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Questions regarding Sklansky bucks calculation...
I was wondering on how to do a Sklansky bucks calculation of a given hold 'em hand but couldn't find good references.
Let's make an example hand up... Assume players A, B & C. Stack sizes irrelevent. Player A post small blind $1, B posts big $2. Hole cards: A: 6[img]/images/graemlins/spade.gif[/img] 7[img]/images/graemlins/spade.gif[/img] B: 2[img]/images/graemlins/club.gif[/img] 2[img]/images/graemlins/diamond.gif[/img] C: A[img]/images/graemlins/heart.gif[/img] A[img]/images/graemlins/diamond.gif[/img] Preflop is C raises to $10, both A & B call. Equities (apx.): A[67]: 20% B[22]: 20% C[AA]: 60% Pot: $30 Common sense says that because C has equity of 60% on a $30 pot ($18), and has invested $10, then C profits $8 and A, B lose $4 each. Is this wrong? Additionaly - do we substract the amount of blind money from money lost for both blinds? Let's take an imaginary flop. 4[img]/images/graemlins/spade.gif[/img] 2[img]/images/graemlins/spade.gif[/img] 9[img]/images/graemlins/heart.gif[/img] A checks, B checks, C bets $20, A & B call $20. Equities, considering all players see turn & river are apx.: A: 30% ($27) B: 65% ($58.5) C: 5% ($4.5) Pot: $90 It is clear that C loses most of his bet. It also seems that A (spade draw) profits 7 sklansky bucks just for calling. But what about B (set)? Considering the fact that C loses most of his bet, A profits a bit, how can B profit more than the left-overs of C's lost bet? Should the money already in the pot change the calculation and if so is it correct to simply say that for this street B profits 38.5 sklansky bucks? This seems akward to me so I'd like to hear your opinions/corrections on that. The imaginary turn... 4[img]/images/graemlins/spade.gif[/img] 2[img]/images/graemlins/spade.gif[/img] 9[img]/images/graemlins/heart.gif[/img] Q[img]/images/graemlins/spade.gif[/img] A checks, B checks, C checks. Equities: A: 75% ($65~) B: 25% ($25~) C: 0% On this street no money is invested and player A with newly made flush does not bet to protect his hand. To protect it, player A has to bet at least half the pot ($45). The question is: how much sklanksy bucks does player A lose? Computing sklansky dollars for last street decisions seems simple (should diregard money already in pot IMO, correct me if I am wrong though). Does this method compute or am I way off here? Also, and more importantly - what are good references for this subject? Thanks in advance. |
#2
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Re: Questions regarding Sklansky bucks calculation...
Bump. Anyone?
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