Theory: Gigabet's "bands" and "The Finch Formula" Grand Unification

[ QUOTE ]
[ QUOTE ]
I'm not sure I'm reading your formula correctly. Is that ln|e^((S-Q)/Q)|?

[/ QUOTE ]

No, that would evaluate merely to (S-Q)/Q.

It's ln(e * (|S-Q|)/Q)

-AF

[/ QUOTE ]

Okay, good. How'd you decide on that?

If I were in math class, I'd write the whole thing as

Ps = Q/T * (1 + sign(S-Q)(1 + ln|(S-Q)/Q|)

To make it easier for people who are less familiar with math notation, maybe two formulas for the different signs?

If you program, you may wanna try writing a program that calculates your actual EV with various chip stacks in various sized tournies, using the trick that your odds of coming in second are the sum of your odds of coming in first in a tourny without player i times the odds that player i wins.

All you need to know is when to lay it down and you will win plenty, no need for this math stuff.

[AtticusFinch]

[ QUOTE ]

Okay, good. How'd you decide on that?


[/ QUOTE ]

The logarithmic function idea comes from economics. The e factor is just to normalize the logarithmic base so it centers around the probability of the average stack. The |s-q| factor is necessary because ln is undefined for negative values, but I wanted logarithmic decay for below-average stacks as well as logarithmic growth for above-average (Logarithmic decay has been demonstrated by economists studying debt-spending. The deeper in debt you are, the less you care about going another $100 into debt.) So, I just pulled the sign outside the ln, so that you decay from Pq logarithmically as your stack falls farther below average.

Finally, I wanted to model the need for exponential growth centered around the average stack, not your stack. Thus the Q denominator.

[ QUOTE ]

If I were in math class, I'd write the whole thing as

Ps = Q/T * (1 + sign(S-Q)(1 + ln|(S-Q)/Q|)



[/ QUOTE ]

That is a bit cleaner. You could also replace Q/T with Pq, to make it clear that this is a comparison to the probability of the average stack:

Ps = Pq * (1 + sign(S-Q)*(1 + ln(|e*(S-Q)/Q|))

[ QUOTE ]

If you program, you may wanna try writing a program that calculates your actual EV with various chip stacks in various sized tournies, using the trick that your odds of coming in second are the sum of your odds of coming in first in a tourny without player i times the odds that player i wins.

[/ QUOTE ]

That's the next step. That approach is similar to how ICM works, in fact. I'll code it up eventually, but I wanted to get some feedback on my formula first.

I'm a little uncomfortable with deviating from the idea of a tourny as a random walk; that's a very good approximation of an unskilled tourny.

If you find a dataminer who's willing, you can test this and see which is better. I definitely think you should follow through with this, as it would be awesome if we didn't have to keep calculating cEV and then guessing EV.

I dont like this "sign" operand. We can get rid of it, thereby making the equation longer and therefore look more complex (which is really what we're after right?), by making it:
Ps = Q/T * (1 + (-1^(ceiling(S/Q))(1 + ln|(S-Q)/Q|)

This change works as long as S > 0, and if S = 0 then stop doing math and yell "rebuy" as loud as you can.

[mlagoo]

Is it possible to make use of this in the way that SNGPT works for SNGs?

How difficult would it be?

[AtticusFinch]

[ QUOTE ]
Is it possible to make use of this in the way that SNGPT works for SNGs?

How difficult would it be?

[/ QUOTE ]

That is the eventual plan. Just be warned, this formula only estimates your odds of winning. Using it might therefore lead to some -$EV decisions. I don't know.

As far as how to use it goes, just proceed as you would with ICM. Figure out all possible results and their probabilities, then evaluate the Finch Formula for each. Whatever action gives you the highest overall chance of winning is the one to take. (Assuming, like me, you always want to play to win.)

[ QUOTE ]
[ QUOTE ]
Is it possible to make use of this in the way that SNGPT works for SNGs?

How difficult would it be?

[/ QUOTE ]

That is the eventual plan. Just be warned, this formula only estimates your odds of winning. Using it might therefore lead to some -$EV decisions. I don't know.

As far as how to use it goes, just proceed as you would with ICM. Figure out all possible results and their probabilities, then evaluate the Finch Formula for each. Whatever action gives you the highest overall chance of winning is the one to take. (Assuming, like me, you always want to play to win.)

[/ QUOTE ]

I am good with math, but I don't see how applying this formula would change the simpler calculations we usually go through in making decisions in a hand. Could you give an example where you think the former would be a better choice than the calculation that would normally be done? I find this interesting, and I am good at math, so when I'm not studying for programming midterms, I'll go over your formula with more attention than I can devote at the moment.

[Exitonly]

[ QUOTE ]
I welcome all comments. ExitOnly, I'm looking in your direction

[/ QUOTE ]

I missed that line till i read it the 2nd time.. [img]/images/graemlins/smirk.gif[/img]


Alright,

First, completely non-important point, doesnt Giga call them "blocks" not "bands"? Or is that something else?

Ok now to the Finch Formula.

If you were to add all the numbers for everyone in the tournament, would you get 1? i've been trying to type this all into Excel to be able to do this myself (As i've lost my TI-89) but i'm screwing it up somewhere and don't feel up to finding my error right now.

I like this as a base, but somehow gotta add-in all the other cashes, because i dont think your chance of winning doubles w/ an early double up, but it probably doubles your EV.

I gotta go run and eat dinner, i'll look at this some more when i get back.

Nice work though.

[ QUOTE ]

If you were to add all the numbers for everyone in the tournament, would you get 1?

[/ QUOTE ]


Can't believe I didn't check that

If you have a stack of 2Q with a field of n players and they all have equal stacks, then they all have (T-2Q)/(N-1) = Q(N-2)/(N-1). Thus, each has a probability of coming in first of

P = Q(1 +/- ln(e(|Q(N-2)/(N-1)-Q|)/Q)/T
= Q/T * (1 - ln(e/(N-1)) = Q/T*ln(N-1)

Thus, the odds that we come in first should be 1-(N-1)Q/T * ln(N-1)

Ps = Q(1 +/- ln(e(|2Q-Q|)/Q)/T
= Q/T * (2+ln(2)).

I think you need to normalize.