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

[curtains]

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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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I just dont get this formula at all? Why should it be true at all in a winner take all format. If you double up the first hand, your chances should double exactly, from a theroetical standpoint. If you are tripled up, your chances should triple. What reason in the world is there to say that they haven't tripled? That's like saying that when you win 100% of the chips you have a 99.2% chance of winning the tournament. Either I'm completely misreading this theory (likely) or it makes zero sense to me.

I'm sorry, but unless someone convinces me otherwise, I see no convincing data in the original post.

[kamrann]

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If you double up the first hand, your chances should double exactly, from a theroetical standpoint. If you are tripled up, your chances should triple. What reason in the world is there to say that they haven't tripled?

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Where does this come from? It hasn't been proved beyond doubt that your probability of winning is exactly equal to your stack size as a proportion of the chips in play, has it? That's been used as a reasonable assumption for various models, such as ICM, but I see no reason to take it as fact.

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That's like saying that when you win 100% of the chips you have a 99.2% chance of winning the tournament.

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This does not have to follow, for example if there was an inflexion point in the curve, it could reconverge to a 100% chance of winning at 100% of the chips, right?

Also, on a totally unrelated note:
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Ps = Q(1 +/- ln(e(|S-Q|)/Q)/T
Let's have a look at how this plays out. First of all, notice that when S = Q, Ps evaluates to Q/T

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Forgive me if this is a dumb question, my math is rusty, but if S = Q then surely you are taking the log of zero, which is undefined? It seems you are missing a closing bracket in the formula though, so maybe this is leading to my confusion.

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This nitpick has now reached its conclusion. [img]/images/graemlins/smirk.gif[/img]


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Ok, I agree 100%, I just should have worded it better. As I said, the formula should be used in conjunction with judgments about your table conditions.


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This is all I meant to say as well. In particular, I was asking whether the same formula would help to calculate the value of your stack relative to others at your table. In other words, I didn't mean to imply that simply considering the other stacks at your table would estimate your stack's worth overall, but rather that a player might want to make two calculations when faced with a decision, calculating first his equity using the Q for the whole tournament, second using the Q of his table alone.

However, I think that was based on a misuderstanding of the formula. As I understand it now, it calculates your chance of winning first, but not necessarily your equity overall?

[Paragon]

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Where does this come from? It hasn't been proved beyond doubt that your probability of winning is exactly equal to your stack size as a proportion of the chips in play, has it? That's been used as a reasonable assumption for various models, such as ICM, but I see no reason to take it as fact.

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There is some mathematical basis to assume your chance of winning is exactly proportional to your stack and the total number of chips in the tournament (especially for a single table).

Take a specific example with 10 or fewer players at a table all with various stacks sizes. Now have every player go allin every hand regardless of cards until someone wins. The chance you win is precisely that proportion in this example. Extending this method you can also find the exact odds of finishing 2nd-10th as well. The algorithm for ICM merely does this, then checks the percentage of the prize pool awarded for each place, and then gives you the value of your stack in terms of real dollars. I actually wrote a program that simulated this (everyone allin) and everything all converges nicely to ICM results. In any case, although I would hesitate to say anything is a fact when theorizing about poker, this proportion everyone constantly refers to is more than a wild guess at least.

Anyway, this would help explain Che's example with the two table scenario. Suppose you have Table A where everyone has stack size X, and Table B where one person has stack size X and all others have less than X. If you freeze the tournament at this point and everyone starts repeatedly allining, my intuition thinks the one player on Table B would have the greatest +$EV, assuming you get the table balancing algorithms identical to PartyPoker or whoever. I never tested anything like this.

In my opinion, the greatest shortcoming to ICM is that it does not factor in fold equity of future hands. I'm not sure how you could analytically describe that though.

[justT]

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I also suspect a factor of the number of players in the field will have to be fit in somewhere.

