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Don’t be a Sheep

Game

At my local bar trivia spot, there’s a game called Don’t be a Sheep. For the game, a question such as

Name a US President that was formerly a Vice President

is asked, and the goal is to choose an answer different from other teams’ answers. You get

Typically these questions have around 10-15 answers with around 15-20 teams participating.

Strategy

An intuitive strategy is to put the most obscure answer you can think of. For the above question, answers such as Martin van Buren, Millard Fillmore, and Calvin Coolidge might check that box. But more often that not, these answers don’t get 3 points and may even get 1 point, because it turns out people at trivia are quite smart and usually another team guesses it thinking they also found the answer no one else thought of.

What about then guessing the obvious answer for reverse psychology? From the trivia nights I have been to, this has worked a few times but has failed just as many when teams caught on. For example, one week the question was

Name 1 of the 10 last U.S. states to be admitted to the Union

An obvious answer to this question was Hawaii and it ended up being put by multiple teams.

Game Theory

Don’t be a Sheep is a game theory problem and it has a game theory optimal strategy i.e. Nash equilibrium. In the case where every team knows every answer, the equilibrium is for every team to choose the answer uniformly randomly. This seems to be a decent strategy in practice and it saves your team a lot of stress. However, if some teams don’t know an answer, how does the strategy change?

Let’s take a simplified scenario where there are only 3 teams and 3 answers. Additionally, let’s simplify the point structure so that you get 1 point if you have the unique answer but 0 points otherwise. In this situation, teams 1 and 2 know all 3 answers, but team 3 only knows answers 1 and 2. What is the Nash equilibrium of this game?

In the above diagram, each ? represents the probability that a team \(T\) chooses an answer \(A\). As probability sums up to 1, the rows should sum up to 1.

For the Nash equilibrum, symmetry makes things easier to compute. In this game, because \(A_1\) and \(A_2\) are identical from the POV of \(T_3\), they should choose \(A_1\) and \(A_2\) each with probability 1/2. Similarly, \(T_1\) should choose \(A_1\) and \(A_2\) with the same probability, but with what probability is unknown so we denote it with \(x\). For the probability that \(T_1\) chooses \(A_3\), that will be \(1 - 2x\) utilizing the fact that the row should sum up to 1. Finally, \(T_1\) and \(T_2\) should have the same strategies given they know the same answers, so \(T_2\) should have the same probabilities.

For the best value of \(x\), it is the one that maximizes the expected value of the payoff for \(T_1\) / \(T_2\). The expected value of the payoff for \(T_1\) is

\[\begin{align*} E[\text{Payoff for $T_1$}] &= 1 * P(\text{$T_1$ has unique answer}) + 0 * P(\text{$T_1$ does not have unique answer}) \\ &= P(\text{$T_1$ has unique answer}) \end{align*}\]

For the probability that \(T_1\) has a unique answer, it is illustrated in the diagram and can be calculated as

\[\begin{align*} P(\text{$T_1$ has unique answer}) &= \sum_{i=1}^3 P(\text{$T_1$ guesses $A_i$ and no on else guesses $A_i$}) \\ &= x(1-x)(1/2) + x(1-x)(1/2) + (1-2x)(2x) \\ &= x(1-x) + (1-2x)(2x) \\ &= x - x^2 + 2x - 4x^2 \\ &= 3x - 5x^2 \end{align*}\]

\(3x - 5x^2\) is a downwards-facing parabola, and its maximum can be found by setting the derivative to 0. The answer is \(x = 0.3\).

Conclusion

In the version above, the ideal strategy was to select answers 1 and 2 30% of the time and answer 3 40% of the time. The uniform random strategy would be to select each answer 33.3% of the time, but the ideal strategy biases towards the obscure answer a bit. So what game theory tells us is that we should be random but bias towards obscure answers.

Despite this though, I still like uniformly at random. You can be wrong about what’s obscure and also teams in general will be over-biased towards obscure answers. Going more uniformly random counteracts that. Trust the dice!