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

[AtticusFinch]

A while back I proposed coming up with a model for valuing chip stacks in MTTs. Sort of like a larger-scale ICM. I've arrived at a preliminary candidate. Interestingly, not only does it quantify the need to grow your stack exponentially, it also goes a long way toward explaining the "band" concept Gigabet posted about a while back.

My thinking was as follows: Clearly the more chips you have the better, but it can't be a linear progression. MTTs are games where, if you expect to win, you must grow your stack exponentially. As such, if you want to evaluate the value of your present stack, the proper function to use is logarithmic.

This makes intuitive sense, as all economists know that money has logarithmic value. Having $1 million is a staggering difference over having $0, but having $6 million is not much better than having $5 million. And if you have $100 million, well, what's a million or two between friends? Similarly, early in a tourney, getting 1000 chips is quite valuable. But later on, when the blinds alone are 20k, it's hardly worth noticing.

I'll start off with the assumption that a player with an average stack Q in a field of size n where there are T total chips in play has a Q/T = 1/n chance of winning. I'll call this value Pq. I'll also assume (for starters anyway) that you only care about winning.

If S is the size of your stack, Q is the average stack size, and T is the total number of chips in play, my first cut at a formula for your chances of winning (which I'll call Ps) is:

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

Where the +/- matches the sign of (S-Q). (Apologies for the hard-to-read format.)

Let's have a look at how this plays out. First of all, notice that when S = Q, Ps evaluates to Q/T, exactly as we would expect.

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.

So your odds of winning exactly double, which is pretty close to the accepted value. (I think the actual accepted value is something like 1.95)

How about a triple?

Ps = Q(1 + ln(2e))/T ~= 2.69*Pq

Your odds have gone up over a double, but have not tripled, which makes intuitive sense.

For a quadruple, the result is 3.1* Pq. Once again, this makes intutitive sense. It's better than a triple, but your odds are not four times that of the average stack.

Finally, if you are halved, the result is .69* Pq. Your odds have gone down significantly, but have not been cut in half. This also makes sense, if you think about it. It accounts for the fact that you can grow exponentially in this game, and catch up quickly in just a few hands. A chip and a chair, baby!

As for Gigabet's "bands," Have a look and you'll see that they correspond with areas of the log graph that are less steep. Once you're way above average, you need to gain far more chips to show any real advantage, while losing a small number of chips hurts you much less than the same loss by a smaller stack.

Similarly, once you're in the desparation zone, losing more chips falls along a flatter curve than gaining chips, which rapidly becomes a very steep curve in terms of gain in value. Thus you are more justified in taking risks like pushing trashy hands.

Note also that this formula is all about your relationship to the average stack, just as Gigabet's Bands are.

The "Finch Formula," as I call it, probably needs some minor adjustments, but I think it has potential to become the ICM for MTTs. If it proves valid, it could help greatly in close push/fold/call decisions.

I welcome all comments. ExitOnly, I'm looking in your direction [img]/images/graemlins/wink.gif[/img]

[CardSharpCook]

Nice work Atti. Nice to see in #s what we know intuitively. However, I don't believe that an early double up actually doubles your odds. Do you?

Very interesting, nice post. One thing I'm wondering about: if I recall correctly, in your original post, you hypothesized that the value of stack sizes would be logarithmic in part because players with larger stacks had more moves available to them than those who were shorter. They have chips to back up their bluffs, discourage people from bluffing at them, etc. But those factors have more to do with the other stacks at your table than with the relationship of your stack to the average in the tournament.

Similarly, I would expect that, assuming you were an above average player for your table, chips you earned beyond the size of your next largest opponent would be less valuable since you would not be able to invest them in +cEV situations. I guess what I'm getting at is that this formula may miss a lot of the value that is dependent on the size of your stack relative to others at your table. I don't know enough about math to answer this: would using the same formula, but would substituting the average stack at your table for Q approximate the worth of your stack relative to others at your table?

Thanks again for the interesting post.

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

[AtticusFinch]

[ 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

[AtticusFinch]

[ QUOTE ]
Nice work Atti. Nice to see in #s what we know intuitively. However, I don't believe that an early double up actually doubles your odds. Do you?

[/ QUOTE ]

I didn't at first. But all my research has shown it almost doubles. As I said, the accepted value that I've seen is something in the neighborhood of 1.95. That my formula evaluates to exactly 2 is interesting, but it was not intentional.

[AtticusFinch]

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But those factors have more to do with the other stacks at your table than with the relationship of your stack to the average in the tournament.


[/ QUOTE ]

I don't fully agree. I think it has more to do with the size of the blinds. And that most often directly relates to the size of the average stack.

[ QUOTE ]

Similarly, I would expect that, assuming you were an above average player for your table, chips you earned beyond the size of your next largest opponent would be less valuable since you would not be able to invest them in +cEV situations.


[/ QUOTE ]

While this is a reasonable point, the order of growth is the same: logarithmic. If the center point is slightly different due to table conditions, it's unlikely to modify the results very much, as a logratihmic function is very close to linear over short distances.

[ QUOTE ]

I guess what I'm getting at is that this formula may miss a lot of the value that is dependent on the size of your stack relative to others at your table. I don't know enough about math to answer this: would using the same formula, but would substituting the average stack at your table for Q approximate the worth of your stack relative to others at your table?


[/ QUOTE ]

Considering how often you move tables, and people bust out, being replaced by new folks, I don't think it's appropriate to evaluate your overall tournament expectation based on your single table. Remember, you still have to beat everyone in the field, not just your table-mates.

I'm not suggesting table conditions won't come into play when making an individual decision. I'm just saying they're not an appropriate factor for making a determination of your overall equity in the tourney.

Thanks for the comments.

[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 ]
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.