More than words: the effects of cheap talk in a volunteer’s dilemma

Abstract

We theoretically and experimentally investigate a game in which exactly one person should make a costly effort to achieve a socially efficient outcome. This setting is commonly known as the volunteer’s dilemma. We implement one-way communication by allowing one player to send a message indicating whether she intends to volunteer and investigate the message’s effects on behavior and efficiency in the subsequent game. We theoretically demonstrate that there are asymmetric mixed-strategy equilibria in the volunteer’s dilemma and argue that one of these is likely to emerge through one-way communication. The experimental data support this notion. We find that the actions of both the sender and receiver of the message are crucially affected by the cheap talk stage and that efficiency in the volunteer’s dilemma increases with one-way communication.

Appendix 1: Proof of Statement 2

We have characterized three types of Nash equilibria in the volunteer’s dilemma:

• (1) All players play pure strategies: one player chooses $s^{V}$, and all remaining players choose $s^{NV}$.

• (2) Some players play the pure strategy $s^{NV}$; the remaining k players play symmetric mixed strategies.

• (3) All players play symmetric mixed strategies.

Obviously, if at least one player chooses $s^{V}$, the single best reply is to choose $s^{NV}$, and hence an equilibrium in which some players play the pure strategy $s^{V}$ and the remaining players mix cannot exist.

First, we will show that there are no further types of equilibria in this game. Second, we will show that we have identified all equilibria of the types characterized above.

There could exist two further types of equilibria:

• (4) Some players play the pure strategy $s^{NV}$, while the remaining players mix with asymmetric probabilities.

• (5) All players play mixed strategies, but they mix with asymmetric probabilities.

In the following, we will prove that equilibria as described in (4) and (5) do not exist by showing that, in equilibrium, if $k$ players mix across strategies while the remaining $n - k$ players choose $s^{NV}$ with certainty, the mixing probability must be the same for all $k$ players in subgroup $M$.

Assume that player $m \in M$ chooses $s^{NV}$ with probability $q_{m} , 0 < q_{m} < 1.$ In a mixed strategy equilibrium, player $m$ must be indifferent between $s^{V}$ and $s^{NV}$, and hence the expected payoff $\varPi \left( {s^{NV} } \right)$ of playing $s^{NV}$ must be equal to the sure payoff $\varPi \left( {s^{V} } \right)$ of playing $s^{V}$.

From $\varPi \left( {s^{V} } \right) = V - C$ and $\varPi \left( {s^{NV} } \right) = \mathop \prod \limits_{{i \in M{ \setminus }m}} q_{i} L + \left( {1 - \mathop \prod \limits_{{i \in M{ \setminus }m}} q_{i} } \right)V$, it follows that

$$ \mathop \prod \limits_{{i \in M{ \setminus }m}} q_{i} = \frac{C}{V - L} \forall m \in M. $$

This also holds for any other player $l \in M$ with $l \ne m$; hence, $\mathop \prod \limits_{{i \in M{ \setminus }m}} q_{i} = \mathop \prod \limits_{{i \in M{ \setminus }l}} q_{i} \Rightarrow \frac{{\mathop \prod \nolimits_{i} q_{i} }}{{q_{m} }} = \frac{{\mathop \prod \nolimits_{i} q_{i} }}{{q_{l} }}$.

It follows that $q_{m} = q_{l } \quad \forall m,l \in M$.

The same argument holds if $k = n$, i.e., if all players choose mixed strategies.

Moreover, there could exist further equilibria of the types described in (2) and (3), i.e., players could mix with symmetric probabilities that are different from $\hat{q} (k)$ and $q^{*}$, respectively.

Consider the equilibrium described in (2). As we have shown, for a given player $m \in M$ to be indifferent between $s^{V}$ and $s^{NV}$, the product of the probabilities with which the remaining mixing players $M{ \setminus }m$ choose $s^{NV}$ must equal a constant $\frac{C}{V - L}$. Obviously, if any other player $j \in M$ chose a probability lower (higher) than $\hat{q}\left( k \right)$, this condition could only be met if a third player $r \in M$ chose a probability higher (lower) than $\hat{q}\left( k \right)$. It follows that there are no equilibria in which $k$ players mix with symmetric probabilities that are different from $\hat{q}\left( k \right)$.

Again, the same argument holds if $k = n$; if all players choose mixed strategies, there are no equilibria in which players mix with symmetric probabilities that are different from $q^{*} .$ □