AdShea comments on Swords and Armor: A Game Theory Thought Experiment - Less Wrong

14 Post author: nick012000 12 October 2010 08:51AM

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Comment author: rosyatrandom 12 October 2010 11:03:37AM *  14 points [-]

You mean this table? :)

(This and the one I made below can be seen properly at http://tinyurl.com/lwgttable , along with the ATT vs DEF tables I worked out the outcomes from)

Hmm. Unless this has gone wrong, the best combo is Sword 1 and Armour 4, with Sword 1/Armour 1 being close). But if you bank on people choosing 1/4, then 1/1 will beat them.

NB: Yes, I made a lot of mistakes and edits to get here, and probably have still made some...

VS a1 a1 a1 a1 a2 a2 a2 a2 a3 a3 a3 a3 a4 a4 a4 a4
s1 s2 s3 s4 s1 s2 s3 s4 s1 s2 s3 s4 s1 s2 s3 s4
a1 s1 0.5 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1
a1 s2 1 0.5 1 1 1 0.5 1 1 1 1 1 1 1 1 1 1
a1 s3 1 0 0.5 0 1 1 1 1 0 0 0 0 1 1 1 1
a1 s4 1 0 1 0.5 0 0 0 0 1 1 1 1 0 0 0 0
a2 s1 0 0 0 1 0.5 0 0 1 0 0 0 1 1 1 0 1
a2 s2 1 0.5 0 1 1 0.5 0 1 1 1 1 1 1 1 0 1
a2 s3 1 0 0 1 1 1 0.5 1 0 0 0 0 1 1 1 1
a2 s4 1 0 0 1 0 0 0 0.5 1 1 1 1 0 0 0 0
a3 s1 1 0 1 0 1 0 1 0 0.5 0 1 0 1 0 1 0
a3 s2 1 0 1 0 1 0 1 0 1 0.5 1 0 1 0 1 0
a3 s3 1 0 1 0 1 0 1 0 0 0 0.5 0 1 1 1 0
a3 s4 1 0 1 0 0 0 1 0 1 1 1 0.5 0 0 0 0
a4 s1 0 0 0 1 0 0 0 1 0 0 0 1 0.5 0.5 0 1
a4 s2 0 0 0 1 0 0 0 1 1 1 0 1 0.5 0.5 0 1
a4 s3 0 0 0 1 1 1 0 1 0 0 0 1 1 1 0.5 1
a4 s4 0 0 0 1 0 0 0 1 1 1 1 1 0 0 0 0.5
_ _ 10.5 1 6.5 9.5 9.5 4 6.5 9.5 8.5 7.5 8.5 9.5 11 9 7.5 9.5
Comment author: AdShea 12 October 2010 03:08:01PM 2 points [-]

That looks like what I got. It'd be interesting to do a cycle analysis like you do with Non-transitive Dice.

Comment author: Jordan 13 October 2010 12:50:46AM 2 points [-]

Thanks for the mini mind blow with the non-transitive dice link.

Can you elaborate on what cycle analysis is?

Comment author: AdShea 13 October 2010 03:04:10PM *  3 points [-]

Cycle analysis is basically drawing out a graph (nodes and edges) of what beats what. For standard Rock-paper-scissors you get a graph that looks like this:

ROCK -------> SCISSORS -------> PAPER

^-----------------------------------------------|

In systems that aren't balanced like the sword and armor problem here you can use it to decide what choice to make by giving each node a probability value based on how many people in game use that and then the best choice would be the node with the greatest (sum of probabilities on outgoing nodes) - (sum of probabilities on incoming nodes).

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