The Mathematics of Semibluffing

This is one thing that I've never managed to take in and I feel it's something extremely important for my game.

How does one calculate the EV of a semi-bluff?

Let's say we hold 9 outs to the nuts on the turn, there's 40 in the pot, our opponent bets 20 and we shove for 80 more, how do we calculate how often our opponent needs to fold for this move to be breakeven+?

If someone could take me (and other readers) through the equation quickly that would be great.

Cheers.

This is something I'm trying to learn so I'd like to try before anyone answers. With 9 clean outs on the turn that is only 18% on catching our nut hand. So 4:1. Your betting $80 into $60 meaning if he calls you lose $80 x 4 $320 and if he folds you win $60. So the way I see it. In 5 pushes you:

Lose $80
Lose $80
Lose $80
Lose $80
Win $220 if he calls your $80 push

That's$320 lost, $220 won so it's -EV $100 so you would need him to fold 2 out of 5 times to be up $20

How'd I do?

Edit: Something I just noticed, if you only 3x the RR, It's only a -$20 EV so you would only need him to fold once to profit.

I think I screwed the pooch on the win $$ amount. Forget what I said. I would need to rethink the whole thing.

After your opponents bet there is 60 in the pot.

Say opponent folds y% of the time and calls (1 - y)% of the time

y% of the time you win 60.
(1-y)% of the time you bet 80 to win 120 (60 in pot plus his 60 to call)

Assuming this is on the flop so that you meant we can hit 9 outs on turn (if we're on the turn you adjust this bit), then we're a 2-1 dog to hit our flush.
So lets say that means we hit 33% of the time (rounding down)

So EV of bet when called is 0.33*120 - 0.67*80 = 39.6 - 53.6 = -14

So EV of move is
y*60 - 14*(1-y)
which equals 0 if and only if
60*y = 14 - 14*y
74*y = 14
y = 19% (rounded up)

Originally Posted by gingerwizard

After your opponents bet there is 60 in the pot.

Say opponent folds y% of the time and calls (1 - y)% of the time

y% of the time you win 60.
(1-y)% of the time you bet 80 to win 120 (60 in pot plus his 60 to call)

Assuming this is on the flop so that you meant we can hit 9 outs on turn (if we're on the turn you adjust this bit), then we're a 2-1 dog to hit our flush.
So lets say that means we hit 33% of the time (rounding down)

So EV of bet when called is 0.33*120 - 0.67*80 = 39.6 - 53.6 = -14

So EV of move is
y*60 - 14*(1-y)
which equals 0 if and only if
60*y = 14 - 14*y
74*y = 14
y = 19% (rounded up)

How in the hell do you do this while playing???

Originally Posted by Trainer_jyms

This is something I'm trying to learn so I'd like to try before anyone answers. With 9 clean outs on the turn that is only 18% on catching our nut hand. So 4:1. Your betting $80 into $60 meaning if he calls you lose $80 x 4 $320 and if he folds you win $60. So the way I see it. In 5 pushes you:

Lose $80
Lose $80
Lose $80
Lose $80
Win $140 if he calls your $80 push

That's$320 lost, $140 won so it's -EV $180 so you would need him to fold 3 out of 5 times to be Up $60

There I retried the numbers.

jym,

you know the answer here...you bring your calculator into the casino...with your green visor, HOH book, and your glasses turned upside down. oh, and your ipod loaded with music...lol.

in all truthfullness, you practice the situation over and over. commit it to memory. and, you will be quick enough at the table to do it...in time.

oh, but i cant. i dont practice.

Even the best players in the world cant do the math on spot, i consider myself a very good mathmatician and i can only estimate within a few percentage points. Doing the math actually isnt that important during the hand, but its useful to look at outside of a hand to think about if some semi bluffs are actually +EV or not.

Ill give you guys a quick tip...

When you want to know what % of the time you need to get a fold for your bet to be BE:

X = your bet
Y = pot

0 EV = X / X+Y

Example: You bet 20 into a 30 pot, and you want to know how often he needs to fold for your bet to be profitable?

20 / 20 + 30 = 2/5 or 40%, so if he folds 40% you are BE, if he folds more than 40 you are + EV.

Also its good to note that these %'s never change.ie if you always bet 2/3 of the pot you will always need 40 % folds to be BE.

