If the dropout state is an even number $s_* \ge 4$, then $L_2<V_2 \le V_3-2\delta$.

Proof

Since $\pi _{\beta , s_*-1}=0$ (Theorem 1.iv), $M_{s_*-1} \le L_2$. Thus, since $\pi _{\gamma , s_*-2}=1$ (Lemma 2.ii), $V_{s_*-2}$ is a convex combination between $M_{s_*-1}$, which is less than $L_2$, and $M_2$, which is a convex combination between $V_3-2\delta$ and $L_2$, as $\pi _{\gamma , 2}=1$. Thus $V_{s_*-2}$ is between $L_2$ and $V_3-2\delta$. Consequently, $M_{s_*-3}$, a convex combination between $L_2$ and $V_{s_*-2}-2\delta$ (since $\pi _{\gamma , s_*-3}=1$), is between $L_2$ and $V_3-2\delta$. Repeating this reasoning, with $s_*$ being an even number, we eventually reach $2=s_*-2m$ for some integer $m \ge 1$, and obtain the fact that $V_2$ is a number between $L_2$ and $V_3-2\delta$. Thus, $L_2<V_3-2\delta$, otherwise the fact $L_2<V_2$ by (8) would be contradicted. Hence $L_2<V_2 \le V_3-2\delta$. $\square$

1.4.1 B.4.1 Bidding at even states

Lemma 8

If the dropout state is an even number $s_* \ge 4$, then $\pi _{\beta , s}=1$ for any even number s such that $2 \le s \le s_*-2$.

Proof

First, by Lemma 7, $L_2<V_3-2\delta$. Thus at state $s=2$ the $\beta$ player strictly prefers to bid, i.e., $\pi _{\beta , 2}=1$. Second, pick any even number s such that $4 \le s \le s_*-2$ and suppose, to the contrary of the lemma, that $\pi _{\beta , s}<1$, which means that the $\beta$ player at state s does not strictly prefer to bid. Thus $M_s \le L_2$ (as the transition $M_s \rightarrow 0$ is ruled out by the fact $\pi _{\gamma , s}=1$). Consequently, $V_{s-1}$, a convex combination between $M_s$ and $M_2$, is weakly less than $M_2$, as $L_2 \le M_2$ by Lemma 5. Furthermore, $M_{s-2}$, a convex combination between $V_{s-1}-2\delta$ and $L_2$, is less than $M_2$, and that in turns implies $V_{s-3} \le M_2$. Repeating this reasoning, with s an even number, we eventually obtain the conclusion that $V_3 \le M_2$, which contradicts Lemma 5. Thus, $\pi _{\beta , s}=1$. $\square$

If the dropout state is an even number $s_* \ge 4$, since $\pi _{\gamma , s}=1$ for all $s \le s_*-2$ (Lemma 2.ii), Eq. (2) and the equal-probability tie-breaking rule together imply

$$\begin{aligned} \forall s \in \{2, 3, 4, \ldots , s_*-2\}: \left[ \pi _{\beta , s}=1 \Longrightarrow q_{\beta , s}=q_{\gamma , s}=1/2 \right] . \end{aligned}$$

(25)

By Lemma 8,

$$\begin{aligned} 2 \le s \le s_*-2 \text{ and } s \text{ is even } \Longrightarrow q_{\beta , s}=q_{\gamma , s}=1/2. \end{aligned}$$

(26)

1.4.2 B.4.2 Bidding at odd states

In the following, we extend the summation notation by defining, for any sequence $(a_k)_{k=1}^\infty$,

$$\begin{aligned} i>j \Longrightarrow \sum _{k=i}^j a_k := 0. \end{aligned}$$

(27)

In particular, $\sum _{k=1}^0 a_k=0$ according to this notation.

Lemma 9

If the dropout state is an even number $s_* \ge 4$, $M_2>V_2+\delta /2$.

Proof

Let $m:=\min \{k \in \{0, 1, 2 \ldots \}: V_{2k+4}-2\delta \le L_2\}$. Note that m is well-defined because $s_*/2-2$ belongs to the set, as $V_{s_*}-2\delta =0 \le L_2$ (Eq. (10)). At any odd state $2k+1 \le 2m+1$ (hence $k-1<m$) we have $V_{2k+2}-2\delta = V_{2(k-1)+4)}-2\delta > L_2$, with the last inequality due to the definition of m; hence by (6) in the state $s=2k+1$ the $\beta$ player bids for sure, i.e. $\pi _{\beta , 2k+1}=1$. Thus, (25) implies that $q_{\beta , s}=q_{\gamma , s}=1/2$ at any such odd state. Coupled with (26), that means the transition at every state s from 2 to $2m+2$ is that the current $\beta$ and $\gamma$ players each have probability 1/2 to become the next $\alpha$ player. Thus,

$$\begin{aligned} V_2=M_2\sum _{k=0}^m 2^{-2k-1}+L_2\left( \sum _{k=0}^m 2^{-2k-2}+2^{-2m-2}z_m \right) -2\delta \sum _{k=1}^{m}2^{-2k}, \end{aligned}$$

(28)

where $z_m := 1$ if $2m+2<s_*-2$, and $z_m:=2\pi _{\gamma , s_*-1}-1$ if $2m+2=s_*-2$; and the last series $\sum _{k=1}^m$ on the right-hand side uses the summation notation defined in (27) when $m=0$.

