PDA

View Full Version : formula for calling bets based on outs

imported_bingobazza
07-31-2005, 05:48 AM
Hi guys,

Im a bit of a donk when it comes to maths problems, but Ive been trying to write a formula for the following, assuming we are on the flop

I have 'a' chance of hitting my outs on the turn (written as 0.2 rather than 20%, for example)
The pot size is 'P'
'n' is the number of callers i think Im going to get.

Im trying to solve the equation for the max bet I can make to give me the correct odds for my chance of hitting my outs, but Im a tiny bit out with this, and I cant correct for it.

Im sure this is a clumsy, and rudimentary post for many of you to read, but any help is appreciated.

Bingo

LesWormMurphy
07-31-2005, 01:01 PM
You shouldn't just count on your outs to decide between folding or calling. If the pot is 15BB on the flop and it costs you one bet to see the turn, even if you only have 2 outs, you better be in there.

But you must also be careful when you count your outs because sometimes an out won't give you the best hand.

For example, if you have Q /images/graemlins/diamond.gif A /images/graemlins/diamond.gif and the flop is 5 /images/graemlins/diamond.gif 9 /images/graemlins/club.gif J /images/graemlins/spade.gif-- you have 6 outs to pair either your Queen or Ace to give you top pair. In addition you have a backdoor flush draw and even a gutshot-backdoor straight draw. All together you have 9 outs, but you should probably lower the value because top pair won't win often, and because the flop is slightly coordinated. So give yourself 6-8 outs, depending, and act accordingly.

Hope it helped.

imported_bingobazza
07-31-2005, 05:01 PM

If you could also show my error here for calculating the max bet (X).....

X = 1 + (2*%chance of hitting out)*pot size*% chance of hitting out

Then that would be really helpful as well.

Cheers

Bingo

spaminator101
07-31-2005, 05:26 PM
are you going to leave implied odds out of the question id say that on the flop implied odds are more important than pot odds

imported_bingobazza
07-31-2005, 06:17 PM
No implied odds, just actual pot odds would be great for now.

imported_bingobazza
08-01-2005, 11:00 AM
Cant anyone do this....maybe Im not as stupid as I thought?

Bingo

shday
08-01-2005, 12:56 PM
[ QUOTE ]
If you could also show my error here for calculating the max bet (X).....

X = 1 + (2*%chance of hitting out)*pot size*% chance of hitting outs

[/ QUOTE ]

...I couldn't figure out how you came up with that.

Here is what I get (where the pot size includes the max bet):

max bet = (pot size)(chance of hitting outs)

or, if you do not include the max bet in the pot size (what you seem to want):

max bet = (pot size)(chance of hitting outs)/(1 - chance of hitting outs)

TaintedRogue
08-02-2005, 09:37 AM
I think we are making it harder than it is.......
If you have 5 outs after the flop......
I use my multiplication table, which I've known by memory since grade school.
9*5 outs =45 and there are 47 unseen cards, so 2 left over divided by 5 = 4 because 5*4 = 20. 5 goes into 47 9.4 times, of which I'll make my hand once, so the odds of me hitting my hand on the turn are 8.4:1
I can do this in my head, at the table, if I've forgotten the odds of a 5 out'r.
I use to have to do that, but not anymore, I've memorized them all.
So, on a 5 out'r, if there are 8 bets in the pot and you believe you will get at least one call on the turn, I'll peel one if I'm drawing to the nutz when I can expect one call, because my implied pot odds are greater than my chances of making my hand.
You need to know how to discount outs though, such as two overcards. See Small Stakes Hold'Em for that.

imported_bingobazza
08-02-2005, 11:55 AM
[ QUOTE ]
I think we are making it harder than it is.......
If you have 5 outs after the flop......
I use my multiplication table, which I've known by memory since grade school.
9*5 outs =45 and there are 47 unseen cards, so 2 left over divided by 5 = 4 because 5*4 = 20. 5 goes into 47 9.4 times, of which I'll make my hand once, so the odds of me hitting my hand on the turn are 8.4:1
I can do this in my head, at the table, if I've forgotten the odds of a 5 out'r.
I use to have to do that, but not anymore, I've memorized them all.
So, on a 5 out'r, if there are 8 bets in the pot and you believe you will get at least one call on the turn, I'll peel one if I'm drawing to the nutz when I can expect one call, because my implied pot odds are greater than my chances of making my hand.
You need to know how to discount outs though, such as two overcards. See Small Stakes Hold'Em for that.