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I think its already there, hidden in the Q/T

[justT]

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I really don't understand? Doubling up oin the first hand in a 100 person tournament should increase your chances of winning by 100%, and thus make them 2%. This will however not increase your EV by 100%, assuming the tournament isn't winner take all. No one else in the tournament should gain any chance of winning the event just because you busted another player. Yes they gain EV if it pays multiple spots, but they don't suddenly have a greater chance of winning first place, that would make no sense.

Just as if you busted everyone but one player on the first hand, your chances to win would be 99%, and your one opponent would have the same chance to win the event as when they first bought in, despite being up against one opponent. Thus your opponent would have gained absolutely from you eliminating all other 98 players, assuming this was a winner take all format. (I know its impossible but just giving random example)


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I was struggling with the same problem last night. I worked the probabilities for a double up in a field of 10, 100, and 1000. I lost the exact numbers but they were something like 1.99, 1.98, 1.95 respectively. The first question is why would it go down for larger fields? I think it makes sense that the more poker there is to be played, the less meaningful your double up is. I suspect it has to do with someone getting extremely lucky (like hitting 17 reds in a row at the roulette table) and/or with you getting extremely unlucky (betting red 17 times in a row only to have it come up black each time). The less poker there is to play, the less that these extreme cases come into play and the closer a double up in chips is to a double up in probability. It doesn't prove anything, but I think that's where it's hidden.

Okay, second question using your example. If I've eliminated 98 players, how can my chance of winning be less than 99%? Or a different way of putting it, how can my beating 98 players increase the one remaining players probability of winning. Okay, back to the string of really lucky events, cause that's what the one remaining player is going to need to beat you. If you had eliminated only 97 players, then there would be one other player who could get extremely lucky and beat both of you. By eliminating that player, you take away the chance of that happening and thus actually increase the probability, albeit VERY slightly, for your one remaining opponent to win. IOW, he ONLY needs to worry about hitting his 17 reds in a row, he doesn't have to worry about the case where he hits his 17 reds in a row but then other short stack then proceeds to hit his 17 reds in a row to beat him.

I gotta go kill a chicken or something.

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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.

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It's pretty necessary, as the example that I gave returns negative probabilities at N>4. It's pretty cool that you got a formula that fits 3 of the four criteria though. I'll try and normalize it when I get a chance (it's parents' weekend at school right now.)

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.

[AtticusFinch]

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I really don't understand? Doubling up oin the first hand in a 100 person tournament should increase your chances of winning by 100%, and thus make them 2%.


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Not quite. Your stack is not exactly 2Q, as Q has increased by the elimination of one player. The new Q is T/(N-1) where T is the total chips in play, and N is the starting number of players. Your stack after doubling is 2T/N, which is slightly smaller than 2Q after the elimination.
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This will however not increase your EV by 100%, assuming the tournament isn't winner take all.


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As I've said repeatedly, I'm not calculating EV. Just estimating odds of winning.

[AtticusFinch]

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I just dont get this formula at all? Why should it be true at all in a winner take all format. If you double up the first hand, your chances should double exactly, from a theroetical standpoint. If you are tripled up, your chances should triple. What reason in the world is there to say that they haven't tripled?


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There are a number of reasons. The simplest one is that the other players benefit from the elimination of a player. The more important reasons are:

1) Blinds go up exponentially. You must keep up with the blinds, so you must grow your stack exponentially.
2) The minimum of the size of your stack and the size of your opponents' stacks determines how much you can gain at any one time.
3) As the blinds go up, people drop out, and the average stack (q) grows.

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That's like saying that when you win 100% of the chips you have a 99.2% chance of winning the tournament. Either I'm completely misreading this theory (likely) or it makes zero sense to me.


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Run my formula. You will find that when you have all the chips, it
returns 100%, and when you have 0 chips, it returns 0. Furthermore, when you have the average number of chips, it returns Q/T.

Every major poker book out there says the same thing. Your next X chips is not worth as much as your first X chips. But it is worth something. That pretty much describes logarithmic growth.

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I'm sorry, but unless someone convinces me otherwise, I see no convincing data in the original post.

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There's no data, only theory. [img]/images/graemlins/smile.gif[/img]