To understand the term for $M_2$ on the right-hand side, note that $M_2$ enters the calculation of $V_2$ at the even states $s=2, 4, 6, \ldots , 2m-2$, and upon entry at state s and in every round transversing from states s to 2, the $M_2$ is discounted by the transition probability 1/2. The term for $L_2$ is similar, except that $L_2$ enters at the odd states $s=3, 5, 7, \ldots , 2m-1$, and that the transition probability for the $L_2$ at the last state $2m-1$ is equal to one if $2m-1<s_*-1$, and equal to $\pi _{\gamma , s_*-1}$ if $2m-1=s_*-1$. That is why the last two terms within the bracket for $L_2$ are

$$\begin{aligned} 2^{-2m-2}+2^{-2m-2}z_m=\left\{ \begin{array}{ll} 2^{-2m-2}+2^{-2m-2}=2^{-2m-1} &{} \hbox { if}\ z_m=1\\ 2^{-2m-2}+2^{-2m-2}\left( 2\pi _{\gamma , s_*-1}-1\right) =2^{-2m-1}\pi _{\gamma , s_*-1} &{} \hbox { if}\ z_m=2\pi _{\gamma , s_*-1}-1. \end{array}\right. \end{aligned}$$

The term for $-2\delta$ is analogous to that for $M_2$.

With $s_* \ge 4$, $V_2-4\delta \ge 0$. Thus, by the above-calculated transition probabilities,

$$\begin{aligned} L_2=\frac{1}{2}(L_3+V_2-3\delta ) \le \frac{1}{2}(V_2-4\delta +V_2-3\delta )=V_2-\frac{7}{2}\delta . \end{aligned}$$

This, combined with Eq. (28) and the fact $z_m \le 1$ due to its definition, implies that

$$\begin{aligned} V_2\le & {} M_2\sum _{k=0}^m 2^{-2k-1} +\left( V_2-\frac{7}{2}\delta \right) \left( \sum _{k=0}^m 2^{-2k-2}+2^{-2m-2} \right) -2\delta \sum _{k=1}^{m}2^{-2k} \\< & {} M_2\sum _{k=0}^m 2^{-2k-1} +V_2\left( \sum _{k=0}^m 2^{-2k-2}+2^{-2m-2} \right) -\frac{7}{8}\delta . \end{aligned}$$

Thus, the lemma is proved if

$$\begin{aligned} 1-\left( \sum _{k=0}^m 2^{-2k-2}+2^{-2m-2} \right) =\sum _{k=0}^m 2^{-2k-1}, \end{aligned}$$

(29)

as the left-hand side of this equation is clearly strictly between zero and one. To prove (29), we use induction on m. When $m=0$, (29) becomes $1-2^{-2}-2^{-2}=2^{-1}$, which is true. For any $m=0, 1, 2, \ldots$, suppose that (29) is true. We shall prove that the equation is true when m is replaced by $m+1$, i.e.,

$$\begin{aligned} 1-\left( \sum _{k=0}^{m+1} 2^{-2k-2}+2^{-2(m+1)-2} \right) =\sum _{k=0}^{m+1} 2^{-2k-1}. \end{aligned}$$

(30)

The left-hand side of (30) is equal to

$$\begin{aligned}&1-\left( \sum _{k=0}^{m} 2^{-2k-2}+2^{-2m-2} \right) +2^{-2m-2}-2^{-2(m+1)-2}-2^{-2(m+1)-2} \\&\quad = \sum _{k=0}^{m} 2^{-2k-1} +2^{-2m-2}-2^{-2(m+1)-1} \quad \text{(the } \text{ induction } \text{ hypothesis) } \\&\quad = \sum _{k=0}^{m} 2^{-2k-1} +2^{-2m-3}, \end{aligned}$$

which is equal to the right-hand side of (30). Thus (29) is true in general, as desired. $\square$

Lemma 10

If the dropout state is an even number $s_* \ge 4$, then $\pi _{\beta , s}=1$ at any odd number state s such that $1 \le s \le s_*-2$.

Proof

Pick any odd number s such that $s \le s_*-2$. It suffices to prove that $V_{s+1}-2\delta > L_2$. Since $s+1$ is even, it follows from (26) that

$$\begin{aligned} V_{s+1}=\frac{1}{2}\left( M_2+M_{s+2}\right) \ge \frac{1}{2}\left( M_2+L_2\right) , \end{aligned}$$

with the inequality due to the fact $M_{s+2} \ge L_2$, which in turn is due to the fact that the $\beta$ player at state $s+2$ can always secure the payoff $L_2$ through not bidding at all. Thus,

$$\begin{aligned} V_{s+1}-2\delta -L_2\ge & {} \frac{1}{2}\left( M_2+L_2\right) -2\delta -L_2 \\= & {} \frac{1}{2}M_2-\frac{1}{2}L_2-2\delta \\= & {} \frac{1}{2}M_2-\frac{1}{2}\left( \frac{1}{2}L_3+\frac{1}{2}(V_2-3\delta ) \right) -2\delta \\\ge & {} \frac{1}{2}M_2-\frac{1}{2}\left( \frac{1}{2}(V_2-4\delta )+\frac{1}{2}(V_2-3\delta ) \right) -2\delta \\= & {} \frac{1}{2}M_2-\frac{1}{2}V_2-\frac{1}{4}\delta , \end{aligned}$$

with the second inequality due to the definition of $L_s$ and the fact $V_2-4\delta \ge 0$ ($s_* \ge 4$). Since $\frac{1}{2}M_2-\frac{1}{2}V_2-\frac{1}{4}\delta >0$ by Lemma 9, $V_{s+1}-2\delta -L_2 >0$, as desired. $\square$

