@thinkingpoker @Tzen1 @Wizard0fAhhs @frosty012 (The thing that has less variance is the size of your stack at the time you would late-reg.)
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Replying to @NateMeyvis
@NateMeyvis@thinkingpoker @Wizard0fAhhs@frosty012 but the avg variance of the field is neutral so mathematically it's neutral?1 reply 0 retweets 0 likes -
Replying to @Tzen1
@Tzen1@thinkingpoker @Wizard0fAhhs@frosty012 Don't know what you mean by "average variance," but I don't think it could be zero.1 reply 0 retweets 0 likes -
Replying to @NateMeyvis
@NateMeyvis@thinkingpoker @Wizard0fAhhs@frosty012 less or more variance doesnt impact EV. On average the outcome is the same.1 reply 0 retweets 0 likes -
Replying to @Tzen1
@Tzen1@thinkingpoker @Wizard0fAhhs@frosty012 delta EV(stack) = 0 does not entail that delta EV(dollar value of stack) = 0.2 replies 0 retweets 0 likes -
Replying to @NateMeyvis
@NateMeyvis@thinkingpoker @Wizard0fAhhs@frosty012 other words. ICM = chipchop if stack is avg and no one busted yet.1 reply 0 retweets 0 likes
@Tzen1 @thinkingpoker @Wizard0fAhhs @frosty012 Whatever ICM says, under standard assumptions you lose $ if you experience 0-EV variance.
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