Originally Posted by Jager

Ill give you guys a quick tip...

When you want to know what % of the time you need to get a fold for your bet to be BE:

X = your bet
Y = pot

0 EV = X / X+Y

Example: You bet 20 into a 30 pot, and you want to know how often he needs to fold for your bet to be profitable?

20 / 20 + 30 = 2/5 or 40%, so if he folds 40% you are BE, if he folds more than 40 you are + EV.

Also its good to note that these %'s never change.ie if you always bet 2/3 of the pot you will always need 40 % folds to be BE.

But that is not true! Because you can win sometimes when he calls. This may serve as a useful upper bound though.

I was taking your cards out of it...

We could do some general calculations for standard situations to have some sort of guideline. And by following that up with the difference having more/less outs, or a different stack/pot ratio, makes should give us a nice feeling about the FE needed.

I hope someone is willing to add the math, because I'm certain I don't do them right:

Example 1:
100BB effective stacks
We have a clean flush draws (9 outs), standard HU pot;

Preflop: raise 4BB, call. With blinds, let's say pot is 10BB afterwards.
Flop (10BB); bet 8BB, call.
Turn (26BB); bet 18BB.......hero pushes

The moment hero pushes the pot is 44BB. villain has 70 behind, hero has 88 to push.
What FE do we need here?

Example 2:
Same one, but now we raise flop.

Preflop: raise 4BB, call. With blinds, let's say pot is 10BB afterwards.
Flop (10BB); bet 8BB, raise 24BB, call.
Turn (58BB); bet 20BB.......hero pushes

Pot is 78, villain has 52 behind, hero has 72 to push.
What FE do we need here?

Originally Posted by gingerwizard

So EV of bet when called is 0.33*120 - 0.67*80 = 39.6 - 53.6 = -14

This I can follow and it makes sense to me.

Originally Posted by gingerwizard

So EV of move is
y*60 - 14*(1-y)
which equals 0 if and only if
60*y = 14 - 14*y
74*y = 14
y = 19% (rounded up)

This I can't follow at all. But I want to understand it. Can you explain it for a non mathematician please?

Originally Posted by Pants_101

Originally Posted by gingerwizard

So EV of move is
y*60 - 14*(1-y)
which equals 0 if and only if
60*y = 14 - 14*y
74*y = 14
y = 19% (rounded up)

This I can't follow at all. But I want to understand it. Can you explain it for a non mathematician please?

Ok, shouldnt be too much of a problem.

The first line here is the same as the one you understood. The standard EV equation. We work out our EV of being called by a winning hand, which was -14. And we we know our EV if the guy folds. So EV of the play is
$EVIfHeFolds*probabilityHeFolds + $EVifHeCalls*probabilityHeCalls =
60*y - 14*(1-y)
So if there is 20% chance he folds, EV is 60*0.2 - 14*0.8.

Now we want to find y for which this move is breakeven (as the function is increasing in y then any bigger y makes the move +EV)
So the next three lines are me solving the equation for breakeven (EV=0) and calculating y in that case. When I did it was 0.189189 so if he folds 19% the move is +EV?

Is that ok?

Originally Posted by minSim

We could do some general calculations for standard situations to have some sort of guideline. And by following that up with the difference having more/less outs, or a different stack/pot ratio, makes should give us a nice feeling about the FE needed.

I hope someone is willing to add the math, because I'm certain I don't do them right:

Example 1:
100BB effective stacks
We have a clean flush draws (9 outs), standard HU pot;

Preflop: raise 4BB, call. With blinds, let's say pot is 10BB afterwards.
Flop (10BB); bet 8BB, call.
Turn (26BB); bet 18BB.......hero pushes

The moment hero pushes the pot is 44BB. villain has 70 behind, hero has 88 to push.
What FE do we need here?

Example 2:
Same one, but now we raise flop.

Preflop: raise 4BB, call. With blinds, let's say pot is 10BB afterwards.
Flop (10BB); bet 8BB, raise 24BB, call.
Turn (58BB); bet 20BB.......hero pushes

Pot is 78, villain has 52 behind, hero has 72 to push.
What FE do we need here?

Not sure how these are different. You can change the numbers anytime but the calculation is the same as I did above (although you are 4-1 dog not 2-1). You plug the numbers in as I did and solve for y.

