maths help please

to daviddem or some other mathemagician out there, please help me determine what is going on in this equation. this was provided to me by someone WITH all the processes done. it is designed to determine the EV of a 5bet shove (or more precisely, an unexploitable 5b shove range, hence EV is set to 0 - the worst hand in our shoving range has 0EV) i need to know what the processes were. i'll also outline what i think has been done with each step to help clarify for myself/to show that i actually tried.

here it is in its entirety before i break down what i think has happened. just so it reads a bit more fluently etc.

0 = F(36.5) + (1-F)(y)(-90) + (1-F)(1-y)(111.5)
0 = 36.5F + (1-F)(-90y) + (1 - F - y + Fy)(111.5)
0 = 36.5F - 90y + 90Fy + 111.5 - 111.5F - 111.5y + 111.5Fy
0 = -75F - 201.5y + 201.5Fy + 111.5
75F - 201.5Fy = -201.5y + 111.5
F(75 - 201.5y) = -201.5y + 111.5
F = (-201.5y + 111.5)/(75 - 201.5y)


and here's what i think's going on, and where i begin to have no idea what's going on.

0 = F(36.5) + (1-F)(y)(-90) + (1-F)(1-y)(111.5)

so this is just our starting equation. F is villain's fold percentage, and y is the equity of the worst hand in our shoving range.

0 = 36.5F + (1-F)(-90y) + (1 - F - y + Fy)(111.5)

so we've multiplied F by 36.5 to get 36.5F.
we have multiplied y by -90 to get (1-f)(-90y) in the middle bit
and we have then multiplied:
1 by 1,
-F by 1,
1 by -y,
and -F by -y to get:
(1 - F - y + Fy)

0 = 36.5F - 90y + 90Fy + 111.5 - 111.5F - 111.5y + 111.5Fy

here we have multiplied 1 by -90y and -F by -90y
also we have multiplied 1, -F, -y, and Fy by 111.5

0 = -75F - 201.5y + 201.5Fy + 111.5

here we have done:
36.5F - 111.5F
-90y - 111.y
90Fy + 111.5 Fy
it's at about this point that i get confused.

75F - 201.5Fy = -201.5y + 111.5

have we simply added 75F and subtracted 201.5Fy to/from both sides of the equation here? ie making sure the equation stays "balanced" or whatever the term is?

F(75 - 201.5y) = -201.5y + 111.5

no idea what's happened at this point.

F = (-201.5y + 111.5)/(75 - 201.5y)

no idea whats happened here because i need to know what happened in the last step.

any help appreciated.

75F - 201.5Fy = -201.5y + 111.5

"have we simply added 75F and subtracted 201.5Fy to/from both sides of the equation here?"
Yes, but not for the reason you stated. The entire purpose of this exercise is to solve for F, so we isolate it by moving both terms with F on the left side of our equation.

F(75 - 201.5y) = -201.5y + 111.5

"no idea what's happened at this point."
The reason we did the last step was so that we can pull out the common term F in order to solve for it. we are just dividing 75F and 201.5Fy by F, and then multiplying them by it. You can see that these expressions are equal by multiplying F times both terms.

F = (-201.5y + 111.5)/(75 - 201.5y)
Since our goal is to solve for F, after we isolate it we divide both sides of the equation by (75-201.5y) because this gives us an expression with F on one side and everything else on the other. This is called a function of F in terms of y.

so like to finish up the math, if our villain was 4betting QQ+, AK, and 78s, then folding just the 78s his fold F would be (4/38) (78s combos/all combos) and our y would be the pokerstove result of the equity of the hand we were considering against QQ+, AK.

One thing to note is that if villain is going to call your 5bet off with a relatively wide range (one including AQ) small PPs become the nuts due to their large equity against these ranges.

Quote:

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75F - 201.5Fy = -201.5y + 111.5

have we simply added 75F and subtracted 201.5Fy to/from both sides of the equation here? ie making sure the equation stays "balanced" or whatever the term is?

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Yes, it's correct the ultimate goal is to isolate F so we do what you said to get all the terms containing a F on one side of the equation and all the ones that don't on the other side.

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F(75 - 201.5y) = -201.5y + 111.5

no idea what's happened at this point.

