Text results appended to pokerstove.txt

197,388,576 games 0.248 secs 795,921,677 games/sec

Board:
Dead:

equity win tie pots won pots tied

Hand 0: 72.790% 72.29% 00.50% 142686684 991662.00 { AA }
Hand 1: 07.276% 06.77% 00.50% 13370796 991662.00 { AQo }
Hand 2: 19.934% 19.81% 00.13% 39098916 248856.00 { 43s }



When there is a battle beetween 3 players, what is the meaning of "pots tied"?

When there is pot splitted the split is possible in this way:

split beetween player 1 and 2
split beetween player 2 and 3
split beetween player 1 and 3
split beetween player 1, 2 and 3.

My question:

from the answer of pokerstove.txt, is possibile to restrive:

split beetween player 1 and 2
split beetween player 2 and 3
split beetween player 1 and 3
split beetween player 1, 2 and 3
?



Thanks a lot for the patient and the attection...I hope you can help me

What do you mean by "restrive"?

I know this:

equity win tie pots won pots tied

Hand 0: 72.790% 72.29% 00.50% 142686684 991662.00 { AA }
Hand 1: 07.276% 06.77% 00.50% 13370796 991662.00 { AQo }
Hand 2: 19.934% 19.81% 00.13% 39098916 248856.00 { 43s }

But I would know the percentage of these split

split beetween player 1 and 2
split beetween player 2 and 3
split beetween player 1 and 3
split beetween player 1, 2 and 3

assuming that the main pot is splitted.

For example the equity of split in the example for hand 0 is 0.5%.

This 0.5% is the sum of :

split beetween hand0 and hand1
split beetween hand0 and hand2
split beetween hand0 and hand1 and hand2

How can i calculate these value?

Not possible using pokerstove as far as I know.
My guess is you would have to start programming.

Why do you need these specific numbers?

Yea, as far as I can tell, calling that number 'pots tied' is a bit misleading since that's not what it is, even in the 2-player case. In the 2-player case, it's number that they add 0.5 to each time there's a split pot, i.e. it's half the number of split pots. So, you can find the equity by just going

(pots won + "pots tied") / (total number of games)

In the 3-player case, I imagine the "pots tied" number is something they add 0.3333 to when there's a 3-way split and 0.5 to when there's a two way split...

Doesn't it come down to simultaneous equations?

Let
a = the number of pots split between players 1 and 2
b = the number of pots split between players 2 and 3
c = the number of pots split between players 1 and 3
d = the number of pots split between players 1,2, and 3

Then
(a/2) + (c/2) + (d/3) = 991662 (looking at player 1)
(a/2) + (b/2) + (d/3) = 991662 (player 2)
(b/2) + (c/2) + (d/3) = 248856 (player 3)
a+b+c+d = 2232180 (total split pots)

If that's correct, presumably there's a website out there somewhere that will solve a set of simultaneous equations?

EDIT: In fact, I think we've only got 3 equations, but 4 unknowns (add the first three equations together, and you get the fourth, so we don't actually have four equations, which we'd need for a unique solution. Is there an equation I'm missing??)

No, you're right, system is not invertible.

Ah, right, thanks. I solved manually and got d=0, which is clearly nonsense because they'll all end up playing the board sometimes. Only then realised we had just 3 equations.

hm can you do least squares? A^T A (x*) = A^T (b)

or is that only when there are more equations than variables

Thank you guys for the support...

I want this information because when someone want to analize an EV spot in a sitngo or in a tournament, in my opinion is not perfect to consider:

equity of player 1
equity of player 2
equity of player 3

but:

percentage of winner of player 1
percentage of winner of player 2
percentage of winner of player 3
percentage of splitted pot beetwen player 1 2 3
percentage of splitted pot beetwen player 1 2
percentage of splitted pot beetwen player 1 3
percentage of splitted pot beetwen player 2 3

This new scenario is different from the early one...specially in stg tournament...

So in your opinion is not possible to obtain the percentage of split?

I agree that sort of thing makes a difference for icm type calculations but I'd imagine its almost always very small

Not sure.
God, I feel old sometimes.
Looooong time since I went anywhere near this sort of stuff.