Proof of Claims (ii) and (iii) of Theorem 1

Claim (iii) of the theorem follows directly from combining Lemmas 2.ii, 8 and 10. With Claim (iii) established, every non-$\alpha$ player in the state $s=1$ (i.e., in the second round) bids for sure, hence (4) implies that the transition $V_1 \rightarrow M_2$ happens for sure. Thus, the payoff from becoming the initial frontrunner is equal to $M_2-\delta$. By contrast, in failing to become the initial frontrunner a player gets the payoff $M_1$, which by (5) and Claim (iii) of the theorem is equal to $\frac{1}{2}\left( V_2-2\delta +L_2\right)$. Thus, the net gain from becoming the initial frontrunner is equal to

$$\begin{aligned}M_2-\delta -\frac{1}{2}(V_2-2\delta +L_2)>& V_2+\delta /2-\delta -\frac{1}{2}(V_2-2\delta +L_2) \\\quad =& \frac{1}{2}\left( V_2-L_2+\delta \right) >0, \end{aligned}$$

with the first “>” due to Lemma 9, and the last “>” due to (23). Thus, every player at the initial round strictly prefers being the frontrunner. Hence Claim (ii) of the theorem follows.

1.5 B.5 Lemma 3

All lemmas in this subsection assume the hypotheses in Lemma 3, that $s_* \ge 4$ is an even number and a strategy profile $\left( \pi _0, \pi _1, \left( \pi _{\beta , s}, \pi _{\gamma , s}\right) _{s=2}^\infty \right)$ according to Theorem 1 is given, with the associated value functions $\left( V_s, M_s, L_s\right) _s$ derived from (6)–(8) and Eq. (14).

Lemma 11

For any positive integer m such that $2m+1 \le s_*-1$, if $V_{2m+1}-2\delta \le L_2$ then $V_3-2\delta < L_2$.

Proof

Pick any m specified by the hypothesis such that $V_{2m+1}-2\delta \le L_2$. Suppose, to the contrary of the lemma, that $V_3-2\delta \ge L_2$. Thus, the law of motion (6) in the case $s=2$, with $\pi _{\gamma , 2}=1$, implies that $M_2$ is between $L_2$ and $V_3-2\delta$, hence $V_3-2\delta \ge M_2 \ge L_2$. By the law of motion (7) in the case $s=2m$, $M_{2m}$ is a convex combination among zero, $V_{2m+1}-2\delta$ and $L_2$. Thus the hypothesis implies that $M_{2m} \le L_2$. Consequently, the law of motion (6) in the case $s=2m-1$, together with $\pi _{\gamma , 2m-1}=1$ and $M_2 \ge L_2$, implies that $V_{2m-1} \le M_2$ and hence $V_{2m-1}-2\delta \le M_2-2\delta$. Then (7) in the case $s=2m-2$ implies $M_{2m-2} \le L_2$. Repeating this reasoning backward, with 3 being odd, we eventually reach state $s=3$ and obtain $V_3 \le M_2$. But since $V_3-2\delta \ge M_2$, we have a contradiction $V_3-2\delta \ge M_2 \ge V_3$. $\square$

Lemma 12

Denote $x:=\pi _{\gamma , s_*-1}$. For any integer m such that $1 \le m \le s_*/2-1$,

$$\begin{aligned} M_{s_*-(2m-1)}= & {} -\delta \sum _{k=1}^{m-1}2^{-2k+2}+M_2\sum _{k=1}^{m-1}2^{-2k} +L_2\left( \sum _{k=1}^{m-1}2^{-2k+1}+2^{-2(m-1)}x \right) , \end{aligned}$$

(31)

$$\begin{aligned} V_{s_*-2m}= & {} -\delta \sum _{k=1}^{m-1}2^{-2k+1}+M_2\sum _{k=1}^m 2^{-2k+1}+L_2\left( \sum _{k=1}^{m-1}2^{-2k}+2^{-2m+1}x\right) , \end{aligned}$$

(32)

$$\begin{aligned} V_{s_*-(2m-1)}= & {} -\delta \sum _{k=1}^{m-1}2^{-2k+1}+2^{-2(m-1)}(1-x)v+L_2\sum _{k=1}^{m-1}2^{-2k} \\ \nonumber&+M_2\left( \sum _{k=1}^{m-1}2^{-2k+1}+2^{-2(m-1)}x \right) , \end{aligned}$$

(33)

$$\begin{aligned} M_{s_*-2m}= & {} -\delta \sum _{k=0}^{m-1}2^{-2k}+2^{-2m+1}(1-x)v+L_2\sum _{k=1}^m 2^{-2k+1} \\ \nonumber&+M_2\left( \sum _{k=1}^{m-1}2^{-2k}+2^{-2m+1}x\right) , \end{aligned}$$

(34)

$$\begin{aligned} L_2= & {} \delta \left( s_*-4+2^{-s_*+3}\right) . \end{aligned}$$

(35)

Proof

First, we prove Eqs. (31) and (32). When $m=1$, Eq. (31), coupled with the summation notation defined in (27), becomes $M_{s_*-1}=xL_2=\pi _{\gamma , s_*-1}L_2$, which follows from (7) and the fact that $V_s=2\delta$ and $M_s=0$ for all $s \ge s_*$, due to Theorem 1. This coupled with Eq. (14) implies that

$$\begin{aligned} V_{s_*-2}=(M_{s_*-1}+M_2)/2=M_2/2+xL_2/2, \end{aligned}$$

which is Eq. (32) when $m=1$ (using again the summation notation in (27)). Suppose, for any integer $m'$ with $1 \le m' \le s_*/2-2$, that Eqs. (31) and (32) are true with $m=m'$. By the induction hypothesis of (32) and Eq. (14),

