raising with draws - how much FE needed?

[Matt Flynn]

< applause >

nice!

you can get by with a couple memorized numbers. but this is more fun.

someone make an SPR script already and plug it into PAHUD with this fold equity formula.

[Galwegian]

Here is the general formula. I used the rule of 4 approximation for calculating equity on the flop.

If
P = pot size on the flop
S = stack size
B = opponents bet size
n = number of outs that you have

let c = ((25-2n)S-P)/((25-2n)S+(25-n)P+25B)

If your opponent will fold to an all in reraise more than c of the time then the all in reraise is better then folding. You can easily rewrite this in terms of spr, as long as you replace bet size by bet-to-pot ratio. Of course this formula is beyond most people's real time arithmetic skills (certainly is for me anyway).

Something that is interesting to note about these formula is that the amount of fold equity required gets smaller as the size of the all in reraise gets smaller. This seems a bit counterintuitive at first, until you observe that for small all in reraises your opponent will be getting good pot odds to call, so the amount of FE that you actually have gets smaller as your stack gets smaller. To find the right time to use this play you must balance the two things. Your stack must be big to get FE but also must be small enough to ensure that you don't need too much fold equity for the all in to be +ev. The formula in this thread suggest that an spr of between 3 and 7 (spr above will often mean that the villain will have to fold more than 50% of the time) is the optimal range for this all in reraise play to succeed. Of course this depends very much on the precise situation.

[Galwegian]

Sorry correction. Should have read

c = ((25-2n)S-nP)/((25-2n)S+(25-n)P+25B)

[binions]

[ QUOTE ]
Binions , just replace P=B in the original formula and divide top and bottom by P .

x> (SPR-1)/(SPR+5)


For 25% , use the formula I provided and again replace P=B

x>(2S-P)/(2S+3P+4B)
x> (2SPR-1)/(2SPR +7)

For 20% , it is

x> (3S-P)/(3S+4P+5B)
x> (3SPR-1)/(3SPR+9)

If villain bets 2/3 of the pot , then you can replace 2/3P=B

for 33% it is :

x> (SPR-1)/ (SPR+4)

Pretty cool , huh ?

[/ QUOTE ]

The formula for 40% equity facing a pot bet is (1/2SPR - 1) / (1/2SPR + 4)

At 50%, you need no fold equity to make the push breakeven.

[Tictac]

How can you guys do these calculations in the small time we have to act? Is it merely practice practice practice or we all need to have a good mathematics background?

[jay_shark]

Hey Tic , go back to my formula in simplified form .
It's not too difficult to work out 6/(SPR+5) .Anyone can do it and it's going to be the case most of the time so it's worthwhile that you memorize it .

Also remember that this only applies when you have a 33% equity and your opponent bets the pot . Otherwise , you will have to make slight adjustments to the formula .

[binions]

[ QUOTE ]
How can you guys do these calculations in the small time we have to act? Is it merely practice practice practice or we all need to have a good mathematics background?

[/ QUOTE ]

There's actually some symmetry to it:

For 20% equity facing a pot sized bet, villain needs to fold more than:

3SPR-1/3SPR+9

For 25%:

2SPR-1/2SPR+7

For 33%:

SPR-1/SPR+5

For 40%

(.5*SPR)-1/(.5*SPR)+4

For 50%:

You need zero fold equity to break even.

When you face a 2/3 pot bet, you simply add a little less to the denominator (ie at 33% its SPR+4 instead of SPR+5).

I have a feeling this will become almost as easy to remember and apply at the table as using the rule of 4.

After all, 3SPR (+9) at 20%, 2SPR (+7) at 20%, 1SPR (+5) at 33% and 0.5SPR (+4) at 40% is a clear pattern. Every time you jump the a categories, you lose 1 SPR and +2 off the demominator, except when you jump from 33 to 40, you lose half SPR and +1 off the denominator. But its the same ratio. If SPR = a +2 loss, then half SPR = +1 loss.