ICM with some improvment could be an invaluable method to study situation like "cEv" in cash game.

I am agree with someone who think "the difference is small". But when there are ICM calculation there are A LOT of operation who are not super accurate.

lot operation not super accurate * small error = not so small error.

If we resolve this problem of splitted pot and with a couple of improvment, ICM will be perfect!

Plz someone resolve this problem of splitted pot!!!

Haha. I just finished a course in linear algebra so I'm an expert (in my own mind)...using the normal equation above, I got

a= 1,734,408.000
b= 248,796.000
c= 248,796.000
d= 177.948

which appears to be correct if we plug these values back into the system. d not equal to zero so the board will be played sometimes...very little.

^^^ Good stuff.

(When I graduated, personal computers were unheard of. Long time ago. )

Did we come up with another equation? Otherwise, it's definitely an underdetermined system...

While I don't know of the formula that answers your question, for this exact example you can figure out the answer. Since Hand 2 can never split with only Hand 0 or only Hand 1 given they have 43s and any straights shared, such as a wheel, will be with both of the other players, so the # of pots tied for Hand 2 (248,856) can be deducted from the pots tied for the other 2 hands (991,662). The difference of the 2 will be when Hand 0 and Hand 1 split.

The bolded part above can't be correct. Since the 43s (player 3 or Hand 2) can't split with ONLY one of the other players since the only possible splits are either wheels or sharing the board. So the 248,856 represents the # of hands that ALL 3 tie. The difference between 991,662 and 248,856 would represent the # of times that player 1 and player 2 tie.

If you add the first three equations together, you get the bolded one. (If I'd realised that to start with, I wouldn't have included the bolded one - it adds nothing.)

So if the bolded one is incorrect (and I don't think it is) there must be something wrong with at least one of the first three.

^^^ but I agree that in this particular case, it looks as though b=c=0 (3rd hand cannot split with just one of the 1st and 2nd hands)

so the solution is

a=1485612
b=0
c=0
d=746568


(Not sure if I'm getting more, or less, confused as this thread goes on )

If I have well understand wanderswaag says that is possible to find a solution to my question.

It's right?

What is the final answer?

I think I can give a good support to ICM improvment with your help!

This is correct for the ranges given. 43s plays the board when it ties against AA or AQ. It can't beat any hand when it plays the board.

For more complicated ranges, I think (but I'm not sure) if you could run the hands separately and get 3 more equations for:
(a/2) + (d/2) = x
(b/2) + (d/2) = y
(c/2) + (d/2) = z

Those equations combined with the a+b+c+d = w equation would always be solvable.

I don't have access to pokerstove right now. But if anybody wants to confirm that

{AA} against {AQo} has 2232180/2=1116090 pots tied
{AA} against {43s} has 746568/2=373284 pots tied
{AQo} against {43s} has 373284 pots tied

then my idea might have some merit. If the numbers are different I'm either completely wrong or made a mistake calculating.

This is still incorrect. You are double and triple counting the number of tie hands. If each player splits the pot 3 way 248,856 times, that is 248,856 hands ONLY not 3x 248,856. The number of times that player 1 and player 2 split is the difference between 248,856 and 991,66, not 2x the difference. Each hand that they split shouldn't be counted as additional hands (i.e if x, y & z split 5 hands and a and b split 12 between themselves, that's a total of 17 split hands 5+12, not 5+5+5+12+12).

There were 197 388 576 hands in total.

According to the Pokerstove results in OP, the first player won outright 142 686 684 of them.
The second player won outright 13 370 796 of them.
The third player won outright 39 098 916 of them.

That's 195 156 396 hands won outright.

That leaves 197 388 576 - 195 156 396
= 2 232 180 hands that were tied.

Each of those tied hands falls into one (and only one) of the four categories which I labelled a,b,c,d.

The solution I posted above is as follows:


Those values lead to the following equity distribution:



(for example, the equity that the first hand gets from ties = (1485612/2) + (746568/3) = 991662)

which agrees with the Pokerstove results in OP.