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Old 10-14-2005, 06:10 PM
AtticusFinch AtticusFinch is offline
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Join Date: Feb 2005
Posts: 620
Default Theory: Gigabet\'s \"bands\" and \"The Finch Formula\" Grand Unification

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]
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