Inside Buchanan's Samaritan's Dilemma: altruism, strategic courage and ethics of responsibility

Abstract

The Samaritan’s Dilemma has largely been investigated, frequently by assuming that Samaritans help recipients out of altruism. Yet, Buchanan did not make any behavioral assumption regarding the Samaritan’s motives. In this paper, we explicitly introduce this assumption in Buchanan’s original model and analyze how it changes the nature of the game. We show that altruism alone does not explain the Dilemma. A parameter that captures the disutility the Samaritan feels when helping someone who does not reciprocate her benevolence must be introduced to make sense of the different version of Buchanan’s Samaritan’s Dilemma. We also show that the Samaritan’s Dilemma is an evolutionary stable outcome, which confirms Buchanan’s intuitions. Finally, a third important point put forward in the paper is that the more altruistic are the Samaritans, the less likely it is that they will show the kind of strategic courage envisaged by Buchanan, which is one of the most important traits Samaritans should display to avoid being trapped in a Dilemma.

Appendix 1

Proof of Lemma 2

The $\left( {x^{*} ,y^{*} } \right)$ equilibrium exists when $x^{*}$ and $y^{*}$ both $\in \left( {0,1} \right)$. This, in turn, requires that two of the following conditions are simultaneously satisfied:

$$2\rho - 1 > \rho - \left( {1 - e} \right) > 0$$

(4′)

$$2\rho - 1 < \rho - \left( {1 - e} \right) < 0$$

(5′)

$$\left( {2\rho - 1} \right)\alpha - c > \rho \alpha - h - c > 0$$

(6′)

$$\left( {2\rho - 1} \right)\alpha - c < \rho \alpha - h - c < 0$$

(7′)

Rearranging condition (4′) yields condition (1); rearranging conditions (6′) and (7′) yields, respectively, conditions (8) and (9). The concluding statement in Lemma 1 follows from the fact that conditions (1) and (9) are simultaneously satisfied when the game is a Passive Samaritan’s dilemma, since both are implied by condition (3), while the parametrization defined in condition (1) is inconsistent with any of the possible combinations across conditions (4′)–(7′), so that $\left( {x^{*} ,y^{*} } \right)$ never exists when the game is an Active Samaritan’s dilemma.□

Proof of Proposition 1

The Jacobian matrix of the system is given by:

$$J = \left( {\begin{array}{*{20}c} {{\text{d}}\dot{x}/{\text{d}}x} & {{\text{d}}\dot{x}/{\text{d}}y} \\ {{\text{d}}\dot{y}/{\text{d}}y} & {{\text{d}}\dot{y}/{\text{d}}y} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {\left( {1 - 2x} \right)\left[ {U\left( 1 \right) - U\left( 0 \right)} \right]} & {x\left( {1 - x} \right)\left( {2\rho - 1} \right)} \\ {y\left( {1 - y} \right)\left[ {\left( {2\rho - 1} \right)\alpha - c} \right]} & {\left( {1 - 2y} \right)\left[ {V\left( 1 \right) - V\left( 0 \right)} \right]} \\ \end{array} } \right)$$

As it is well-known, a stationary point is attractive if both the eigenvalues of the Jacobian matrix evaluated at that equilibrium have negative real parts. In addition, when either of the partial derivative ${\text{d}}\dot{x}/{\text{d}}y$ and/or ${\text{d}}\dot{y}/{\text{d}}x$ is = 0—which happens whenever $x = 0$ or $x = 1$ and/or $y = 0$ or $y = 1$—the eigenvalues exactly correspond to the partial derivatives ${\text{d}}\dot{x}/{\text{d}}x$ and ${\text{d}}\dot{y}/{\text{d}}y$. With these facts in mind, the topological properties of the four vertices $\left( {0, 0} \right)$, $\left( {0, 1} \right)$, $\left( {1, 0} \right)$ and $\left( {1, 1} \right)$ can be straightforwardly checked by evaluating the sign of ${\text{d}}\dot{x}/{\text{d}}x$ and ${\text{d}}\dot{y}/{\text{d}}y$ at each of the four vertices, since the partial derivatives ${\text{d}}\dot{x}/{\text{d}}y$ and ${\text{d}}\dot{y}/{\text{d}}x$ are always = 0 at each of these points.