$$\begin{aligned}&M_{s_*-(2m'+1)} = \frac{1}{2}\left( V_{s_*-2m'}-2\delta +L_2\right) \\&\quad = -\delta \left( 1+\frac{1}{2}\sum _{k=1}^{m'-1}2^{-2k+1}\right) +\frac{M_2}{2}\sum _{k=1}^{m'}2^{-2k+1}\\&\qquad +\frac{L_2}{2}\left( 1+\sum _{k=1}^{m'-1}2^{-2k}+2^{-2m'+1}x\right) , \end{aligned}$$

which is Eq. (31) when $m=m'+1$. By the above calculation of $M_{s_*-(2m'+1)}$ and Eq. (14),

$$\begin{aligned} V_{s_*-(2m'+2)}= & {} \frac{1}{2}\left( M_{s_*-(2m'+1)}+M_2\right) \\= & {} -\frac{\delta }{2}\left( 1+\frac{1}{2}\sum _{k=1}^{m'-1}2^{-2k+1}\right) +\frac{M_2}{2}\left( 1+\sum _{k=1}^{m'}2^{-2k+1}\right) \\&+\frac{L_2}{4}\left( 1+\sum _{k=1}^{m'-1}2^{-2k}+2^{-2m'+1}x\right) , \end{aligned}$$

which is Eq. (32) in the case $m=m'+1$. Thus Eqs. (31) and (32) are proved.

Next we prove Eqs. (33) and (34). When $m=1$, Eq. (33), coupled with the notation $\sum _{k=1}^0a_k=0$, becomes $V_{s_*-1}=(1-x)v+xM_2$, which is true by definition of x and the fact $\pi _{\beta , s_*-1}=0$ (Theorem 1). Then by Eq. (14)

$$\begin{aligned} M_{s_*-2}=\left( V_{s_*-1}-2\delta +L_2\right) /2 =\left( (1-x)v+xM_2-2\delta +L_2\right) /2, \end{aligned}$$

which is Eq. (34) when $m=1$ (again using the notation $\sum _{k=1}^0a_k=0$). Suppose, for any integer $m'$ with $1 \le m' \le s_*/2-2$, that Eqs. (33) and (34) are true with $m=m'$. By the induction hypothesis and Eq. (14),

$$\begin{aligned} V_{s_*-(2m'+1)}= & {} \frac{1}{2}\left( M_{s_*-2m'}+M_2\right) \\= & {} -\frac{\delta }{2}\sum _{k=0}^{m'-1}2^{-2k}+2^{-1}2^{-2m'+1}(1-x)v+\frac{L_2}{2}\sum _{k=1}^{m'} 2^{-2k+1} \\ \nonumber&+M_2\left( 2^{-1}+2^{-1}\sum _{k=1}^{m'-1}2^{-2k}+2^{-1}2^{-2m'+1}x\right) , \end{aligned}$$

which is Eq. (33) in the case $m=m'+1$. By the above calculation and Eq. (14),

$$\begin{aligned} M_{s_*-(2m'+2)}= & {} \frac{1}{2}\left( V_{s_*-(2m'+1)}-2\delta +L_2\right) \\= & {} -\delta \left( 1+\frac{1}{2}\sum _{k=1}^{m'}2^{-2k+1}\right) +2^{-1}2^{-2m'}(1-x)v \\&+L_2\left( \frac{1}{2}+2^{-1}\sum _{k=1}^{m'}2^{-2k}\right) +\frac{M_2}{2}\left( \sum _{k=1}^{m'}2^{-2k+1}+2^{-2m'}x \right) , \end{aligned}$$

which is Eq. (34) in the case $m=m'+1$. Hence Eqs. (33) and (34) are proved.

Finally we prove Eq. (35). Applying Eq. (14) to (8) recursively we obtain, for any integer $s_* \ge 4$, that

$$\begin{aligned} L_2= & {} \frac{1}{2}\left( V_2-3\delta +\frac{1}{2}\left( V_2-4\delta +\frac{1}{2}\left( \cdots +\frac{1}{2}\left( V_2-(s_*-1)\delta \right) \right) \right) \right) \\= & {} \frac{\delta }{2}\left( s_*-3+\frac{1}{2}\left( s_*-4+\frac{1}{2}\left( \cdots +\frac{1}{2}\cdot 1\right) \right) \right) \\= & {} \delta \left( \frac{1}{2}(s_*-3)+\frac{1}{2^2}(s_*-4)+\frac{1}{2^3}(s_*-5)+\dots +\frac{1}{2^{s_*-3}} \right) , \end{aligned}$$

which is equal to the right-hand side of (35). In the above multiline calculation, the first and second lines are due to $V_2=s_*\delta$ (Theorem 1.iv). $\square$

Lemma 13

$V_{s_*-2}-2\delta \ge L_2 \Longrightarrow \forall m \in \{1, \ldots , s_*/2-1\}: V_{s_*-2m}-2\delta \ge L_2$.

Proof

By the law of motion and Eq. (14), Eqs. (31), (32), (33), (34) and (35) hold. Denote

$$\begin{aligned} \mu (m):= & {} 2^{-2m+1}, \\ \mu _*:= & {} 2^{-s_*+3}. \end{aligned}$$

With the fact $\sum _{k=1}^{m-1}2^{-2k}=(1-2^{-2m+2})/3$, Eq. (32) becomes

$$\begin{aligned}V_{s_*-2m} = & -\delta \cdot \frac{2}{3}(1-2\mu (m))+M_2\left( \frac{2}{3}(1-2\mu (m))+\mu (m)\right) \\& +L_2\left( \frac{1}{3}(1-2\mu (m))+\mu (m)x \right) . \end{aligned}$$

