A rock paper scissors lizard spock tournament

In class today we gave a quick summary of what we have seen so far, ran a Rock Paper Scissor Lizard Spock tournament (Tom took my candy again) and talked over the support enumeration algorithm which is the important part of the Nash equilibrium chapter.

A recording of the class is available here.

Summary of the first week

I asked you all to talk amongst yourselves and remind each other what we saw last week. We came up with:

A rock paper scissors lizard tournament

After this we moved on to class rock paper scissors lizard spock tournament.

The first time I heard of this game (a variation of rock paper scissors) was in an episode of the Big Bang Theory. You can find a clip of it here: YouTube and a summary of the rules are available here.

Before we started I asked us to modify the rules so that you could only choose from 2 actions:

This was not interesting: everyone just picked Lizard. Adding Rock (or indeed any of the other actions) adds a best response.

After everyone played their tournaments (a knock out with a final) following that there was a “last player standing” session which lead to the grand final. Tom was victorious here and we had a conversation about what made someone be good at Rock Paper Scissors Lizard Spock.

Quite quickly we got to the idea of being unpredictable as opposed to knowing something about your opponent (Peter mentioned that apparently we are all more likely to play Rock first).

The Support Enumeration Algorithm

I talked over the Support Enumeration algorithm in the Nash equilibrium chapter. This is an algorithm that uses the Best Response Condition from the previous chapter to systematically check all places a Nash equilibrium could be.

The first step of the algorithm is to consider all potential pairs of supports of a strategy: this corresponds to considering all possible allow actions.

In the case of Rock Paper Scissors where there are 3 actions overall this gives: \(3 + 3 + 1=7\) actions (3 support with just one action, 3 supports with one omitted action and 1 support with all actions). The algorithm requires us to consider all pairs of potential supports which is technically \(7 ^ 2 = 49\) although practically, as we will see, we will do much less.

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