A Method to Compute Required Folds For Breakeven Draw Pushing (Long)

[Panthro]

this is good stuff.

nh sir.

[Montezuma21]

great post Grunch. Thanks a lot.

One point: Shouldn't it be $145 instead of $140 btw?

[Grunch]

[ QUOTE ]
Another thing, too, is that this type of calculation assumes that we will win 100% of the time we hit one of our 8 outs and that we will lose 100% of the time otherwise.

I wonder how much of an impact this actually has on our figures. I'm going to guestimate 2-3% but that's coming from experience and my knowledge alone and isn't proven at all. How would we go about adding these gray areas to our calculations? They may actually make what we think is a small 0.5BB edge into a slightly -EV play.

[/ QUOTE ]

The 100% assumption does make a signifigant difference in deciding if pushing is the optimal play -- in fact, it makes a big difference even in determining the breakeven point. But we can adjust our computations in a very simple way to account for a more reaslistic estimate of our equity.

The adjustment is performed by discounting our outs up front. Say we have 8 outs, and we think that if we get there our hand will win 75% of the time. This is probably much more realistic than 100%, and it also gives us a way to accomodate for redraws by our opponents. If 8 outs will win for us 75% of the time, then we can say that our hand is "worth" 6 outs (8*.75=6); at least for the purpose of this computation.

Let's look at an example of this in practice. We'll use the same hand, but we'll add in the idea that when we get there we'll win only 75% of the time. Here's the relevant data:

As of this moment, the bet is $10 to call, there is $30 in the pot, and there is $125 more behind. You have 8 outs, and you think they are clean. When we make our straight, we think we'll win 75% of the time, so our hand is "worth" 8*.75=6 outs. According to the rule of 2/4 then, our hand will win 25% of the time when we push.

Now we simply build our continuum the same way as before.. The only difference this time is that when we push & he calls we're a 3:1 dog instead of a 2:1 dog:

+140 - 125 - 125 - 125 + ...

Reduced to 'units' this becomes:

+4.5 - 4 - 4 - 4 = -7.5

The -7.5 represents the gap that we have to fill in by winning without a showdown. We can see already that we have a lot of ground to make up, but let's carry this through. He can't fold 7.5 times, so we round up to 8:

+4.5 - 4 - 4 - 4 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

That's a lot of folds. In this continuum, we won once by making our straight, lost 3 times by either missing or getting there and losing anyway, and won 8 times when he folded to our push. That's a total of 12 plays, 8 of which are folds. He has to fold 8/12=2/3 or 66% of the time now in order for pushing to be breakeven.

Let's figure out f by using the precise method to verify our results:

<font class="small">Code:</font><hr /><pre>0 = 30f + (1-f)[.25(125-15+30) + .75(-125)]
0 = 30f + (1-f)(35 - 94)
0 = 30f - 59 + 59f
89f = 59
f = 59/89 = .66</pre><hr />

So the continuum method still works.

One note about how I use this method in actual play. Usually I'm not doing this continuum building to directly determine if pushing is the best play. Typically I'm just looking to see if it is either obviously +EV or obviously -EV. Normally I'll do this computation real quick just to get a ballpark idea where f is. Does he have to fold closer to 10% of the time or 80% of the time? Usually the number I come up with is well within either the "definitely yes" or "definitely no" ranges. If it is close, then I rely on other means to figure out what the right play is. Typically mathematical methods like this can only take you so far. They can get you to the ballpark. Other things like intuitive reasoning, metagame issues or even just reads will guide you towards home plate.

[Grunch]

[ QUOTE ]
One point: Shouldn't it be $145 instead of $140 btw?

[/ QUOTE ]

Yes, you're right. My mistake, I had hero raising to 125 instead of 130. Since there's 125 behind and we have already bet 5, if we push we are making it 130 to go.

[Dire]

-incorrect approximation

[joeblack]

I am skeptical to observing a high FE in SSNL...

In other words do you think there is a difference in Fold Equity per levels?

(ie: Do you think a SSNL player more likely to call your push than a HSNL player?)

[orange]

lol this is all just jumble to me. i cant read it. &lt;---math retard.

[Chaos_ult]

Obviously fold equity is based entirely on situational experience and observation of your opponents.

[tubasteve]

[ QUOTE ]
Hey good post, clearly explined. Question..

He has to fold about 58% to be 0EV. Is folding to his raise equilivent to 0EV as well?

What I am saying is, if god came down and told us that he WILL fold EXACTLY 58% of the time (or the % that guarantees 0EV long term), is it better to simply fold? The same profit is made (0EV) but we skip the short term variance altogether.

If so, this arises a question. I know its correct to make as many +EV moves as we can. But I've heard green plastic in some of his videos, and some other people say, that they are willing to give up very small EV edges with huge variance simply because theyd prefer to pay that small price to avoid huge swings.

So, if this is true, what the max % that we as serious poker plays SHOULD give up to avoide huge variance plays with little return, if any?

[/ QUOTE ]

to me, if a play like this is 0EV i think you should gamble since if you win you get the chance to play deepstacked

also, NINJA BUMP HIIIIIIIIIYA

[godofPOPOV]

[ QUOTE ]
Tickner, that's a huge &amp; very good question.

Yes, folding is 0EV by definition. I guess you could say it's better to fold in this spot because of the rake, but let's ignore that for now.

I don't know the answer to your question. I suspect it's the kind of thing we can chalk up to being a matter of style, but something tells me it's not. If you want a silly analogy, you could think of passing null eges to avoid variance almost like an implementation of a Martingale system. How do you feel about a Martingale system?

I do know one thing. I learned poker mostly through 2p2. And if 2p2 has engrained anything in to my psyche, it's this: passing up null edges is for people who hate money or ninnies. Now I know where this is coming from. But I also know that there is no never in poker, and in order to grow in to a great player, we have to learn to think for ourselves.

In short, I suspect you are correct. Passing up variance might be the right thing to do sometimes.

[/ QUOTE ]

i think passing up variance in an absolute situation is very important because of the negative effects that tilt can have on some peoples emotions and the affect of their level of play tilt can have.