@St_Rev I have a math question - what is the reason why, for there to be a mixed strategy in a Nash situation, the strategies being mixed...
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Replying to @sarahdoingthing
@St_Rev ...have to be equal in expected value when considered independently?1 reply 0 retweets 0 likes -
Replying to @sarahdoingthing
@St_Rev (even though they are often mixed at a ratio other that 50/50)?1 reply 0 retweets 0 likes -
Replying to @sarahdoingthing
@sarahdoingthing Not sure. Here's a wild guess though: suppose they're at 60/40.1 reply 0 retweets 0 likes -
Replying to @St_Rev
@sarahdoingthing You can model THAT in turn as an 80% chance to have a 50-50 strategy, and a 20% chance to choose option 1.2 replies 0 retweets 1 like -
Replying to @St_Rev
@sarahdoingthing Then if you play against option 1, you have a 20% chance of winning and an 80% chance of being 50-50.1 reply 0 retweets 0 likes -
Replying to @St_Rev
@sarahdoingthing This is a completely wild guess, as I am not sure what a Nash strategy is.1 reply 0 retweets 0 likes -
Replying to @sarahdoingthing
@sarahdoingthing Can you give a concrete example? This seems strange.1 reply 0 retweets 0 likes -
Replying to @St_Rev
@St_Rev this is one of the many explanations that I have failed to deeply grok http://oyc.yale.edu/sites/default/files/mixed_strategies_handout_0.pdf …2 replies 0 retweets 0 likes
@sarahdoingthing In conclusion, I don't know the answer but 800 other people do.
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. Banned in Sweden. SubGenius, Zhuangist, white-hat troll. Defrocked mathematician. Brain problems.