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

[Grunch]

Here's the situation (do not comment on the play of this hand, it is totally irrelevant and was contrived for the math):

Full Tilt Poker - No Limit Hold'em Cash Game - $0.50/$1 Blinds - 5 Players - (LegoPoker HH Converter)

<font color="black">Preflop:</font> Hero is dealt 6[img]/images/graemlins/club.gif[/img] 8[img]/images/graemlins/diamond.gif[/img] (5 Players)
UTG folds, CO calls $1.00, BTN folds, <font color="red">Hero raises to $4.50</font>, BB folds, CO calls $3.50

<font color="black">Flop:</font> ($10) Q[img]/images/graemlins/heart.gif[/img] 7[img]/images/graemlins/diamond.gif[/img] 9[img]/images/graemlins/club.gif[/img] (2 Players)
<font color="red">Hero bets $5.00</font>, <font color="red">CO raises to $15.00</font>, Hero ... ?

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. That is, you have a 32% (according to the rule of 2/4) chance to improve to a straight, and when you do you will win 100% of the time. Assume these numbers to be correct. Ignore the equity of backdoor draws.

You are considering pushing. Every time you push and the opponent calls, you are losing money long-term because you only have 32% equity. In other words, 68% of the time you will lose your stack when he calls. But that's not the whole story. Pushing also has Fold Equity. Fold Equity means that there is a certian ammount of value in pushing because the opponent may fold, and you will win the $30 that's in the pot right now. When you push and he folds then, it is a long-term +EV play. So the Net EV of pushing becomes the EV when he folds plus the EV when he calls. (It so happens that the EV when he calls is negative.) Our challenge is to figure out how often he has to fold for pushing to be breakeven.

There is a simple mathematical method to compute this, but it is too cumbersome to use at the table. Here is how to compute how often he must fold in order for pushing to be breakeven. Let f = a number from 0 to 1, representing the probability he folds:

</pre><hr />EV = EVfold + EVcall
0 = 30f + (1-f) [ .32(125-15+30) + .68(-125) ]
0 = 30f - (1-f)(44.-85)
0 = 30f - (-41) - f(-41)
0 = 30f + 41f - 41
0 = 71f - 41
41 = 71f
41/71 = f = .58</pre><hr />

So in order for pushing to be 0EV, the opponent must fold 58% of the time.

How do you figure this out at the table? The math above is too complex for most people to do in thier head, especially on the 15 second timers common in internet poker. But there is a simpler way.

The key to understanding the total equity in pushing is to envision a continuum of all the wins &amp; losses you get from when he calls, and fill in the gaps with wins when he folds. I'll explain this by example.

When you push and he calls, you are a 2:1 dog. So for every 3 plays, you'll lose twice. Here's how that looks on our continuum (remember, we're looking for breakeven, or EV=0):

+140 - 125 - 125 + ...

The '...' represents the gap that we're going to fill in.

For me, doing '140 - 125 - 125' is still to complicated for table math; I need smaller numbers. If we look at the size of our push in comparison to the size of the pot, we see that we are pushing 4x the $30 in the pot. So to make the numbers simpler, reduce all the numbers to "units" where 1 unit is the size of the pot, and round them to something easy for you to handle. For me, that's 0.5 units:

+4.5 - 4 - 4 + ...

This is approximate, but it's very close. Good enough to make good decisions at the table.

Now things are pretty simple. I can do the math "4.5-4-4" in our heads, even under pressure:

+4.5 - 4 - 4 = -3.5

So we see we lose when he calls. How many times does he have to fold to make EV = 0? The answer is simple, it's 3.5, or practically speaking, 4. So this is what we need our continuum to look like in order for the push to be breakeven:

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

Note that everything is still broken out. I didn't reduce '4.5 - 4 - 4' to just -3.5. This is important, because the last step in figuring out how often he has to fold is to count up all the plays in our continuum. In order to do that we need to not reduce the expressions, since each expression is one play. In our continuum there are 7 expressions, and so there are 7 plays: We won 4.5 once, lost 4 twice, and win 1 four times. So we can see how often he needs to fold: 4 out of 7 times.

Now again I can't do "4/7" in my head precisely, but I can get close enough. I know that 4/8 is exactly 0.5, so 4/7 is a little bigger than that. He needs to fold a little more than half the time in order for pushing to be breakeven.

If you actually do the math on a calculator, you see that 4/7 = 57% which is extremely close to what we came up with above when we computed the number precisely. In fact the two methods are based on the same principal, they are just 2 different ways of looking at it. The error comes from the rounding of the numbers to 'units'. With no rounding, this method is precise.

Hope this helps some of you. I know it helped me when I figured out how to do this.

[FishSticks]

I don't think it's sound strategy to raise 68o out of position preflop with a limper in the pot. Just fold preflop and you won't have to worry about the math.

<font color="white"> lol jk obv - this is a handy post; I know how to work it out longhand but I can never apply it at the table (outside of some vague estimate) </font>

[Grunch]

[img]/images/graemlins/smile.gif[/img] Ahh, to be green again.

Just kidding.

[MrMysterious]

nice post

[holyfield5]

whats cliff notes? push is bad?

[delta k]

[Chaos_ult]

This post was actually very helpful to me Grunch.

Thank you for simplifying it.

&lt;3 &lt;3 &lt;3

[Grunch]

This is a shameless self-bump. But I'll make it a rich bump, and add Cliff Notes.

There's a simple mathematical method to compute the Fold Equity you need in order for it to be breakeven when you push a draw. But that method is too cumbersome to use at the table. I present another method that is very close to accurate which, with practice, can be used to compute breakeven Fold Equity at the table.

[Tickner]

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?

[Grunch]

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.