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Again the goal is to isolate F and this is just stemming from the distributive property of multiplication:
A*(B+C) = A*B + A*C
or with numbers for example:
you can see that 3*(5+4) = 3*9 = 27, right.
But that is also the same as 3*5 + 3*4 = 15 + 12 = 27, right?
Spoon also explains it here (part 5): http://www.flopturnriver.com/pokerfo...ad-180192.html

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F = (-201.5y + 111.5)/(75 - 201.5y)

no idea whats happened here because i need to know what happened in the last step.

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Well explained by Mr Bucket above, we want to get F alone on one side, so we divide both sides by 75 -201.5y

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Originally Posted by daviddem

Again the goal is to isolate F and this is just stemming from the distributive property of multiplication:
A*(B+C) = A*B + A*C
or with numbers for example:
you can see that 3*(5+4) = 3*9 = 27, right.
But that is also the same as 3*5 + 3*4 = 15 + 12 = 27, right?

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i understand the above. but i don't understand how that process is related to what we have done with the equation.

75F - 201.5Fy = -201.5y + 111.5
F(75 - 201.5y) = -201.5y + 111.5

namely i don't understand how we can do/why we do the following

"we are just dividing 75F and 201.5Fy by F, and then multiplying them by it. You can see that these expressions are equal by multiplying F times both terms."

forgive my stupidity.

wait! i think i see. we haven't changed anything, have we?

ie we start with 75F - 201.5Fy
and to separate F, we just turn that into F (75 - 201.5y), which can be simplified (or whatever the term is) to 75F - 201.5Fy, but we don't want it written like that because we want to isolate F. am i on the right track?

Yes that's it, maybe you are confused because there are three terms. Even when you have something complicated like
a*F*x*y^2*z + d*k*(j+2)*m*F, you can always extract/isolate that F and put it in front of the rest, which you then enclose in brackets:
F*(a*x*y^2*z + d*k*(j+2)*m)

goodo. thankyou kind sirs

Now you just do that kinda math with your brain while 24-tabling, and you got yourself some sick winrate.

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Originally Posted by rpm

i understand the above. but i don't understand how that process is related to what we have done with the equation.

75F - 201.5Fy = -201.5y + 111.5
F(75 - 201.5y) = -201.5y + 111.5

namely i don't understand how we can do/why we do the following

"we are just dividing 75F and 201.5Fy by F, and then multiplying them by it. You can see that these expressions are equal by multiplying F times both terms."

forgive my stupidity.

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Lets put it in a simple perspective, because you still seem a bit confused (nhf, i know lots of ppl hate maths).

I think in english its a math term called distribution.

This is what happens:

AxB + AxC = Ax (B+C)

Or lets try an example with some real numbers.

15 + 25 = 40
15 + 25 = 3x5 + 5x5 = 5x (3+5) = 5x 8 = 40

:)

i'm back.

so i now understand all of the above. and my understanding is that

F = (-201.5y + 111.5) / (75 - 201.5y)

if we give this thing a y value, it should give us the corresponding F value which makes it = 0 (because that was our starting premise). the reason it should = 0 is because we are looking to find what the worst hand in our 5b shoving range can be based on certain amounts of fold equity. (F is villain's fold %, y is the equity of the worst hand in our shove range).

so i went ahead and started messing aboot with some different y values. in this particular case, i chose 0.35.

F = (-201.5 * 0.35) + 111.5 / 75 - (201.5 * 0.35)
F = -70.525 + 111.5 / 75 - 70.525
F = 40.975 / 4.475
F = 9.156

so thus i assumed that if y = 0.35, F = 0.915 (because F has to be between 0.00 and 1.00, i assume i should take that to mean 9% of the time, or 0.09)

so i then plugged both back into the original equation, to see if they equalled zero.

0 = 36.5F + (1-F)(-90y) + (1 - F - y + Fy)(111.5)
0 = 33.21 + 0.909 * 31.5 + 0.59 (111.5)
0 = 3.321 - 28.633 + 65.875
0 = 40.563

0 obviously does not equal 40.563. and thus i have either fucked up the calculations. or completely missed the point of this equation.

so my questions for you maths people out there are:
- is there a mistake in my working out?
- am i wrong in assuming if i solve the first one, the F and y variables should be able to be plugged into the original equation, and give an outcome of 0?

