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

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

[AtticusFinch]

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


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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.

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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|)



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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|))

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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.

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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.

[Exitonly]

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

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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.

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.

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

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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.

[AtticusFinch]

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

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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.

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I figured I would. This is just a first cut. There are 3 things I like about it that I'd like to maintain:

1) If S = Q, Ps = Pq

2) It's continuous throughout

3) Ps hits 0 precisely when S = 0

Which should be doable.

[justT]

<font class="small">Code:</font><hr /><pre> Now, Let's say you double up early, and have ~2q chips. Your chance to win now is:

Ps = Q(1 + ln(e(2Q - Q)/q))/T = 2Q/T = 2*Pq.
</pre><hr />

It looks like something is screwed up in your formula. Or maybe I'm screwing it up. ln(e(2Q - Q)/q) = 1 for any value for Q?

[Exitonly]

Math has officially gone over my head... last math class i took was AP Calc in highschool (5 on the test ::flips collar:: )

so yea, i'm not gonna be much help w/ the math of this, but i'll keep looking it over anyway. Maybe i'll find some theory errors or something.

[Exitonly]

are you doing e^(2Q-Q) or e * (2q-q).. that could be your problem.

[AtticusFinch]

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


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Err, check it over. ln(e(|2Q-Q|)/Q) = ln(eQ/Q) = 1

Ps thus evaluates to 2Q/T = 2Pq.

Your earlier calculation is correct, however.

I'll think about how to normalize it, but in the meantime, ponder whether it's really necessary. This doesn't have to be a actual probabilistic calculation to be useful. It only needs to model the relationship between probabilities of different stacks well enough.

If you come up with a normalized version, by the way, I'd love to see it.