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I am having some difficulty in solving the following problem. I was wondering whether someone would be kind enough to sketch a solution or even better to solve the whole game. Thanks

Suppose that 100 people live in a village of whom 51 support the conservative candidate and 49 support the liberal candidate. Villagers get a payoff of 10 if their candidate gets elected, -10 if the other one wins. Voting costs villagers 1 unit. Those who do not vote evade these costs but get the payoff related to the winning candidate ( +/- 10).

  • Find a Nash Equilibrium (in mixed strategies) in which all conservatives use the same strategy and all liberals use the same strategy. In case of equal number of votes toss of a coin decides. (So for instance in the case of equal number of votes, if villager i is voting, then its payoff would be of 1/2*(10-1); while if villager i is NOT voting, then its payoff would be of 1/2*10)
  • What is the expected number of villagers who will vote in this equilibrium?
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  • $\begingroup$ Dear Paolo. Welcome to math.SE. You can use Mathjax to make your formulas such as 1/2*(10-1) look like $\frac{1}{2}(10-1)$. In this case you would need to enclose the formula between two dollar signs and add a little bit of code : \$\frac{1}{2}(10-1)\$. For more on using Mathjax, see meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Martin Van der Linden Dec 11 '13 at 19:08
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    $\begingroup$ On MSE (math stack exchange), it is good practice to give some explications about where exactly you are having difficulties in solving the problem. This helps a lot in getting useful answers. Without knowing what you have tried, I would remind you the crucial property of mixed Nash equilibria : if a mixed strategy is an equilibrium one, then it must be that, given the probability that others assign to their action (here vote of not vote), the agent is indifferent between playing either of his actions. Does that help or does it have nothing to do with your difficulties in solving the problem? $\endgroup$ – Martin Van der Linden Dec 11 '13 at 19:17
  • $\begingroup$ Thanks for the reply. So I guess from the code you posted that I can use what little of LaTeX markup I know. Regarding the game, I can tell you what I've tried but it isn't really much help. My "guess" is that a NE equilibrium in mixed strategies of this specific game will look like: all 49 villagers who support the liberal candidate will vote, while the 51 villagers who support the conservative candidate will randomize. But I cannot find a probability distribution for the "conservatives" over the action space \{Vote,Not Vote\} such that as you said no one has profitable deviations. $\endgroup$ – Paolo Dff Dec 12 '13 at 17:26
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    $\begingroup$ @PaoloDff The tie payoffs $\frac{1}{2}(10-1)$ and $\frac{1}{2}10$ don't make sense to me. Are you sure you copied this right? $\endgroup$ – Keep these mind Dec 12 '13 at 17:41
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    $\begingroup$ Indeed the tie payoffs make no sense. If a fair coin decides the winner in case of a tie, then the payoff for a tie should be the average of the payoffs for the two coin results. That would be $0$ for a non-voter and $-1$ for voter. $\endgroup$ – joriki Sep 27 '15 at 15:42
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sketch a solution

Suppose that the assumption (made in a comment) that all liberals (with probability $q=1$) vote is correct.

Then a conservative will be indifferent between voting and non-voting if the cost and the gain from voting balance each other out. The cost of voting is $1$. The gain of voting is either $0$ or $10$. It is only $10$ if his vote is either the tying or winning vote. (Note that I am ignoring the question's statements regarding payoffs in a tie; I think they are wrong and I've taken a more logical route.)

Let $p$ be the (independent) probability that a conservative votes. The probability that his/her vote would be the tying one is $\binom{50}{48}p^{48}(1-p)^2$. The probability that his/her vote would be the winning one is $\binom{50}{49}p^{49}(1-p)$.

So, we look to solve $$1=10\left[\binom{50}{48}p^{48}(1-p)^2+\binom{50}{49}p^{49}(1-p)\right].$$

Being lazy, we plug this into WolframAlpha, and lo and behold, we get two solutions: $p\approx0.898255$ and $p\approx0.997891$.

This is somewhat unexpected. However, I am almost completely sure that with $p\approx0.898255$, the liberals' strategy $q=1$ wouldn't be optimal anymore. So, we can ignore that case.

This leaves us with showing that, given $p\approx0.997891$, $q=1$ is indeed Nash. For this we need to show that one liberal will not be better off (in expectation) by non-voting.

And (more laziness) that's where this half of the answer ends... It requires more work, but my guess is that it would turn out just fine. (And the expected number of voters in this, yet to be established, Nash equilibrium would be $51p+49\approx99.892441$.)

I'm actually not that sure. Anyway, perhaps commenters may provide some further insight.

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  • $\begingroup$ Regarding the payoffs, I re-checked the text that I have and they are what I posted. I can re-use your solution @Transmissionfrom but the problem was from a game theory exam which I have to solve by hand. So I am still wondering if there is a "trick" to further simplify the problem. My assumption that all liberals will vote is arbitrary and maybe not the "correct" one. $\endgroup$ – Paolo Dff Dec 14 '13 at 19:08
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    $\begingroup$ @PaoloDff In that case, you needn't accept my answer (as you did). I think you can de-accept it. $\endgroup$ – Keep these mind Dec 14 '13 at 19:12
  • $\begingroup$ I think your calculation is correct (assuming the more sensible tie payoffs $0$ for non-voters and $-1$ for voters). But you picked the wrong solution. For $p\approx0.997891$, there's an almost $90\%$ chance an individual conservative vote doesn't count because the conservatives win anyway, whereas for $p\approx0.898255$ there's an almost $90\%$ chance an individual conservative vote doesn't count because the conservatives lose anyway. In the former case, the almost certain vote of the $51$-th conservative makes it $99.5\%$ likely that the conservatives win, so the liberals wouldn't vote. $\endgroup$ – joriki Sep 27 '15 at 19:14
  • $\begingroup$ By contrast, in the latter case, the probability for $48$ conservatives to vote is $\approx0.127147$, and the probability for $49$ conservatives to vote is $\approx 0.0687257$, for a total of almost $20\%$ probability that a single missing liberal vote would matter, so here we have a Nash equilibrium with conservatives voting with probability $p\approx0.898255$ and all liberals voting (with approximately $94.8$ voters expected to vote and the result being a liberal win with $90.3\%$, a tie with $6.9\%$ and a conservative win with $2.8\%$ probability). $\endgroup$ – joriki Sep 27 '15 at 19:15
  • $\begingroup$ Thus, somewhat paradoxically, the liberals are likely to win because they have fewer voters. $\endgroup$ – joriki Sep 27 '15 at 19:15

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