Contingent fees and endogenous timing in litigation contests

Abstract

This study examined the contingent fee arrangements and adversarial systems applied in the United States. In the American context, a plaintiff (or a defendant) pays a contingent fee (an hourly fee) to their lawyer. In this adversarial system, lawyers can either be the first or the second mover. Solving the American practice with endogenous timing of litigation efforts, we obtained the following results: (i) if the defendant’s total hourly fee rate is not sufficiently high, the plaintiff’s lawyer is the first mover, with the plaintiff then being the underdog and (ii) if the rate is sufficiently high, the lawyer is the second mover, with the plaintiff then being the favorite. We demonstrated that these results are not ideal for the plaintiff. The equilibrium order of effort increases the equilibrium contingent fee and total legal effort in the trial, making reaching a settlement difficult. However, this improves the justice achieved through litigation if the total hourly fee rate is not significantly high. We suggest that the American practice of contingent fees with endogenous timing of effort is not economically efficient but is suitable for achieving justice.

Appendix

Comparison of expected payoffs for Delegate 1 and expected losses for Player 2 in the second stage of the endogenous timing game.

We sequentially compare Delegate 1’s expected payoffs with Player 2’s expected losses in the second stage of the endogenous timing game. Before we consider the comparison, we note the followers’ participation constraints in the 1 M and 2 M subgames. In the 1 M subgame, Delegate 2’s participation constraint is not violated if ch < 2 (see Eq. 9). In the 2 M subgame, the constraint that Delegate 1 participates under ch > 1/2 (see Eq. 12). Thus, the constraint that the two delegates participate under the trial is 1/2 < ch < 2.

First, we compare the expected payoffs for Delegate 1. Using Lemmas 1, 3 and 4, we obtain the following:

$$\begin{gathered} \pi_{{1}}^{S} < \pi_{{1}}^{{{1}M}} {\text{for}}\quad ch \ne {1}, \hfill \\ \pi_{{1}}^{S} = \pi_{{1}}^{{{1}M}} {\text{for}}\quad ch = { 1}, \hfill \\ \end{gathered}$$

(A1)

wherein π₁^S − π₁^1M = – c²h(1 − ch)²v/{4(1 + ch)²}. Furthermore, we find the following:

$$\begin{gathered} \pi_{{1}}^{S} > \pi_{{1}}^{{{2}M}} {\text{for 1}}/{2 } < ch < { 1}, \hfill \\ \pi_{{1}}^{S} = \pi_{{1}}^{{{2}M}} \;{\text{for}}\quad ch = { 1}, \hfill \\ \pi_{{1}}^{S} < \pi_{{1}}^{{{2}M}} {\text{for 1 }} < ch < { 2}, \hfill \\ \end{gathered}$$

(A2)

wherein π₁^S − π₁^2M = (1 − ch)(4c²h² + ch − 1)v/{4ch²(1 + ch)²}.

Next, we compare the expected losses of Player 2. Using Lemmas 1, 3 and 4, we find that:

$$\begin{gathered} L_{{2}}^{S} > L_{{2}}^{{{1}M}} {\text{for 1}}/{2 } < ch < { 1}, \hfill \\ L_{{2}}^{S} = L_{{2}}^{{{1}M}}\;{\text{for}}ch = { 1}, \hfill \\ L_{{2}}^{S} < L_{{2}}^{{{1}M}} {\text{for 1 }} < ch < { 2}, \hfill \\ \end{gathered}$$

(A3)

wherein L₂^S − L₂^1M = – ch(1 − ch)(c²h² − ch − 4)v/{4(1 + ch)²}. Furthermore, we find the following:

$$\begin{gathered} L_{{2}}^{S} > L_{{2}}^{{{2}M}} {\text{for}}\quad ch \ne {1}, \hfill \\ L_{{2}}^{S} = L_{{2}}^{{{2}M}} \;{\text{for}}\quad ch = { 1}, \hfill \\ \end{gathered}$$

(A4)

wherein L₂^S − L₂^2M = (1 − ch)²v/{4ch(1 + ch)²}.

Finally, we solve for the subgame perfect equilibria in the second stage of the full game. Delegate 1 and Player 2 have two possible strategies: first or second. First represents choosing the first period, and second, choosing the second period. Table

Table 1 Expected payoffs and expected losses depending on the timing of the effort

1 presents Delegate 1’s expected payoffs and Player 2’s expected losses based on Lemmas 1, 3 and 4. Combining (A1)–(A4) and Table 1, we obtain the subgame perfect equilibria in the second stage:

$$\begin{gathered} \left( {First,Second} \right) {\text{for 1}}/{2 } < ch < { 1}, \hfill \\ \left( {First,First} \right) \, \left( {First,Second} \right) \, \left( {Second,First} \right) \, \left( {Second,Second} \right) {\text{for}}\quad ch = { 1}, \hfill \\ \left( {Second,First} \right) {\text{for 1 }} < ch \le {2} \hfill \\ \end{gathered}$$