At $\left( {0,0} \right)$ we have that:

$${\text{d}}\dot{x}/{\text{d}}x = 1 - e - \rho$$

$${\text{d}}\dot{y}/{\text{d}}y = h + c - \rho \alpha$$

Hence, we have three cases:

• 1. if $1 - e - \rho < 0$ and $h + c - \rho \alpha < 0$, $\left( {0,0} \right)$ is a sink (asymptotically stable).

• 2. If $1 - e - \rho > 0$ and $h + c - \rho \alpha > 0$, $\left( {0,0} \right)$ is source (asymptotically unstable).

• 3. If $1 - e - \rho < 0$ and $h + c - \rho \alpha > 0$, or $1 - e - \rho > 0$ and $h + c - \rho \alpha < 0$, $\left( {0,0} \right)$ is a saddle (asymptotically unstable).

Given Lemma 1, when the game is either an Active or a Passive Samaritan’s Dilemma, $\left( {0,0} \right)$ is always a sink.

At $\left( {1,1} \right)$ we have that:

$$\begin{aligned} {\text{d}}\dot{x}/{\text{d}}x & = e - \rho \\ {\text{d}}\dot{y}/{\text{d}}y & = \alpha \left( {1 - \rho } \right) - h \\ \end{aligned}$$

Hence, we have three cases:

• 1. if $e - \rho < 0$ and $\alpha \left( {1 - \rho } \right) - h < 0$, $\left( {1,1} \right)$ is a sink (asymptotically stable).

• 2. If $e - \rho > 0$ and $\alpha \left( {1 - \rho } \right) - h > 0$, $\left( {1,1} \right)$ is source (asymptotically unstable).

• 3. If $e - \rho < 0$ and $\alpha \left( {1 - \rho } \right) - h > 0$ or $e - \rho > 0$ and $\alpha \left( {1 - \rho } \right) - h < 0$, $\left( {1,1} \right)$ is a saddle (asymptotically unstable).

Given Lemma 1, when the game is an Active Samaritan’s Dilemma, $\left( {1,1} \right)$ is always a saddle; when the game is a Passive Samaritan’s Dilemma, $\left( {1,1} \right)$ is always a sink.

At $\left( {1,0} \right)$ we have that

$$\begin{aligned} {\text{d}}\dot{x}/{\text{d}}x & = \rho + e - 1 \\ {\text{d}}\dot{y}/{\text{d}}y & = h - \left( {1 - \rho } \right)\alpha \\ \end{aligned}$$

Hence, we have three cases:

• 1. if $\rho + e - 1 < 0$ and $h - \left( {1 - \rho } \right)\alpha < 0$, $\left( {1,0} \right)$ is a sink (asymptotically stable).

• 2. If $\rho + e - 1 > 0$ and $h - \left( {1 - \rho } \right)\alpha > 0$, $\left( {1,0} \right)$ is source (asymptotically unstable).

• 3. If $\rho + e - 1 < 0$ and $h - \left( {1 - \rho } \right)\alpha > 0$, or $\rho + e - 1 > 0$ and $h - \left( {1 - \rho } \right)\alpha < 0$, $\left( {0,1} \right)$ is a saddle (asymptotically unstable).

Given Lemma 1, when the game is an Active Samaritan’s Dilemma, $\left( {1,0} \right)$ is always a saddle; when the game is a Passive Samaritan’s Dilemma, $\left( {1,0} \right)$ is always a source.

At $\left( {0,1} \right)$ we have that

$$\begin{aligned} {\text{d}}\dot{x}/{\text{d}}x & = \rho - e \\ {\text{d}}\dot{y}/{\text{d}}y & = \rho \alpha - h - c \\ \end{aligned}$$

Hence, we have three cases:

• 1. if $\rho - e < 0$ and $\rho \alpha - h - c < 0$, $\left( {0,1} \right)$ is a sink (asymptotically stable).