If you're not willing to try the maths then how will you get a feel for it?

Originally Posted by gingerwizard

Originally Posted by minSim

We could do some general calculations for standard situations to have some sort of guideline. And by following that up with the difference having more/less outs, or a different stack/pot ratio, makes should give us a nice feeling about the FE needed.

I hope someone is willing to add the math, because I'm certain I don't do them right:

Example 1:
100BB effective stacks
We have a clean flush draws (9 outs), standard HU pot;

Preflop: raise 4BB, call. With blinds, let's say pot is 10BB afterwards.
Flop (10BB); bet 8BB, call.
Turn (26BB); bet 18BB.......hero pushes

The moment hero pushes the pot is 44BB. villain has 70 behind, hero has 88 to push.
What FE do we need here?

Example 2:
Same one, but now we raise flop.

Preflop: raise 4BB, call. With blinds, let's say pot is 10BB afterwards.
Flop (10BB); bet 8BB, raise 24BB, call.
Turn (58BB); bet 20BB.......hero pushes

Pot is 78, villain has 52 behind, hero has 72 to push.
What FE do we need here?

Not sure how these are different. You can change the numbers anytime but the calculation is the same as I did above (although you are 4-1 dog not 2-1). You plug the numbers in as I did and solve for y.

If you're not willing to try the maths then how will you get a feel for it?

Well I gave it a try. Here are my calculations:

Example 1:
When villain folds, hero wins 44.

When villain calls;
hero wins 0.18 x 114 = 20.52
hero losses 0.82 x 88 = 72.16.
EV of getting called = -51.64

44*y = 51.64 - 51.64*y
95.64*Y = 51.64
y = 54%

Example 2:
When villain folds, hero wins 78.

When villain calls;
hero wins 0.18 x 130 = 23.40
hero losses 0.82 x 52 = 42.64
EV of getting called = -19.24

78*y = 19.24 - 19.24*y
97.24*Y = 19.24
y = 20%

Any good?

Looks about spot on to me. Nice one!

Hope you feel you learned more by doing it yourself than by reading my walk through of it.

Originally Posted by IowaSkinsFan

Even the best players in the world cant do the math on spot, i consider myself a very good mathmatician and i can only estimate within a few percentage points. Doing the math actually isnt that important during the hand, but its useful to look at outside of a hand to think about if some semi bluffs are actually +EV or not.

agreed, but you can tell if it is close enough to give yourself the "green light," cant you?

Originally Posted by Chopper

agreed, but you can tell if it is close enough to give yourself the "green light," cant you?

See Jagers tip above for quick figuring.

How accurate will it be generically to apply a "we make the best hand" ratio to the upper bound Jager mentioned?

(Side note: Jager, how would that shortcut work if your opp had put more money in than you? You have to put your call in as part of the pot and then your raise as the ratio of your call+his raise+initial pot?)

Repeating the above example, assuming the above sidenote is true: Jager's upper bound shows that we're betting 60 to win 80+40 = 60/140 = 3/7 =42%. We're 2:1 to make the best hand, so we adjust in that we only need a fold enough to get it down to 1:1 --> 42% - 42%/2 = 21%. That's pretty close to 19%, and could be calculated on the fly.

Let's try in the other examples:
Push on turn example: betting 70 to win 70+44+18 = 70/132 = 53%. To adjust for the 4:1 draw --> 53% - 53%/4 = 40%


hmm...either I messed up the maths, my side note was wrong, or this doesn't work. Ginger, can you tell me which?

Jeffrey I haven't yet used this tip for when a player has already bet a street and you are plannig a shove. I use it for when I am given an oppurtunity to bet, either oop or opp. Your theory does make sense to me though. When you are dealing with exact scenarios and you know you will win when you hit, you can get very accurate with your bets using your equity.

Example:
Villain has AsKc
You have Ah3h

Preflop 4 bb raise you call, HU flop with Villain, pot is 9.5 bbs.
Flop: Kh 6h 9s
You have 9 outs to the nuts, and unless you get running 3's you need a heart.

Villain bets 8 bbs.
You can call or raise and have 35% equity.