[Dismas]

[ QUOTE ]
I'm sure that many of the more mathematically inclined posters might have
worked this out for themselves, but I thought that some people might
find the following useful. If you don't like math, you probably
don't want to read this. If you are a math geek (like me), you might
check my calculations to see if I am right. I worked this out while
I was analyzing a play that I had made in a live game last night and
I came up with a nice little formula that people might find helpful.
The basic situation is the following.

In NL holdem you have a 8/9 out draw on the flop and your single
opponent bets into you. How much fold equity do you need to make it
correct to reraise all in?

Say that

P = pot on the flop

S = effective stack

B = opponents bet


Now E(fold) = S. That is, the expected value of your stack given
that you fold is S. We want to compare this with E(all in) - the
expected value of your stack if you raise all in. Say that your
opponent will fold x of the time and call 1-x of the time (0<= x <=
1). We want to know how big x must be in order to make raising all
in at least as good as folding.

I will assume that if he calls, you have a 1 in 3 chance of winning
the hand. This is approximately true for many common draws (e.g
flush draw, oesd with a backdoor flush draw)

Then

E(all in) = x(P+B+S)+1/3(1-x)(2S+P)

since if opponent folds, your stack will be P+B+S and if he calls,
your stack will on average be 1/3(2S+P). Therefore raising all in is
better than folding if

x(P+B+S)+1/3(1-x)(2S+P) > S

Rearranging this yields

x>(S-P)/(S+2P+3B)

The nice thing about this is that this formula is just about useable at the table (there is only one division involved - my real time arithmetic skills struggle to cope with most formulae that involve lots of divisions).

As an example, here is the hand that prompted me to work this out.

I was at a final table of the local cardroom tourney and I picked up
Th9h in the big blind. There was 12K in the pot and the flop was
QhJs5c. I had 29K behind and I checked to my opponent who bet 7K
into the pot (he had me covered). So S=29, P = 12 and B=7. Therefore
(S-P)/(S+2P+3B) = 17/74 which is about 0.23.

Thus I needed him to fold at least 1/4 of the time to make the all
in raise better than folding (at least in terms of tournament
chips). Given my knowledge of this particular opponent, I think that
he won't fold 1/4 of his range here so I should have folded. In the
actual event, I reraised all in and busted from the tourney when he
called with QT.

This formula only works if you have close to a 1 in 3 chance of
winning when called, but it would not be hard to modify to other
drawing situations as well. Any comments?

[/ QUOTE ]

I don’t know if this is any easier but this is what I came up with a while back to calculate required fold equity.
Called = total pot if called.

W = Stack if (called * Equity) - Initial Stack
Fold % required = (W) / (W - Current Pot Size)

[Galwegian]

[ QUOTE ]
I don’t know if this is any easier but this is what I came up with a while back to calculate required fold equity.
Called = total pot if called.

W = Stack if (called * Equity) - Initial Stack
Fold % required = (W) / (W - Current Pot Size)

[/ QUOTE ]

I'm not sure I understand. It seems to me that the numerator is bigger than the denominator here, so you would require that the opponent folds more than 100% of the time????

I might just be misinterpreting your formula. Can you give an example of how you would use this formula to calculate required FE?

[Dismas]

[ QUOTE ]
[ QUOTE ]
I don’t know if this is any easier but this is what I came up with a while back to calculate required fold equity.
Called = total pot if called.

W = Stack if (called * Equity) - Initial Stack
Fold % required = (W) / (W - Current Pot Size)

[/ QUOTE ]

I'm not sure I understand. It seems to me that the numerator is bigger than the denominator here, so you would require that the opponent folds more than 100% of the time????

I might just be misinterpreting your formula. Can you give an example of how you would use this formula to calculate required FE?

[/ QUOTE ]

Sure.

You’re stack size is 150 (opponent has you covered).
Pot size is 50.
You’re equity if called is 20%.

W = ((150 + 50 + 150) * .20) – 150
W = (350 * .2) – 150
W = -80
-80 / (-80 ) – 50 = -80 / -130 = 0.6153846

You need him to fold 62% of the time for this to be chip EV neutral.