• 2. If $\rho - e > 0$ and $\rho \alpha - h - c > 0$, $\left( {0,1} \right)$ is source (asymptotically unstable).

• 3. If $\rho - e < 0$ and $\rho \alpha - h - c > 0$, or $\rho - e > 0$ and $\rho \alpha - h - c < 0$, $\left( {0,1} \right)$ is a saddle (asymptotically unstable).

Given Lemma 1, when the game is an Active Samaritan’s Dilemma, $\left( {0,1} \right)$ is always a saddle; when the game is a Passive Samaritan’s Dilemma, $\left( {0,1} \right)$ is always a source.

To prove the stability properties of the $\left( {x^{*} ,y^{*} } \right)$ equilibrium, we study its Trace and Determinant and see that:

$${\text{TR}}\,J_{{(x^{*} ,y^{*} )}} = 0\,{\text{and}}\,{\text{Det}}\,J_{{(x^{*} ,y^{*} )}} = - x^{*} (1 - x^{*} )y(1 - y^{*} )[(2\rho - 1)\alpha - c](2\rho - 1)$$

From Lemma 1, we already know that $\left( {2\rho - 1} \right)\alpha - c > 0$ and $2\rho - 1 > 0$ are simultaneously satisfied when condition (3) is satisfied. In this case, ${\text{Det }}J_{{\left( {x^{*} ,y^{*} } \right)}} < 0$, which proves that $\left( {x^{*} ,y^{*} } \right)$ is a saddle when the game is a Passive Samaritan’s dilemma. In addition, there exists parametrizations for which ${\text{Det }}J_{{\left( {x^{*} ,y^{*} } \right)}} \ge 0$, which completes the Proof of Proposition 1.□

Proof of Proposition 2

From the results in Proposition 1, it is straightforward to derive the following set of results. First, the existence conditions of the four monostable regimes are as follows:

• 1. The stationary point $\left( {0,0} \right)$ is the unique attractor iff $e > \max \left\{ {\rho ,1 - \rho } \right\}$ and $\alpha > \max \left\{ {\frac{h}{1 - \rho },\frac{h + c}{e}} \right\}$.

• 2. The stationary point $\left( {1,1} \right)$ is the unique attractor iff $e < \min \left\{ {\rho ,1 - \rho } \right\}$ and $\alpha < \min \left\{ {\frac{h}{1 - \rho },\frac{h + c}{e}} \right\}$.

• 3. The stationary point $\left( {1,0} \right)$ is the unique attractor iff $e < \min \left\{ {\rho ,1 - \rho } \right\}$ and $\alpha > \max \left\{ {\frac{h}{1 - \rho },\frac{h + c}{e}} \right\}$.

• 4. The stationary point $\left( {0,1} \right)$ is the unique attractor iff $e > \max \left\{ {\rho ,1 - \rho } \right\}$ and $\alpha < \min \left\{ {\frac{h}{1 - \rho },\frac{h + c}{e}} \right\}$

Second, the existence conditions of the two bistable regimes are as follows:

• 1. The stationary points $\left( {0,0} \right)$ and $\left( {1,1} \right)$ simultaneously attract iff conditions (1) and (9) are simultaneously satisfied.

• 2. The stationary points $\left( {1,0} \right)$ and $\left( {0,1} \right)$ simultaneously attract iff simultaneously attract iff conditions (7) and (8) are simultaneously satisfied.

Third, the existence conditions of the two cyclical regimes are as follows:

• 1. The system exhibits cyclical behavior with counterclockwise oscillations around $\left( {x^{*} ,y^{*} } \right)$ iff conditions (7) and (9) are simultaneously satisfied.

• 2. The system exhibits cyclical behavior with clockwise oscillations around $\left( {x^{*} ,y^{*} } \right)$ iff conditions (1) and (8) are simultaneously satisfied.

Observe that when the game is an Active Samaritan’s Dilemma, the monostable regime featuring $\left( {0,0} \right)$ as the only attractor results; when the game is a Passive Samaritan’s Dilemma, the bistable regime featuring $\left( {0,0} \right)$ and $\left( {1,1} \right)$ as attractors result.□