Sizing bet sizing to plan next street size. - Poker Theory

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Sizing bet sizing to plan next street size.

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02-19-2011 , 03:19 PM

KodoBeast00

newbie

Join Date: Feb 2011 Posts: 20

I assume everyone know that if you want to bet All-In next street for a pot size bet.

You take the smallest stack minus the pot size divided by 3.

But now lets say I wanna bet diffent amount on the river that put my opponent or myself all-in..
Pot is 20$ I want to bet 3/4Pot All-In next street,

Pot is 20$ I want to bet 1/2Pot All-In next street,
How do I proceed?

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02-19-2011 , 03:34 PM

powder_8s

temp-banned

Join Date: Mar 2010 Posts: 3,782

I think you can simplify your thinking if you use the effective stack size at the start of the hand.

-If you want a pot size bet on the river. You need to have 1/3 of the effective stacks in before the river. Ex. $100 effective stacks. $33 from each stack in on the turn puts $66 in the pot leaves a full pot size bet to get it all in on the river.

-For 3/4 pot size bet you need to have 40% in on the turn. $40 +$40 = $80 in the pot and leaves a $60 or 3/4 pot size bet.

-For 1/2 pot you need to have 1/2 of effective stacks in on the turn. $50 + $50 = $100 with a $50 bet left for the river.

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02-19-2011 , 03:40 PM

KodoBeast00

newbie

Join Date: Feb 2011 Posts: 20

That doesnt seem to work..

If the pot is 100$

effective stack sizes are 300$

If we follow your logic we have to bet 1/3 of the effective stack size to bet pot on next street. but if we bet 100$ here, the pot willbe 300$ with 200$ left to bet for only a 2/3pot bet.

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02-19-2011 , 03:58 PM

powder_8s

temp-banned

Join Date: Mar 2010 Posts: 3,782

I said,"use the effective stack size at the start of the hand"

Your example you did not use starting stack sizes. In you example starting stack sizes would have been $350. If it is heads up and the pot is $100 after the flop you need to bet $125 on the turn to have put in a total of $175 or 50% of effective starting stacks. This leaves a 1/2 pot size bet for the river.

This is a meant as a simple formula that will get you close. It assumes its heads up and there is no dead money in the pot. If you cant remember the starting stack sizes. You can get close enough by, dividing the pot but the number of opponents and add that to their stack.

Last edited by powder_8s; 02-19-2011 at 04:14 PM.

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02-19-2011 , 04:13 PM

KodoBeast00

newbie

Join Date: Feb 2011 Posts: 20

Oh I get it.. I guess thats a good way to find a good bet size.

But thats only practical if you remember the starting stacks sizes. That can be difficult if you did bet the flop and now you try to figure out what size to bet on the turn.

Maybe its actually easier then an algebra formula

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02-19-2011 , 04:20 PM

powder_8s

temp-banned

Join Date: Mar 2010 Posts: 3,782

Quote:

Originally Posted by KodoBeast00

Maybe its actually easier then an algebra formula

Thats why I came up with this. I was struggling to use more complex formulas. I wanted something I could do easily in my head. I am not worried if the math is not perfect. Lets say there is some dead money in the pot. I might leave 79% of the pot bet instead of 75%, on the river. Big deal. Its close enough for me.

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02-19-2011 , 05:14 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 5,047

Here is the general equation:

Bet = (S-Pot*f)/(2f-1),

where S =effective stack at time of bet
Pot = pot size at time of bet
f = fraction of new pot to bet next street

If S = 100 and Pot = 50

For f = 1/2, Bet = 37.5
For f= 3/4, Bet = 25
For f = 1.0, Bet = 16.7 [Stack- Pot]/3
For f = 1.5, Bet = 6.25

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02-19-2011 , 05:15 PM

nonunique

enthusiast

Join Date: Feb 2011 Posts: 80

Edit: scooped by statmanhal.

The algebra (heads up) is:
pot + 2*bet = x*(stack - pot), where x is the inverse of the fraction of the pot you want to be shoved on the NEXT street (1/2 pot makes x = 2, 1/3 pot makes x = 3, etc). Solve for bet.

x = 1: bet = (stack - 2*pot)/2 = 0.5*stack - pot
x = 2: bet = (2*stack - 3*pot)/2 = stack - 1.5*pot
x = 3: bet = (3*stack - 4*pot)/2 = 1.5*stack - 2*pot
x = 4: bet = (4* stack - 5*pot)/2 = 2*stack - 2.5*pot

or

bet = (x/2)*stack - ((1+x)*pot)/2

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02-23-2011 , 12:06 AM

King Midas III

enthusiast

Join Date: Feb 2011 Posts: 79

Guys none of this math can exist in a vacuum. The truth is that psychology is all that matters in deciding whether or not to go all in. Math has nothing to do with such a decision.

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02-23-2011 , 08:05 PM

#10

MU MUF MUFC OK

grinder

Join Date: Jan 2010 Posts: 438

Quote:

Originally Posted by King Midas III

Guys none of this math can exist in a vacuum. The truth is that psychology is all that matters in deciding whether or not to go all in. Math has nothing to do with such a decision.

op isn't trying to decide whether or not to go all in. and what do you mean by math can't exist in a vacuum? it dosn't make any sense.

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