formula for calling bets based on outs

imported_bingobazza · #1 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 · #2 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 [img]/images/graemlins/diamond.gif[/img] A [img]/images/graemlins/diamond.gif[/img] and the flop is 5 [img]/images/graemlins/diamond.gif[/img] 9 [img]/images/graemlins/club.gif[/img] J [img]/images/graemlins/spade.gif[/img]-- 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 · #3 07-31-2005, 05:01 PM

Helpful....

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 · #4 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 · #5 07-31-2005, 06:17 PM

No implied odds, just actual pot odds would be great for now.

imported_bingobazza · #6 08-01-2005, 11:00 AM

Cant anyone do this....maybe Im not as stupid as I thought?

Bingo

shday · #7 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 · #8 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 · #9 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 ]

Your maths is great...but your reading could use some work. Detention after school.

Bingo

imported_bingobazza · #10 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