Hence

$$\begin{aligned} V_{s_*-2m}-2\delta -L_2= & {} -\delta \left( \frac{2}{3}(1-2\mu (m))+2\right) +M_2\left( \frac{2}{3}(1-2\mu (m))+\mu (m)\right) \\&-L_2\left( 1-\frac{1}{3}(1-2\mu (m))-\mu (m)x \right) \\= & {} -\frac{4}{3}(2-\mu (m))\delta +\frac{1}{3}(2-\mu (m))M_2 \\&-\left( s_*-4+\mu _*\right) \delta \left( \frac{2}{3}(1+\mu (m))-\mu (m)x\right) , \end{aligned}$$

with the second equality due to (35). Thus, $V_{s_*-2m}-2\delta \ge L_2$ is equivalent to

$$\begin{aligned} \frac{1}{3}(2-\mu (m))M_2 \ge \delta \left( \frac{4}{3}(2-\mu (m)) +\left( s_*-4+\mu _*\right) \left( \frac{2}{3}(1+\mu (m))-\mu (m)x\right) \right) , \end{aligned}$$

i.e.,

$$\begin{aligned} \frac{M_2}{\delta } \ge 4+\frac{2(1+\mu (m))-3\mu (m)x}{2-\mu (m)}(s_*-4+\mu _*). \end{aligned}$$

(36)

Since $s_*-4 \ge 0$ by hypothesis, and

$$\begin{aligned}&\frac{d}{d\mu (m)}\left( \frac{2(1+\mu (m))-3\mu (m)x}{2-\mu (m)}\right) \\&\quad = \frac{(2-\mu (m))(2-3x)+2(1+\mu (m))-3\mu (m)x}{(2-\mu (m))^2} \\&\quad = \frac{6(1-x)}{(2-\mu (m))^2} \ \ge \ 0, \end{aligned}$$

the right-hand side of (36) is weakly increasing in $\mu (m)$, which in turn is strictly decreasing in m. Thus the right-hand side of (36) is weakly decreasing in m. Consequently, $V_{s_*-2m}-2\delta -L_2 \ge 0$ is satisfied for all m if the inequality holds at the minimum $m=1$, i.e., if $V_{s_*-2}-2\delta -L_2 \ge 0$, as claimed. $\square$

Proof of Lemma 3

Let $s \in \{1, 2, \ldots , s_*-2\}$. If s is even and $V_3-2\delta \ge L_2$, then Lemma 11 implies $V_{s+1}-2\delta >L_2$; thus, by (7) and by the fact that $\pi _{\gamma , s}=1$ due to the strategy profile specified in Theorem 1, the $\beta$ player at s gets $L_2$ if he does not bid, and $\frac{1}{2}(V_{s+1}-2\delta )+\frac{1}{2}L_2$ if he does. Hence bidding is the unique best response for $\beta$ at s. If s is odd and $V_{s_*-2}-2\delta \ge L_2$, then Lemma 13 implies that $V_{s+1}-2\delta \ge L_2$; thus, by the same token as in the previous case, the $\beta$ player at s weakly prefers to bid. $\square$

1.6 B.6 Theorem 2

Lemma 14

For any even number $s_* \ge 4$, if Eqs. (14) and (35) hold and $M_2 \ge V_2=s_*\delta$, then at the initial and second rounds each player strictly prefers to bid.

Proof

First, consider the second round, which means $s=1$. For each non-$\alpha$ player, becoming the next $\alpha$ player gives him an expected payoff $V_2-2\delta =(s_*-2)\delta$ by the hypothesis $V_2=s_*\delta$, whereas staying put gives payoff $L_2$, which is less than $(s_*-3)\delta$ by Eq. (35). Thus, each non-$\alpha$ player strictly prefers to bid at state one, hence $s=2$ occurs for sure given $s=1$. Second, consider the initial state. Based on the analysis of the previous step (from $s=1$ to $s=2$), becoming the first $\alpha$ yields the expected payoff $-\delta +M_2$, whereas staying put yields $\frac{1}{2}(V_2-2\delta +L_2)$. Since $M_2 \ge V_2$ by hypothesis and $V_2-2\delta >L_2$ by the previous analysis, each player strictly prefers to become the first $\alpha$ player. $\square$

Lemma 15

Any integer $s_* \ge 3$ constitutes an equilibrium if $s_*$ is an even number and there exists $(M_2, x, L_2) \in \mathbb {R}_+^3$ such that—

• a. $(M_2, x, L_2) \in \mathbb {R}_+ \times [0, 1] \times \mathbb {R}_+$ and it solves simultaneously Eq. (32) in the case $m=s_*/2-1$, Eq. (34) in the case $m=s_*/2-1$ such that $V_2=s_*\delta$, and Eq. (35);

• b. $M_2 \ge s_*\delta$ (which is equivalent to $V_3-2\delta \ge L_2$, cf. (12));

• c. Ineq. (36) is satisfied in the case $m=1$.

Proof

Pick any even number $s_* \ge 4$ and assume Conditions (a)–(c). Consider the strategy profile such that everyone bids in the initial round and, in any future round, acts according to the strategy profile specified in Theorem 1. This strategy profile implies Eq. (14), which allows calculation of the value functions $(V_s, M_s, L_s)_{s=2}^{s_*}$ via the law of motions. By Conditions (a) and (b), $M_2 \ge V_2=s_*\delta$, hence Lemma 14 implies that bidding at the initial round is a best response for each player, and bidding at second rounds a best response for each non-$\alpha$ player. The incentive for each player to abide by the strategy profile at any state $s \ge s_*$ is the same as in the surplus-dissipating subgame equilibrium. At the state $s_*-1$, bidding with probability x is a best response for the $\gamma$ player because he is indifferent about bidding, since $V_2-s_*\delta =0=L_{s_*}$, and not bidding at all is the best response for the $\beta$ player because $V_{s_*}-2\delta =0<L_2$. At any state s with $2 \le s \le s_*-2$, bidding is the best response for the $\gamma$ player because $V_2-(s+1)\delta >L_{s+1}$ (by Eq. (8)); Condition (c) by Lemma 13 suffices the incentive for the $\beta$ player at every odd state to bid. To incentivize the $\beta$ player at every even state $s \le s_*-2$ to bid, Lemma 11 says that it suffices to have $V_3-2\delta \ge L_2$, which is equivalent to $M_2 \ge L_2$ since, by the law of motion and Eq. (14), $M_2$ is the midpoint between $V_3-2\delta$ and $L_2$. Since $L_2<s_*\delta$ by Eq. (35), the condition $M_2 \ge L_2$ is guaranteed by Condition (b), $M_2 \ge s_*\delta$. $\square$