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so this is just our starting equation. F is villain's fold percentage, and y is the equity of the worst hand in our shoving range.

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First it looks to me that there is a problem with your original equation. If I understand correctly, the pot before your play is 36.5. You shove your remaining 90. If villain calls, your equity is y (mind that this is your equity against his calling range). Let's look at what can happen:
- villain folds, your profit is 36.5
- villain calls and you win, your profit is 90+36.5=126.5
- villain calls and you loose, your loss is 90

So the equation should be:
0=F(36.5) + (1-F)(y)(126.5) + (1-F)(1-y)(-90)

No?

Let's first fix that, then we'll look at the rest.

that was what i thought at first too. but we are not finding the EV of our shove, we are finding what the worst hand we can shove is that gives us 0EV (because in an unexploitable 5b shove range, the worst hand we shove will have an EV of 0).

to be honest i don't completely understand how this equation tells us this. but it was given to me by spoon, who knows his shit. so i've just accepted that this particular equation does what i've been told it does.

i assume that the equation uses "y" for the branch covering "he calls we win" and "1-y" for "he calls, we lose" because we are basically looking at worst case scenario - we are looking to find the lowest F values which give us 0EV based on certain y values. thus i guess we are singling out the worst two card combination in our shoving range, and thus we win 126.5 * the equity of the WORST hand in our range (namely "y") when he calls. and we then obviously lose 90 "1-y" (converted to %) of the time. i could be wrong here. regardless. i'd say it's safe to assume that the equation is right, and that i've fucked up somewhere in my working. or i just completely misunderstood what it's for.

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Originally Posted by daviddem

First it looks to me that there is a problem with your original equation. If I understand correctly, the pot before your play is 36.5. You shove your remaining 90.

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this is true.

Quote:

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Originally Posted by daviddem

If villain calls, your equity is y (mind that this is your equity against his calling range).

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we're not trying to determine the EV of shoving as such. more specifically (as best i understand), we are looking to set up an equation into which we can put certain y values, and it will give us the corresponding F values which = 0, for reasons outlined above. it is for this reason that "y" is is the equity of the worst hand in our shoving range against villain's calling range.

Quote:

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Originally Posted by daviddem

Let's look at what can happen:
- villain folds, your profit is 36.5
- villain calls and you win, your profit is 90+36.5=126.5
- villain calls and you loose, your loss is 90

So the equation should be:
0=F(36.5) + (1-F)(y)(126.5) + (1-F)(1-y)(-90)

No?

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i don't think so. for reasons stated above. though i guess an incredibly long way around my problem here would be to do a million of these and find which y and F values = 0 through trial and error. though the shortcut version seems far more appealing, if i can figure out how the f to do it.

The problem you are trying to solve is vastly more complicated than that, but it is solvable using computers and an iterative process.

The solution to this problem is shown in Mathematics of Poker, starting page 123 with [0,1] toy games and progressing to show how to solve the no limit hold'em problem with an iterative algorithm and computers. The [0,1] toy games are easy to solve on paper and give you a very good idea of the process involved. If you care you can read that part of Maths of Poker, and report here if you have problems with the math.

This solution has been implemented in software such as SNG Wizard. You can download a free trial version of it (and there are also "workarounds" to the trial version expiry).

However, the demonstration in Mathematics of Poker and implementation in SNG wizard consider the problem when the starting range of each player is "any 2 cards". This is not the case in your 5bet problem, because both players already have non random ranges. However the exact same methodology could be used to solve your 5bet problem, provided you assign decently accurate starting ranges to both players.

so it is not as simple as solving for F? (i hope i don't come across as rude here but) i'm not looking for advice on how to 5bet unexploitably/play optimally vs 4bets or anything broad like that. i'm simply after pointers in where i'm going wrong in the mathematical procedures involved in the above equation. is your above post extrapolating from my original question? or is it genuinely an extremely complex process, to the point of requiring a computer, to figure out F in

F = (-201.5y + 111.5) / (75 - 201.5y)

if we are given a y value?

again, i'm relatively clueless when it comes to this stuff, so forgive me if i'm wrong.

No you don't need a computer to do that. It was my understanding from your previous posts that your were looking for your unexploitable 5bet shoving range. If this is the case, then refer to my previous post and mathematics of poker.