Scenario 1:

If you call and the turn blanks, and he fires again for 20 bbs into a 25.5 bbs pot you would then need to call 20 bbs to win a 45.5 bbs pot or you need to call a ~44% pot bet with only 18% equity. In fact he only needs to bet 6 bbs to eliminate your direct odds (6/31.5 = 19%). If you now shove for your remaining stack, Villain needs to call 68 bbs to win a 133.5 bbs pot, or 68/133.5 for 50.9%. He needs to have ~51% equity to make his call profitable, thus you really have zero fold equity IF villain can put you on your draw.

Scenario 2:

If you raise the flop to 24bbs. Your raise has immdiate results IF you think villain will fold 59%(24/24+17.5). However Villain needs to call 16 bbs with a 41.5 bbs pot for ~39%, or he needs 39% equity to call your bet. We know that villain has 65% equity and should not fold. If villain calls, turn blanks, and you both have 64 bbs left and a 57.5 bbs pot any reasonable bet will commit him and kill your direct odds. Any 3bet/shove on the flop by villain will commit him and again kill your direct odds.

So then why raise? You are trying to get a free card when villain checks turn, because you need to see both the turn and river cards to maximize your draw. As you can see though that playing just a flush draw is not good against even TPTK, unless villain will fold.

Originally Posted by JeffreyGB

How accurate will it be generically to apply a "we make the best hand" ratio to the upper bound Jager mentioned?

(Side note: Jager, how would that shortcut work if your opp had put more money in than you? You have to put your call in as part of the pot and then your raise as the ratio of your call+his raise+initial pot?)

Repeating the above example, assuming the above sidenote is true: Jager's upper bound shows that we're betting 60 to win 80+40 = 60/140 = 3/7 =42%. We're 2:1 to make the best hand, so we adjust in that we only need a fold enough to get it down to 1:1 --> 42% - 42%/2 = 21%. That's pretty close to 19%, and could be calculated on the fly.

Let's try in the other examples:
Push on turn example: betting 70 to win 70+44+18 = 70/132 = 53%. To adjust for the 4:1 draw --> 53% - 53%/4 = 40%


hmm...either I messed up the maths, my side note was wrong, or this doesn't work. Ginger, can you tell me which?

Surely in adjusting we must slash HIS odds of winning when he calls. So in the 2:1 case its
42% - 58%/3 = 26% *he loses 1 in 3 of those times he calls*
And in the 4:1 case roughly
53% - 47%/5 = 44% is

The reason this won't be a great estimation is that of course he doesn't just call 58% or 53% of the time. But on the fly not too bad.

hmmmm, I might have to think about this one more carefully

Originally Posted by Jager

Jeffrey I haven't yet used this tip for when a player has already bet a street and you are plannig a shove. I use it for when I am given an oppurtunity to bet, either oop or opp. Your theory does make sense to me though. When you are dealing with exact scenarios and you know you will win when you hit, you can get very accurate with your bets using your equity.

Example:
Villain has AsKc
You have Ah3h

Preflop 4 bb raise you call, HU flop with Villain, pot is 9.5 bbs.
Flop: Kh 6h 9s
You have 9 outs to the nuts, and unless you get running 3's you need a heart.

Villain bets 8 bbs.
You can call or raise and have 35% equity.

Scenario 1:

If you call and the turn blanks, and he fires again for 20 bbs into a 25.5 bbs pot you would then need to call 20 bbs to win a 45.5 bbs pot or you need to call a ~44% pot bet with only 18% equity. In fact he only needs to bet 6 bbs to eliminate your direct odds (6/31.5 = 19%). If you now shove for your remaining stack, Villain needs to call 68 bbs to win a 133.5 bbs pot, or 68/133.5 for 50.9%. He needs to have ~51% equity to make his call profitable, thus you really have zero fold equity IF villain can put you on your draw.

Scenario 2:

If you raise the flop to 24bbs. Your raise has immdiate results IF you think villain will fold 59%(24/24+17.5). However Villain needs to call 16 bbs with a 41.5 bbs pot for ~39%, or he needs 39% equity to call your bet. We know that villain has 65% equity and should not fold. If villain calls, turn blanks, and you both have 64 bbs left and a 57.5 bbs pot any reasonable bet will commit him and kill your direct odds. Any 3bet/shove on the flop by villain will commit him and again kill your direct odds.

So then why raise? You are trying to get a free card when villain checks turn, because you need to see both the turn and river cards to maximize your draw. As you can see though that playing just a flush draw is not good against even TPTK, unless villain will fold.

great post Jager. This is something ive never fully understood and this makes it much easier. I always get lost in the long-hand math posts (no offense ginger).