Lemma 16

For any $s_* \ge 4$, Condition (c) in Lemma 15 implies Condition (b) in Lemma 15.

Proof

Condition (c) in Lemma 15 is Ineq. (36) in the case $m=1$, i.e., when $\mu (m)=2^{-2m+1}=1/2$. Hence the condition is equivalent to

$$\begin{aligned} \frac{M_2}{\delta } \ge 4+(2-x)(s_*-4+\mu _*). \end{aligned}$$

(37)

To prove that this inequality implies Condition (b), i.e., $M_2/\delta \ge s_*$, it suffices to show

$$\begin{aligned} 4+(2-x)(s_*-4+\mu _*) > s_*, \end{aligned}$$

i.e.,

$$\begin{aligned} (1-x)(s_*-4)+\mu _*(2-x)>0, \end{aligned}$$

which is true because $s_* \ge 4$, $\mu _*=2^{-s_*+3}>0$ and $x \le 1$. $\square$

Lemma 17

Condition (a) in Lemma 15 is equivalent to existence of an $x \in [0, 1]$ that solves Eq. (15).

Proof

Condition (a) requires existence of $(M_2, x, L_2) \in \mathbb {R}_+ \times [0, 1] \times \mathbb {R}_+$ that satisfies Eqs. (32), (34) and (35) in the case of $m=s_*/2-1$ and $V_2=s_*\delta$. Combine (32) with (35) and use the notation $\mu _*:=2^{-s_*+3}$ and the fact $\sum _{k=1}^{m-1}2^{-2k}=(1-2^{-2m+2})/3$ to obtain

$$\begin{aligned}s_*\delta =V_2=&-\delta \cdot \frac{2}{3}(1-2\mu _*)+M_2\left( \frac{2}{3}(1-2\mu _*)+\mu _*\right) \\ \quad &+\underbrace{\delta (s_*-4+\mu _*)}_{L_2} \left( \frac{1}{3}(1-2\mu _*)+\mu _*x\right) , \end{aligned}$$

i.e.,

$$\begin{aligned} \frac{M_2}{\delta }=\frac{1}{2-\mu _*} \left( 3s_*+2(1-2\mu _*)-(s_*-4+\mu _*)(1-2\mu _*+3\mu _*x)\right) . \end{aligned}$$

(38)

By the same token, (34) coupled with (35) is equivalent to

$$\begin{aligned}M_2\left( 1-\frac{1}{3}(1-2\mu _*)-\mu _*x \right) =& -\delta \left( 1+\frac{1}{3}(1-2\mu _*)\right) \\ \quad & +(1-x)\mu _*v+\delta (s_*-4+\mu _*)\left( \frac{2}{3}(1-2\mu _*)+\mu _* \right) , \end{aligned}$$

i.e.,

$$\begin{aligned} \frac{M_2}{\delta }\left( 2(1+\mu _*)-3\mu _*x\right) =\frac{3\mu _*v}{\delta }(1-x) +(2-\mu _*)(s_*-6+\mu _*). \end{aligned}$$

(39)

Plug (38) into (39) and we obtain Eq. (15). $\square$

Lemma 18

For any even number $s_* \ge 4$, suppose that Eq. (38) holds. Then Condition (c) in Lemma 15 is equivalent to Ineq. (16), which is implied by $x \ge 0$ if and only if $s_* \le 6$.

Proof

Condition (c) in Lemma 15 has been shown to be equivalent to Ineq. (37). Provided that Eq. (38) is satisfied, Ineq. (37) is equivalent to

$$\begin{aligned}&4+(2-x)(s_*-4+\mu _*) \\ \le & \frac{1}{2-\mu _*} \left( 3s_*+2(1-2\mu _*)-(s_*-4+\mu _*)(1-2\mu _*+3\mu _*x)\right) . \end{aligned}$$

This inequality, given the fact $1-2\mu _* \ge 0$, is equivalent to

$$\begin{aligned} x \ge \frac{1}{2(1-2\mu _*)}\left( 5-4\mu _*-\frac{3(s_*-2)}{s_*-4+\mu _*}\right) , \end{aligned}$$

i.e., Ineq (16). Given Condition (a) in Lemma 15, which implies $x \ge 0$, Ineq. (16) is redundant if and only if the right-hand side of (16) is nonpositive, i.e.,

$$\begin{aligned} \frac{3(2-\mu _*)}{2(1-2\mu _*)(s_*-4+\mu _*)} \ge 1, \end{aligned}$$

i.e.,

$$\begin{aligned} s_* \le 4-2^{-s_*+3}+\frac{3\left( 2-2^{-s_*+3}\right) }{2\left( 1-2^{-s_*+4}\right) }. \end{aligned}$$

