# Free rider problem in game theory

Suppose a town is building a bridge, and it costs $B$. There are $n$ villagers.

Each village's valuation of the bridge is private information, $v_i$.

It is common knowledge that this valuation is drawn from a uniform distribution $[0,1]$. $B\in[0,1]$.

Villager can only submit $0$ or $B$.

If one villager submits $B$, then the bridge is built and every other villager pays their submission.

If no bridge is built, everyone gets $0$.

How do I construct an expected payoff a village $i$?

What I got down to is having 2 scenarios: $v_i>B$ and $v_i\leq B$.

But in each case, I have two possible payoffs. For the former case,

if every other player submits $0$, $i$ should submit B, because $v_i-B>0$.
if someone pays $B$, $i$ should submit 0, because she gets $v_i$.

You get a similar type of for the other scenario.

But how would I incorporate this into expected payoff of $i$ and how should I go about constructing a social welfare function?

I feel like this is just a variation of all-payout auction with discrete action space for each $i$.

• Is $v_i\le c$ a typo for $v_i\le B$? Also, is the value of $B$ common knowledge? Feb 27, 2018 at 17:52
• Writing down expected payoff is one thing, constructing a social welfare function is an entirely different matter... Feb 27, 2018 at 17:52
• Yes, I am asking both lol. The cost of bridge is common knowledge. Feb 27, 2018 at 18:08
• See my answer for the expected payoff, and a symmetric BNE. A lot more assumptions are needed for the construction of a social welfare function (see Wikipedia for a detailed discussion). Feb 27, 2018 at 20:34

Let $b_i\in\{0,B\}$ be $i$'s strategy. Then $i$'s payoff depends on the strategy profile $(b_i,b_{-i})$, where $b_{-i}=(b_j)_{j\ne i}$.

$$u_i(b_i,b_{-i})= \begin{cases} v_i&\text{if b_i=0 and b_j=B for some j\ne i}\\ 0& \text{if b_i=0 and b_j=0 for all j\ne i}\\ v_i-B&\text{if b_i=B} \end{cases}$$

Thus, $b_i=B$ is a best response if $$u_i(B,b_{-i})\ge u_i(0,b_{-i}) \quad\Leftrightarrow\quad v_i-B\ge (1-\Pr(b_j=0,\;\forall j\ne i))v_i.\tag{1}$$

Since the game is ex ante symmetric, we could further assume that every $i$ adopts a threshold strategy, i.e. $$b_i=\begin{cases} 0&\text{if v_i\le\overline v}\\ B&\text{if v_i>\overline v} \end{cases}\tag{2}$$ where $\overline v$ is some common threshold value. Then, the probability in $(1)$ can be written as $$\Pr(b_j=0,\;\forall j\ne i)=\Pr(v_j\le \overline v,\;\forall j\ne i)=(\overline v)^{n-1},\tag{3}$$ where the last equality is obtained from the assumption that $v_j$'s are i.i.d. and $v_j\sim U[0,1]$.

Consolidating $(1)$ to $(3)$, we can solve for the cutoff value $\overline v=B^{1/n}$.

• Herr, thanks for the response. When you say ex-ante symmetric, how do you mean? And also, in general, when we use the notion of "symmetry" in solution concepts or games, like "symmetric game" or "symmetric Nash eq'm", what do they exactly mean? Feb 27, 2018 at 22:25
• Ex ante symmetry is in the sense that every villager, before observing their private value, faces the same problem (same information, same prior, same strategy space, same payoff function, etc.). So $i$'s problem is the same as $j$'s problem, except the change of the name index. Therefore it's reasonable to assume that they will adopt the same strategy ex ante (as defined in $(2)$). If a profile of such strategies also happens to be mutual best responses, then we have a symmetric (Bayesian) Nash equilibrium. Read more here. Feb 27, 2018 at 23:06
• How did you get the IFF in (1), RHS probability? Feb 28, 2018 at 16:54
• @FrankSwanton: The condition "if others pay 0 and your valuation is $≤B$" is not feasible: this is a simultaneous move game, and you don't get to observe what others have chosen. A payoff function describes the payoffs associated with each possible outcome, but what you're doing is to prescribe what should be done be a player if she were to behave in an optimal way, potentially ignoring outcomes that are suboptimal. This is precisely the confusion I was alluding to. Feb 28, 2018 at 17:20
• @FrankSwanton: Note that in (1) I allow the other players' strategies to be flexible, i.e. $(b_{-i})$ can be anything. Mar 1, 2018 at 0:19