raising with draws - how much FE needed?

[Galwegian]

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?

[jay_shark]

Yup I just checked it over and the formula is correct .

It's a neat way to calculate the percentage of times one needs to fold for a shove to be positive EV .

Here is the formula if you have 6 outs , or about a 25% chance of winning .

x(P+B+S)+1/4(1-x)(2S+P) > S
4x(P+B+S) + 2S+P -x(2S+P)>4S
x[2S+3P+4B] >2S-P
x>(2S-P)/(2S+3P+4B)

[jay_shark]

If you only have a 25% edge, then you need your opponent to fold about 37.7% of the time which is a big jump from 22.97% from the numbers given in the example .

The formula for 5 outs or equivalently about 20% is
x> (3S-P)/(3S+4P+5B)

Again using the numbers provided , you have to assure yourself that your opponent will fold around 44.1% for the play to be profitable .

[Galwegian]

BTW Jay, you can you these formula to help evaluate semibluff lines even when your semibluff is not an all in raise. If you have position on your opponent and you are sure that he will check the to you on the turn if he calls, then these formulae still provide a lower bound for the fold equity that you require. You just have to replace S (stack size) with the size of your raise. This is just a lower bound as it doesn't take into account any extra money that you might make when you make your draw.

[Galwegian]

O yeah, one other thing, since stack-to-pot ratios are all the rage these days. In the original situation (8/9 draw), say that your opponent makes a pot sized bet on the flop. Then the formula simplifies to

(spr-1)/(spr+5)

where spr is the stack-to-pot ratio. Since we have all read PNL [img]/images/graemlins/smile.gif[/img] and we are all keeping track of the spr, this is very easy to calculate quickly.

Cliff notes: You have a 8/9 out draw on the flop and opponent pots it to you. If you think he will fold more than (spr-1)/(spr+5) of the time, then an all in reraise is better than folding.

[binions]

[ QUOTE ]
O yeah, one other thing, since stack-to-pot ratios are all the rage these days. In the original situation (8/9 draw), say that your opponent makes a pot sized bet on the flop. Then the formula simplifies to

(spr-1)/(spr+5)

where spr is the stack-to-pot ratio. Since we have all read PNL [img]/images/graemlins/smile.gif[/img] and we are all keeping track of the spr, this is very easy to calculate quickly.



[/ QUOTE ]

So, assume 100 in the pot going into the flop, villain has 800 left and you cover. Effective SPR is 8:1. He bets pot, and you have a 2:1 draw if you see both cards.

SPR -1 / SPR +5 = 7/13 so he needs to fold 7/13 of the time to a push to break even. Let's see if it works:

7 times he folds and you win 200 = +1400

6 times we get called. We win 2 and lose 4.
2 times you win 900 = +1800
4 times you lose 800 = -3200

Yep. 3200 - 3200 = 0.

Good work. So, we know the shortcut when we have 33% equity and villain bets pot. How does the SPR -1 / SPR +5 formula adjust when:

You have 20%, 25%, 40%, 50% equity and villain bets pot?

You have 20%, 25%, 33%, 40% and 50% equity and villain bets 2/3 pot.

Thanks

[jay_shark]

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 ?

[jay_shark]

I will make some further simplifications to the formula .

(SPR-1)/(SPR+5) = (SPR +5 -6)/(SPR +5) = 1 - 6/(SPR+5)

That is , the push is neutral EV if you expect to be called 6/(SPR+5) .This is actually quite nice as it's very easy to do this calculation on the spot during a difficult situation .

So in the original example , as long as you get called
6 /7.416666 , then the push makes no difference .If you expect to be called less than that number then the push is positive EV .

[binions]

[ QUOTE ]
I will make some further simplifications to the formula .

(SPR-1)/(SPR+5) = (SPR +5 -6)/(SPR +5) = 1 - 6/(SPR+5)


[/ QUOTE ]

I am with you so far. In my SPR=8 example, 1 - 6/(8+5) = 1 - 6/13 = 7/13.

[ QUOTE ]
That is , the push is neutral EV if you expect to be called 6/(SPR+5) .This is actually quite nice as it's very easy to do this calculation on the spot during a difficult situation .

[/ QUOTE ]

Now I see. 6/(8+5) = 6/13. This is breakeven if you think you will be called 6/13.

[binions]

[ QUOTE ]

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 ]

So (SPR +4 - 5)/SPR +4 = 1 or 1 - 5/(SPR +4) so if he calls 5/SPR+4 = breakeven. Did I get that right? (2/3 pot bet and 33% draw equity)

Comparing, in the SPR=8 scenario, when he bets pot if he calls your push less than 6/13 you win. When he bets 2/3 pot, if he calls your push less than 5/12 you win.

The less he bets, the more he needs to fold to your push for you to breakeven (54% v 59%)