This inequality is satisfied when $s_* \in \{4, 6\}$, as its right-hand side is equal to $\infty$ when $s_*=4$, and 61/8 when $s_*=6$. The inequality does not hold, by contrast, when $s_* \ge 8$, as

$$\begin{aligned} s_* \ge 8\Rightarrow & {} 2^{-s_*+2} \le 2^{-6} \ \Rightarrow \ \frac{1-2^{-s_*+2}}{1/4-2^{-s_*+2}} \le \frac{1-2^{-6}}{1/4-2^{-6}}=\frac{63}{15} \\\Rightarrow & {} 4-2^{-s_*+3}+\frac{3\left( 2-2^{-s_*+3}\right) }{2\left( 1-2^{-s_*+4}\right) }<4+\frac{3}{2}\cdot \frac{2}{4} \cdot \frac{63}{15} < 8 \le s_*. \end{aligned}$$

Thus, for all even numbers $s_* \ge 4$, Ineq. (37) follows if and only if $s_* \le 6$. $\square$

Lemma 19

Any $s_* \ge 3$ constitutes an equilibrium if $s_*$ is an even number and—

• i. either $s_* \le 6$ and Eq. (15) admits a solution for $x \in [0, 1]$;

• ii. or $s_* \ge 8$ and Eq. (15) admits a solution for $x \in [0, 1]$ such that Ineq. (16).

Proof

The lemma follows from Lemma 15, where Condition (a) has been characterized by Lemma 17, Condition (b) by Lemmas 16 can be dispensed with, and Condition (c), by Lemma 18, can be dispensed with when $s_* \le 6$ (hence Claim (i) of the lemma) and is equivalent to Ineq (16) when $s_*>6$ (hence Claim (ii) of the lemma). $\square$

Lemma 20

If $x=1$, the left-hand side of (15) is less than the right-hand side of (15).

Proof

When $x=1$, the left-hand side of (15) is equal to $(2-\mu _*)^2(s_*-6+\mu _*)$, and the right-hand side equal to

$$\begin{aligned}&\left( 2(1+\mu _*)-3\mu _*\right) \left( 3s_*+2(1-2\mu _*)-(s_*-4+\mu _*)(1-2\mu _*+3\mu _*)\right) \\&\quad = (2-\mu _*)\left( 2s_*+6-\mu _*-\mu _*s_*-\mu _*^2\right) . \end{aligned}$$

Thus, the lemma follows if

$$\begin{aligned} (2-\mu _*)(s_*-6+\mu _*)<2s_*+6-\mu _*-\mu _*s_*-\mu _*^2, \end{aligned}$$

i.e., $9\mu _*<18$, which is true because $\mu _*=2^{-s_*+3}$. $\square$

Lemma 21

$s_*=4$ constitutes an equilibrium if and only if $v/\delta >35/2$, and $s_*=6$ constitutes an equilibrium if and only if $v/\delta >6801/120$ ($=56.675$).

Proof

By Lemma 19, with $s_* \le 6$ the necessary and sufficient condition for equilibrium is that Eq. (15) admits a solution for $x \in [0, 1]$. By Lemma 20, the left-hand side of that equation is less than its right-hand side when $x=1$. Thus, it suffices to show that the left-hand side is greater than the right-hand side when $x=0$, i.e.,

$$\begin{aligned}&\frac{3\mu _*v}{\delta }(2-\mu _*)+(2-\mu _*)^2(s_*-6+\mu _*) \\ >& 2(1+\mu _*) \left( 3s_*+2(1-2\mu _*)-(s_*-4+\mu _*)(1-2\mu _*)\right) , \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{v}{\delta }(2-\mu _*)>s_*(4+\mu _*)+(6-\mu _*)(2/\mu _*-2-\mu _*). \end{aligned}$$

Since $\mu _*$ is equal to 1/2 when $s_*=4$, and equal to 1/8 when $s_*=6$, the above inequality is equivalent to $v/\delta >35/2$ when $s_*=4$, and $v/\delta >6801/120$ when $s_*=6$. $\square$

Proof of Theorem 2

Claims (i) and (ii) of the theorem are just Lemma 21. To prove Claim (iii), pick any even number $s_* \in \{8, 10, 12, \ldots \}$. By Lemma 19.ii, $s_*$ constitutes an equilibrium if Eq. (15) admits a solution for $x \in [0, 1]$ that satisfies Ineq. (16). By Lemma 20, the left-hand side of (15) is less than its right-hand side when $x=1$. Thus, it suffices to show that the left-hand side is greater than the right-hand side when x is equal to some number greater than or equal to the right-hand side of Ineq. (16). To that end, note from $s_* \ge 8$ that $\mu _*=2^{-s_*+3} \le 1/32$, hence $2-\mu _* \ge 63/32$ and $1+\mu _* <33/32$. Thus, the left-hand side of (15) is greater than

$$\begin{aligned} \frac{3\mu _*v}{\delta }(1-x)\frac{63}{32} +\left( \frac{63}{32}\right) ^2 (s_*-6), \end{aligned}$$

and the right-hand side of Ineq. (16)

$$\begin{aligned} 1-\frac{3(2-\mu _*)}{2(1-2\mu _*)(s_*-4+\mu _*)} < 1-\frac{3 \times 63/32}{2 \times 1 \times (s_*-3)}. \end{aligned}$$

Therefore, it suffices, for $s_*$ to constitute an equilibrium, to have

$$\begin{aligned} \frac{3\mu _*v}{\delta }(1-x)\frac{63}{32} +\left( \frac{63}{32}\right) ^2 (s_*-6) \end{aligned}$$

greater than or equal to the right-hand side of (15) when

$$\begin{aligned} x=x_*:=1-\frac{3 \times 63}{64(s_*-3)}. \end{aligned}$$

To that end, denote $\phi (s_*, x)$ for the right-hand side of (15), i.e.,

$$\begin{aligned}& \phi (s_*, x)\\:=& \left( 2(1+\mu _*)-3\mu _*x\right) \left( 3s_*+2(1-2\mu _*)-(s_*-4+\mu _*)(1-2\mu _*+3\mu _*x)\right) \end{aligned}$$