Although surely we need to factor in implied odds into the equation when figuring out whether or not are semi-bluff is +ev.

...dont we? Or is implied odds to situational to factor into such an equation??

Originally Posted by Geanosssss

Although surely we need to factor in implied odds into the equation when figuring out whether or not are semi-bluff is +ev.

...dont we? Or is implied odds to situational to factor into such an equation??

The problem with implied odds is that we cannot accurately calculate them. It is very tricky to gauge your implied odds post flop and usually you don't have any. This is because most of the unkown information is now known, except the last 2 cards. At this point the equities don't change nearly as much OR as often as they do from preflop to the flop.

Originally Posted by Jager

Ill give you guys a quick tip...

When you want to know what % of the time you need to get a fold for your bet to be BE:

X = your bet
Y = pot

0 EV = X / X+Y

Example: You bet 20 into a 30 pot, and you want to know how often he needs to fold for your bet to be profitable?

20 / 20 + 30 = 2/5 or 40%, so if he folds 40% you are BE, if he folds more than 40 you are + EV.

Also its good to note that these %'s never change.ie if you always bet 2/3 of the pot you will always need 40 % folds to be BE.

This calculation is correct.
The limiting conditions should be taking into consideration though:

1. The calculation is based on 0 outs for hero's cards. The equity of our two cards hand can change the needed FE a lot. In above example, if we hold a flush draw, the FE needed is only 17% (instead of 40%)

2. The calculation does not take into account any betting on this street. (i.e. villain bets, hero wants to push)

I am working on an easy formula to combine effective stacks/pot ratio and equity of our cards, so that nr 1 be covered. I don't know if it's possible though.
Also covering nr2 is only possible with gingerwizards more complicated formula.

Dont want to create new thread so i ask simple question related to this:

Situation:
Guy check and i'm sure i can steal the pot right now
Lets say pot is 14 and i bet 6. Lets say if i get called i have 0% to win to keep this simple
So for this to be +EV how much of time it need to work:
is it:
A. 6 / (14+6) = 0.3 = 30% of time
-OR-
B. 6 / 14 = 0.42 = 42% of time

i think its A but it dont total sense to me to add there my bet somehow...
Anyone bother explaining ? Thank you.

its B.

Of course if you're sure you can steal then neither as he always folds.

Originally Posted by gingerwizard

its B.

Of course if you're sure you can steal then neither as he always folds.

That seems exactly counter to Jager's short cut. Is his short cut wrong?

No it's A, not B.

I'll adjust what you said to make it easier for everyone.

lets say the pot is 14 and you bet 14. Obviously, if he folds 50% of the time, you'll win 14 dollars half the time and you'll lose $14 the other half. therefore if he folds 50% it's a 0EV move.

Lets see if Jagers calc works:

14/(14+14)= 14/28= .5

Ok what happens if we don't add the bet:

14/14 = 1

obviously you don't need to win this pot 100% of the time for the bet to be good. Jagers calc is correct with the 50% number.

thanks for confirmation Massimo.

i actually wrote this down so i post it here:
simple EV calc for that is:
b = our bet
p = pot we are stabbing
x = breakeven point

EV = x*(-b) + (1-x)*(p)
if ev calc is correct than ev=0 is our break even point
0 = -x*b + (1-x)*p
....
....
x=p/(b+p)

Originally Posted by crazycrazy

thanks for confirmation Massimo.

i actually wrote this down so i post it here:
simple EV calc for that is:
b = our bet
p = pot we are stabbing
x = breakeven point

EV = x*(-b) + (1-x)*(p)
if ev calc is correct than ev=0 is our break even point
0 = -x*b + (1-x)*p
....
....
x=p/(b+p)

This is wrong it is: x=b/(b +p)
And yes this is the derivation of my shortcut.

