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[Game Theory] A pedestrian is hit by a car. How many people will help?

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  1. Nov 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider the following social problem. A pedestrian is hit by a car and lies injured on the road. There are [itex]n[/itex] people in the vicinity of the accident. The injured pedestrian requires immediate medical attention, which will be forthcoming if at least one of the [itex]n[/itex] people call for help. Simultaneously and independently, each of the [itex]n[/itex] bystanders decides whether or not to call for help (by dialing 911 on a cell phone or pay phone). Each bystander obtains [itex]v[/itex] units of utility if someone (anyone) calls for help. Those who call for help pay a personal cost of [itex]c[/itex] . That is, if person [itex]i[/itex] calls for help, then he obtains the payoff [itex]v-c[/itex]. If person [itex]i[/itex] does not call but at least one other person calls, then person [itex]i[/itex] gets [itex]v[/itex]. Finally, if none of the [itex]n[/itex] people calls for help, then person [itex]i[/itex] obtains 0. Assume [itex]v>c[/itex].

    1. The purpose of this question is to find the symmetric Nash equilibrium of this [itex]n[/itex]-player game. This equilibrium is in mixed strategies, i.e. such that each person is indifferent between his/her two possible strategies: to call or not to call. Therefore, each player’s payoff must be equal when he/she calls and when he/she does not call.

    a. We already know that player [itex]i[/itex]’s payoff is [itex]v-c[/itex] when he/she calls. Write the payoff of player [itex]i[/itex] when he/she does not call, letting [itex]p[/itex] be the probability that a person does not call for help. Hint: there are [itex]n-1[/itex] players others than player [itex]i[/itex]. Therefore, with probability [itex]p^{n-1}[/itex], no one of the other players will call, and with probability [itex]1-p^{n-1}[/itex] at least one of the other players will call. ​
    b. By setting player [itex]i[/itex]’s payoff equal when he/she calls and does not call, find the probability that a person does not call [itex]p[/itex] in equilibrium (Hint: this will be a function of [itex]c/v[/itex] and [itex]n[/itex]).
    2. Compute the probability that at least one person calls for help in equilibrium [itex] 1-p^n [/itex].
    How does this depend on n? Can you comment? (Hint: to answer the second part of the question you need to differentiate it with respect to [itex]n[/itex]).


    2. Relevant equations



    3. The attempt at a solution
    All I need is to figure out what the payoffs are, and I will be able to solve the rest. For part 1.a the payoff I came up with is [itex](1-p^{n-1})v+p^{n-1}(v-c)[/itex], but I am leaning towards the fact that it is wrong. Help appreciated.
     
  2. jcsd
  3. Nov 10, 2013 #2

    D H

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    Yep. It's wrong. There's a pn-1 probability that no one else calls. What is the payoff in this event?
     
  4. Nov 11, 2013 #3
    If no one calls it is zero. Then, the payoff of the ith player is [itex]p^{n-1}*0+(1-p^{n-1})v[/itex] . So for part b would it be correct to say that [itex]v-c=(1-p^{n-1})v[/itex] ?
     
  5. Nov 11, 2013 #4

    D H

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    Correct. That's your Nash equilibrium.
     
  6. May 25, 2015 #5
    Hmm... he died in a car crash, how ironic.
     
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