(recall that $\mu _*=2^{-s_*+3}$). Note, from $0<x<1$, that $-1/32<\mu _*(2-3x)<1/16$. Hence

$$\begin{aligned} \frac{63}{32}=2-\frac{1}{32}<&2(1+\mu _{*})-3\mu _{*}x < 2+\frac{1}{16}=\frac{33}{16}, \\ \frac{15}{16}=1-\frac{1}{16}<&1-2\mu _{*}+3\mu _{*}x < 1+\frac{1}{32}=\frac{33}{32}. \end{aligned}$$

Thus, the first factor $2(1+\mu _*)-3\mu _*x$ of $\phi (s_*, x)$ is positive for all $x \in (0, 1)$. If the second factor of $\phi (s_*, x)$ is nonpositive when $x=x_*$ then $\phi (s_*, x_*) \le 0$ and we are done, as the left-hand side of (15) is positive. Hence we may assume, without loss of generality, that

$$\begin{aligned} 3s_*+2(1-2\mu _*)-(s_*-4+\mu _*)(1-2\mu _*+3\mu _*x_*) > 0. \end{aligned}$$

Consequently, $\phi (s_*, x_*)$ can only get bigger if we replace its first factor by the upper bound 33/16, and the term $1-2\mu _*+3\mu _*x$ in the second factor by its lower bound 15/16 (note that, in the second factor, $s_*-4+\mu _*>0$ because $s_* \ge 8$). I.e., $\phi (s_*, x_*)$ is less than

$$\begin{aligned} \frac{33}{16}\left( 3s_*+2(1-2\mu _*)-\frac{15}{16}(s_*-4+\mu _*)\right)= & {} \frac{33}{16}\left( \frac{33}{16}s_*+\frac{23}{4}-\frac{79}{16}\mu _*\right) \\< & {} \frac{33}{16}\left( \frac{33}{16}s_*+\frac{23}{4}\right) \\< & {} 5s_*+12. \end{aligned}$$

Therefore, the above observations put together, we are done if

$$\begin{aligned} \frac{3\mu _*v}{\delta }(1-x_*)\frac{63}{32} +\left( \frac{63}{32}\right) ^2 (s_*-6) \ge 5s_*+12 \end{aligned}$$

In other words, it suffices to have

$$\begin{aligned} \frac{3\mu _*v}{\delta }\cdot \frac{3 \times 63}{64(s_*-3)}\cdot \frac{63}{32} +\left( \frac{63}{32}\right) ^2 (s_*-6) \ge 5s_*+12, \end{aligned}$$

i.e.,

$$\begin{aligned} \frac{3^2\mu _*v}{\delta } \ge -2(s_*-6)(s_*-3)+\frac{32\times 64}{63^2}\left( 5s_*+12\right) (s_*-3). \end{aligned}$$

With $\frac{32\times 64}{63^2} \approx 0.516$, the above inequality holds if

$$\begin{aligned} \frac{3^2\mu _*v}{\delta } \ge -2(s_*-6)(s_*-3)+(5s_*+12)(s_*-3), \end{aligned}$$

i.e.,

$$\begin{aligned} \frac{9\mu _*v}{\delta } \ge 3s_*^2+15s_*-72, \end{aligned}$$

which, coupled with $\mu _*=2^{-s_*+3}$, is equivalent to the hypothesis (17) of Claim (iii). $\square$

1.7 B.7 Theorems 3

First, when $s=s_*=1$, clearly $M_s=\pi _{\beta , s_*-1}L_2=\pi _{\beta , s_*-1}\delta /2>0$ and $L_s=L_{s_*-1}=0$. Second, we show that if $s < s_*-1$ then $M_s>0$ and $L_s>0$. By Eq. (8) and Theorem 1, $L_s \ge \frac{1}{2}(V_2-(s+1)\delta ) = \frac{1}{2}(s_*\delta -(s+1)\delta )>0$, with the last inequality due to $s \le s_*-2$. By (7), the follower can secure an expected payoff no less than $L_2$, through staying put at s (while the underdog bids for sure), hence $M_2 \ge L_2>0$. Third, for any state $s \le s_*-1$, we show that $V_s>2\delta$. With $s \le s_*-1$, the frontrunner’s surplus, by (6), is

• i. either $V_s=(M_2+M_{s+1})/2$ (if $s<s_*-1$)

• ii. or $V_s=(1-\pi _{\gamma , s_*-1})v+\pi _{\gamma , s_*-1}M_2$ (if $s=s_*-1$).

In Case (i), his surplus is

$$\begin{aligned} (M_2+M_{s+1})/2>M_2/2>\left( V_2+\delta /2\right) /2 \ge \left( 4\delta +\delta /2\right) /2>2\delta , \end{aligned}$$

with the second inequality due to Lemma 9, and the third inequality due to $V_2=s_*\delta$ and $s_* \ge 4$. In Case (ii), his surplus is

$$\begin{aligned} (1-\pi _{\gamma , s_*-1})v+\pi _{\gamma , s_*-1}M_2>(1-\pi _{\gamma , s_*-1})2\delta +\pi _{\gamma , s_*-1}4\delta >2\delta , \end{aligned}$$

with the first inequality due to $v>2\delta$ by assumption, the fact $M_2>V_2$ by Lemma 9, and the fact $V_2\ge 4\delta$ as in Case (i) $\square$