You have to remember that you will also lose your bet if he doesn't fold. That is why B is not correct in your previous example.

oh ye i copied it wrong from paper anyway good its corect...

interesting thread. been thinking a bit about this subject myself and done some math on it. i have made up a general formula for the calculation of desired fold probability.
maybe it will help or just to confuse people a bit more... whatever...
anyways here it goes:

F = Needed Fold Probability to break even
V = Value of Hand (e.g. 0.33 for above flush draw at flop or 0.18 at turn)
P = $ Pot
C = $ to be called when faced with a raise (facing no raise then C=0)
R = $ Raise

ev-equation then comes to following:

FP + (1-F) (P+R) V - (1-F) (C+R) (1-V) > 0

after some tedious juggling around of things, which i will spare the audience with the full process, this comes to:

F > [ C + R - V (P+C+2R ) ] / [ P + C + R - V (P+C+2R) ]

took me a while to get this correct but pretty sure it is now. tested it with the examples above and it comes to the right results.

actually think i found a little error in above calculation of example 2.
not that it is really important just so people dont wonder why this formula delivers different result for this example...

Originally Posted by minSim

Example 2:
When villain folds, hero wins 78.

When villain calls;
hero wins 0.18 x 130 = 23.40
hero losses 0.82 x 52 = 42.64
EV of getting called = -19.24

78*y = 19.24 - 19.24*y
97.24*Y = 19.24
y = 20%

the line 0.82 x 52 = 42.64 should actually be 0.82 x 72 = 59.04. you forgot to account for the amount needed to call.
EV of call then = -35.64
y then = 31%.
this is then the same result as with this formula.

i understand the equation looks pretty ugly but isnt really as complicated as it might seem at first look.
in case of no amount needed to be called C=0. things then look quite a bit simpler:
F > [ R - V (P+2R ) ] / [ P + R - V (P+2R) ]

cheers

Originally Posted by JeffreyGB

Originally Posted by gingerwizard

its B.

Of course if you're sure you can steal then neither as he always folds.

That seems exactly counter to Jager's short cut. Is his short cut wrong?

Meant to type A.

Jager's argument is obviously correct in the 0% to win case. Proof by example is never a good way to go so i'll prove it mathematically.

Pot is currently y, and you go all in for x. What % of the time must he fold for this to be profitable assuming he always has the best hand when he calls?

Let F be his fold percentage.
EV of x bet is
y*F - x*(1-F). (++)
This is break even if and only if
x*(1-F) = y*F
x = (F*x) + (F*y)
x = F*(x+y)
so F = x/(x+y)
Since (++) is increasing in F then the bet is +EV if and only if F > x/(x+y)


Now lets see how concrete we can make the general case: Now lets assume that you are drawing if he calls so that associated with his call you have W% chance to win. To make this easier to see I'll break it down very slowly.

EV = y*F + (1-F)*{W*(x+y) - (1-W)*x}

Going through the 2nd part carefully if you don't get it, this means that for the (1-F)% of the time we get called, W% of the time we win the pot y, plus his call x and (1-W)% of the time we lose our bet x.

You might want to skip the bit in between the stars here if the maths is getting too tricky.

*****
Firstly we must check that the EV equation is monotonic in F so that the conclusions we draw when setting EV to zero and solving for F can be legitimatised.

d(EV)/dF = y - W(x+y) +x -Wx
= y(1-W) + x(1 - 2W)
which is >= 0 unless winning % is greater than 50 (and you're ahead anyway making this calculation somewhat redundant)
If >=0 then EV is monotonic in F and the solution below will be correct always.

******

So we set EV to 0 and solve for F

y*F + (1-F)*{W*(x+y) - (1-W)*x} = 0 <=>
yF -F{W(x+y)-(1-W)x} = (1-W)x - W(x+y) <=>
F = {(1 - 2W)x - Wy}/{y(1-W) + (1 - 2W)x}
= 1 - y/{y(1-W) + (1-2W)x}

Is this hard? Not really! x and y are known. W you are guessing based on your number of outs and is a number you're practised at getting right. With practise this is totally doable at the table. At least estimatable. Have a play.

Oh and to prove it satisfies Jager's equation (the special case when W = 0) set W = 0, => F = 1 - y/x+y
=x/x+y as required.

since doing calculations for semibluffs on the fly seems somewhat demanding i have made myself a chart for some common values. the values here though do not take into account $ amounts to be called. needed fold percentages would have to be somewhat higher if your semibluff is a reraise. for a plain raise though they may serve as a quickreference.

http://img174.imageshack.us/img174/6290/fvshvbw7.jpg
Fold% = breakeven point
Handvalue% = chance of winning hand when called

cheers