The equation I gave you above describes the EV of 5 bet shoving:
- wihen you hold a hand that has an equity y against opp's calling range
- against an opp who folds to a 5bet shove with a frequency F
- with a pot size of 36.5 and a bet size of 90

The starting equation in your OP does not do that, and I am not quite sure what it does or how the guy who gave it to you obtained it. It'd be interesting to ask him and report here.

So for the sake of it, I will work with my equation. By a similar process to what was done in your OP, my equation can be turned into:

F = (90-216.5y) / (126.5-216.5y)

Now you can plug in equity values y to find the opp's folding frequency F that you need to breakeven (and you can also graph this relationship in Excel).

For example, to answer your question, let's see how much fold frequency we need to shove profitably with a hand that has 20% equity against his calling range: we look for F and we know y=0.2:
F = (90-216.5*0.2) / (126.5-216.5*0.2)
F = (90-43.3) / (126.5-43.3)
F = 46.7 / 83.2
F = 0.561

So we need him to fold at least 56.1% of the time for a shove to be profitable when we hold a hand that has 20% equity against his calling range.

Now let's first find what is the minimum equity we need to shove profitably against an opponent who never folds. So F=0 and the equation becomes:
0 = (90-216.5*y) / (126.5-216.5*y)
And this will be equal to zero whenever
90-216.5*y=0
90=216.5*y
0.416=y

So if you have 41.6% equity or more against his range, even if he never folds, you can always shove profitably.

(sidenote: don't worry about what the equation does for values of y beyond 0.416 because you will always get values of F that are out of scope, either negative or greater than 1)

Now we can graph F against the values of y comprised between 0 and 0.416:
http://www.flopturnriver.com/pokerfo...e81-f-vs-y.png

Hope this helps?

re

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Originally Posted by daviddem

The starting equation in your OP does not do that, and I am not quite sure what it does or how the guy who gave it to you obtained it. It'd be interesting to ask him and report here.

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:

Quote:

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Originally Posted by rpm

to be honest i don't completely understand how this equation tells us this. but it was given to me by spoon, who knows his shit. so i've just accepted that this particular equation does what i've been told it does.
.

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i'll check out the part of your response which requires some thinking when less inebriated. thanks again for taking the time man.

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Originally Posted by daviddem

. It was my understanding from your previous posts that your were looking for your unexploitable 5bet shoving range. If this is the case, then refer to my previous post and mathematics of poker.
?

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i know what an unexploitable 5b shove range is, what i didn't know how to do is determine that range.

an unexploitable shoving range is one in which the very worst hand in that range has an EV of 0. so i was looking to be able to determine which pot/fold equity combinations produce 0EV. spoon gave me that equation which (as best i know, though i've become kind of confused) is designed to determine which F values produce 0EV based on different y values (ie the worse hand in our shoving range). hence telling us what the bottom of our range should be based on different fold equity %'s in order to 5b AI unexploitably.

edit: on that note, your graph seems to show me exactly what i was after. so thanks again for that.

One thing to be wary of here is the idea of "worse hand in your unexploitable shoving range".

For example take the following game:
- starting stacks of 1000
- blinds 0.5 and 1
- you are SB and button
- you can shove or fold, then villain can call or fold

It can be shown that your unexploitable strategy is to shove only AA, and the unexploitable strategy of villain is to call with only AA.

However as the starting stacks decrease relative to the blinds, it will become optimal (unexploitable) for you to start shoving wider than just AA, against villain who still only calls with AA at this stage. You might think that the next hand you will add to your shoving range is KK? Not so. It can be shown that the first hand to add to your shoving range is ATs. With stack sizes of 833.25, you will find that the EV of shoving AA only is the same as the EV of shoving {AA,ATs}, whereas the EV of shoving {AA,X} (with X=any hand other than AA or ATs) is worse. One of the reasons for this is that your fold equity is better with ATs than with hands that do not contain an A because there are only 3 AA combos left for villain to have, whereas there are 6 otherwise. Another reason is that AT offers more straight possibilities than for example AK. If you stove {AA,ATs} vs AA, you will get slightly better equity than with {AA,AKs}.

Again it is very much worth reading the corresponding chapter of MoP. Very insightful once you get past the math.