Sklansky-Chubokov numbers

Jeg har netop skrevet dette spørgsmål til Sklansky på 2+2, så derfor er det på "engelsk" (beklager hvis mit dårlige engelsk er svært at forstå:-)).

Tager jeg fejl i nedenstående?

In "about this forum" you write that if anybody has a question for you they can sneak it in as a reply. I hope it is ok that i do it here.

The question is concerning your book No Limit Hold'em - Theory and Pratice page 216.

You write the following EV equation:
0=(0.935)($3)+(0.0645)[(0.433)($X+$3)+(0.567)(-$X)]

Shouldn't the part [(0.433)($X+$3)+(0.567)(-$X)] have been [(0.433)($X-$1+$3)+(0.567)(-$X)]?

Your opponent has already put a $2 big blind while you have only put the $1 small blind. When you push he therefore only has to call $X-$1.

For hands with high Sklansky-Chubukov numbers, like the AK hand in the example, this difference in the EV equation is of little practical importance, but i would guess that the difference in the Sklansky-Chubukov number for a hand like Q4 would be significantly affected by the extra $1.

Flueknepper ;)

holy shit , lol , så keder man sig vist.......
Mvh Søren

@bestbluff

Umiddelbart vil jeg give dig ret.

Sklansky skriver "[...] your stack is $X after posting small blind [...]".

Dvs. du raiser til $X+1 og da din modstander har lagt big blind skal han altså kun betale $X-1, så din gevinst hvis han kalder er 0.433 * ( $X-1 + $3).

Jørn

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