[/ QUOTE ]

Bingo

imported_bingobazza
08-02-2005, 12:12 PM
[ QUOTE ]

or, if you do not include the max bet in the pot size (what you seem to want):

max bet = (pot size)(chance of hitting outs)/(1 - chance of hitting outs)

[/ QUOTE ]

This error seems to be along the same lines of the error Im making. For example, if there is \$35 in the pot, and the chance of hitting the out is, lets say, 13%, then

Max bet = 35*.13/1-.13
4.55/.87
Max bet is 5.2298.

Assuming there is a caller, then the pot size after the caller is 5.2298*2 + 35 = 45.45

But 13% of 45.45 is 5.9085, not 5.2298. My original formula gets you closer, but is still wrong.

Im trying to make a bet that results in the bet being the identical % of the resultant pot (with n callers) to the % chance of me hitting my outs on the next card.

Bingo

jba
08-02-2005, 01:01 PM
[ QUOTE ]

Im trying to make a bet that results in the bet being the identical % of the resultant pot (with n callers) to the % chance of me hitting my outs on the next card.

[/ QUOTE ]

so you want:

chance of hitting = bet / pot + bet

which is

bet = chance * pot / (1-chance)

so if you have a .25 chance of hitting in a 10 bet pot

bet = .25 * 10 / .75

bet = 3.333

there's 10 in the pot, you bet 3.3333 your opponent calls the pot is now 13.33333. 3.3333 = 13.33333*.25

shday
08-02-2005, 01:20 PM
ok, if you want to factor in the number of callers:

bet = chance*pot/(1 - chance(1 + callers))

obviously, if there are no callers this reduces to the previous equation.

imported_bingobazza
08-02-2005, 02:18 PM
[ QUOTE ]
[ QUOTE ]

Im trying to make a bet that results in the bet being the identical % of the resultant pot (with n callers) to the % chance of me hitting my outs on the next card.

[/ QUOTE ]

so you want:

chance of hitting = bet / pot + bet

which is

bet = chance * pot / (1-chance)

so if you have a .25 chance of hitting in a 10 bet pot

bet = .25 * 10 / .75

bet = 3.333

there's 10 in the pot, you bet 3.3333 your opponent calls the pot is now 13.33333. 3.3333 = 13.33333*.25

[/ QUOTE ]

If I bet 3.3333 into a 10 pot, and the op calls, theres 16.6666666 in the pot....

Bingo

jba
08-02-2005, 02:28 PM
[ QUOTE ]

If I bet 3.3333 into a 10 pot, and the op calls, theres 16.6666666 in the pot....
Bingo

[/ QUOTE ]

are you sure you want to be counting your own bet??

imported_bingobazza
08-02-2005, 02:52 PM
I take your point, but yes, I do want to count my bet, as well as their call(s). Its not my money once its gone into the pot. Lets say I have a 25% chance to hit my outs on the turn. Clearly, in a \$100 pot, having a maximum bet parameter of 25% of 100 (pot) or 125 (pot plus my bet) is incorrect if there is a caller. It should be higher to account for the call, and probably even higher than that if fold equity is considered...but we'll ignore that for now. I appreciate your input here.

Bingo

jba
08-02-2005, 03:05 PM
[ QUOTE ]
I take your point, but yes, I do want to count my bet, as well as their call(s). Its not my money once its gone into the pot. Lets say I have a 25% chance to hit my outs on the turn. Clearly, in a \$100 pot, having a maximum bet parameter of 25% of 100 (pot) or 125 (pot plus my bet) is incorrect if there is a caller. It should be higher to account for the call, and probably even higher than that if fold equity is considered...but we'll ignore that for now. I appreciate your input here.

Bingo

[/ QUOTE ]

well you sorta lost me there, but if that's what you want...

chance of hitting = bet / (pot + 2*bet)

which is

bet = (pot * chance)/(1-2*chance)

so if you have a .25 chance of hitting in a 10 bet pot

bet = .25 * 10 / .5

bet = 5

there's 10 in the pot, you bet 5 your opponent calls the pot is now 20. 5 = .25 * 20

do yourself a favor and do some EV calculations on this system before you put it to work.

imported_bingobazza
08-02-2005, 03:33 PM
Yip, I hit that one too....but it breaks down &gt;50% chance.

Bingo

jba
08-02-2005, 03:52 PM
[ QUOTE ]
Yip, I hit that one too....but it breaks down &gt;50% chance.

Bingo

[/ QUOTE ]

yes it does. do you see why?

how could your bet ever be over half the pot with an opponent matching you dollar for dollar, and existing dead money in the pot?

imported_bingobazza
08-02-2005, 04:12 PM
Ofcourse!!!!! thanks jba, couldnt see the wood for the trees. I appreciate all your help on this one. Have another think about the EV and PM me if you like.

Bingo