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#1
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correct my EV calculations
reviewing a hand, so ignore the flop bet for now... right now I am trying to correct a math blunder for the EV calculation of a turn semibluff.
PokerStars 1/2 Hold'em (9 handed) Hand History Converter Tool from FlopTurnRiver.com (Format: 2+2 Forums) Preflop: Hero is MP3 with 7[img]/images/graemlins/club.gif[/img], A[img]/images/graemlins/heart.gif[/img]. <font color="#666666">4 folds</font>, <font color="#CC3333">Hero raises</font>, <font color="#666666">1 fold</font>, Button calls, <font color="#CC3333">SB 3-bets</font>, <font color="#666666">1 fold</font>, Hero calls, Button calls. Flop: (10 SB) Q[img]/images/graemlins/club.gif[/img], 6[img]/images/graemlins/club.gif[/img], 9[img]/images/graemlins/spade.gif[/img] <font color="#0000FF">(3 players)</font> SB checks, <font color="#CC3333">Hero bets</font>, Button calls, SB calls. Turn: (6.50 BB) 8[img]/images/graemlins/diamond.gif[/img] <font color="#0000FF">(3 players)</font> SB checks, Hero checks, Button checks. Assuming Hero bets again on the turn... what percentage of the time does it have to be breakeven? here's what I did so far let x represent the % of the time that villian folds here. so x % of the time, hero wins 6.5 BB's our equation so far: 6.5x = 0 but 1-x amount of the time, hero will be called. (for simplicities sake, lets magically assume that 1 villian will always fold) that 1-x time hero gets called. lets give hero 9.5 outs to improve (8 for OESD, and lets give 1.5 for A outs, since they aren't completely clean) sooo, ignoring also the next betting round (again, to make this calculations easier), let's calculate the EV of getting called 9.5 outs with 46 unseen cards, he'll win 7.5 BB's, the remainder of the time, he'll lose 1 BB (9.5/46)*7.5 + (36.5/46)(-1) = ~.7554 BB brings our equation now to 6.5x + (1-x).7554 = 0 lets simplify it a bit 5.7446x + .7554 = 0 sooo [img]/images/graemlins/confused.gif[/img] I have a negative answer? |
#2
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Re: correct my EV calculations
You will never be able to solve this, because the way you put it you will win money any way, if he folds you win the 6.5BB and if he calls you have an expectation of 0,755BB, so there isnt a % of folding when this play is breakeven.
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#3
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Re: correct my EV calculations
doh!
ty... that explain it. and in other words, betting is unexploitable here? |
#4
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Re: correct my EV calculations
If all your assumptions are right, yes.
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#5
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Re: correct my EV calculations
And that's a very big IF.
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#6
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Re: correct my EV calculations
How about ignoring the entire pre flop play!! WOW!
LOL - you do know that is REAL loose right? sorry - nothing else to add. |
#7
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Re: correct my EV calculations
[ QUOTE ]
If all your assumptions are right, yes. [/ QUOTE ] what did I miss? even if 1 villian doesn't always magically fold, then it wouldn't change much. if we let x be % of times boht villians fold, and y being the average time each individual folds ( therefore, x = y*y), the second half would just be a little crazy, I just wanted to avoid messy calculations) the right half would go from (1-x).7554 to... lets see y^2 is the amount of times you win 6.5BBs 6.5(y^2) (1-y^2) of the time, you'll get 8.5 BBs (9.5/46) of the time, and lose 1 bb (36.5/46) of the time. (EV of ~ .962) 6.5(y^2) + .962(1-y^2) 2*(1-y) of the time, only one of them will call. (EV of .7554 each case) 6.5(y^2) + .962(1-y^2) + 2*(1-y)*.7554 = 0 simplifies to: 5.538y^2 - 1.5108y + 2.4728 = 0 still never hits zero even if we give ourselves 8 outs instead of 9.5, the simplified equation turns out to: 5.5434y^2 - .9566y + 1.3914 = 0 still g00t |
#8
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Re: correct my EV calculations
What you calculate, 6.5x + (1-x).7554 is the EV of betting. That EV is positive, because of the pot-size. You have to compare it to the EV of checking, (9.5/46)*6.5 (assuming no further action). So for betting to be right:
5.7446x + .7554 > (9.5/46)*6.5 => x > ( (9.5/46)*6.5 - 0.7554 )/5.7446 = 10% The chance that both players fold must be more than 10%. |
#9
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Re: correct my EV calculations
[ QUOTE ]
What you calculate, 6.5x + (1-x).7554 is the EV of betting. That EV is positive, because of the pot-size. You have to compare it to the EV of checking, (9.5/46)*6.5 (assuming no further action). So for betting to be right: 5.7446x + .7554 > (9.5/46)*6.5 => x > ( (9.5/46)*6.5 - 0.7554 )/5.7446 = 10% The chance that both players fold must be more than 10%. [/ QUOTE ] nice! this thread makes me happy in the pants |
#10
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Re: correct my EV calculations
Some additional things to think about: first, you can't count eight outs for your OESD because there's a flush draw out there. You have discounted your ace outs, but you may have to discount them even more because your kicker no longer plays if an ace hits the river, increasing the likelihood of a split. If you bet and it folds a bare ace you have just helped yourself, so a bet doesn't necessarily have to fold both opponents to have merit.
Sushi's analysis seems to make sense. |
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