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

[AtticusFinch]

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<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.
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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?

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ln(e*(2Q-Q)/Q) = ln (e*Q/Q) = ln(e) = 1.

[Che]

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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.

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Contrived Example: While hand-for-hand as the bubble approaches in a Party MTT, player A has a stack = 3*average. Every player at A's table also has 3*average. Player B has a stack = 3*average. Every other player at B's table has .2*average.

Player B's $EV at his current table is clearly higher than Player A's at A's current table due to B's ability to steal the blinds every hand while Player A must be much more conservative. Thus, Player B's overall $EV is also higher since their future $EV's must be assumed to be equal (future table situations are unknown, but should balance out on average in the long run).

I use this example to demonstrate that factoring the current table situation into $EV calculations is appropriate.

However, a precise formula that would account for table differences obviously approaches the impossible in terms of complexity.

What you appear to be looking for here, however, is just a framework that calculates $EV using chip stack ratios assuming all else equal. The skilled player will be able to adjust the $EV estimate up or down if relative chip stacks at current table dictate. Or if any other relevant, but unaccounted for, variable dictates, e.g. relative skill level at current table.

Summary: My claim is that table situation is an *appropriate* variable to consider, but your effort to quantify $EV can provide value without considering it.

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

I hope you keep working on this, AF. I think you're onto something.

Later,
Che

[jcm4ccc]

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I wanted logarithmic decay for below-average stacks as well as logarithmic growth for above-average . . . 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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It seems to me that you created a formula to correspond with how you think that the probability of winning should be. There seems to be no theoretical basis for your formula.

As to your formula, you say that doubling up exactly doubles your chances of winning. So let's take a theoretical tournament with 100 players. It's the first hand. At this point, everybody has a 1% chance of winning.

You double up on the first hand. So now, according to your formula, your chances of winning have doubled to 2%. One player is out of the tournament. There are 98 other players left. These 98 players have a 98% chance total of winning the tournament (since you have a 2% chance of winning), which is exactly what they had before you busted that guy. In other words, the fact that you busted out one player does not in any way help the other 98 players. You accrued all the benefits of busting that guy. That does not seem right to me.

Actually, I believe if you used your formula at this point to calculate the chances of all 99 players (yourself included), you will see that the total adds up to something greater than 100%, I believe.

Now let's assume that you busted out 2 players. Your chances of winning the tournament are about 2.69%. There are 97 players left. They have a 97.31% chance of winning the tournament, which is a little bit better than what they had before you busted those two guys. So busting one guy didn't help them, but busting two guys did. Doesn't seem logical to me.

[justT]

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You double up on the first hand. So now, according to your formula, your chances of winning have doubled to 2%.

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Per his formula, your chances of winning would be a bit less than 2%, he "cheated" when he calculated the double up case here:

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Now, Let's say you double up early, and have ~2q chips.

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you'd actually have a bit less than 2q since your doubling increased Q, the net affect is your probability of winning increases by just a tad under a double

[justT]

Ah, I got it now. I looked at a couple of extreme cases (2 person tourney, and chip &amp; a chair) and it didn't explode. On a first cut basis, it looks reasonable.

[AtticusFinch]

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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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I hope you keep working on this, AF. I think you're onto something.


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Will do. Let me know if you have any ideas.

[AtticusFinch]

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It seems to me that you created a formula to correspond with how you think that the probability of winning should be. There seems to be no theoretical basis for your formula.


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It's an attempt at a model. My theoretical basis is the logarithmic value of money taken from economics. Not that I need one. All that's required it to test to see if it fits the data reasonably well. I designed it so its results fit with most people's intuitive sense of stack values.

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As to your formula, you say that doubling up exactly doubles your chances of winning. So let's take a theoretical tournament with 100 players. It's the first hand. At this point, everybody has a 1% chance of winning.

You double up on the first hand. So now, according to your formula, your chances of winning have doubled to 2%. One player is out of the tournament. There are 98 other players left. These 98 players have a 98% chance total of winning the tournament (since you have a 2% chance of winning), which is exactly what they had before you busted that guy. In other words, the fact that you busted out one player does not in any way help the other 98 players. You accrued all the benefits of busting that guy. That does not seem right to me.



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You're right. You seem to be missing the fact that this is a first cut. I agree the double point is a little high, but the higher points look pretty close.

Also, remember that if you double, you won't have exactly 2Q chips. You'll have slightly less than 2Q, because Q has increased due to the loss of a player. (I used the 2Q approximation for simplicity, assuming a large field) . Thus your odds will end up slightly less than 2Pq, and some small advantage will be distributed across the other players.

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Actually, I believe if you used your formula at this point to calculate the chances of all 99 players (yourself included), you will see that the total adds up to something greater than 100%, I believe.


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Actually I suspect it adds up to less than 100. The logarithmic graph grows more slowly than a linear one.

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Now let's assume that you busted out 2 players. Your chances of winning the tournament are about 2.69%. There are 97 players left. They have a 97.31% chance of winning the tournament, which is a little bit better than what they had before you busted those two guys. So busting one guy didn't help them, but busting two guys did. Doesn't seem logical to me.

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Yep. Some adjustments need to be made. I'm sure some constant factors will have to be applied to make it fit better. I also suspect a factor of the number of players in the field will have to be fit in somewhere. I'm busily working on the next rev of the formula.

Set aside these details for a moment and just evaluate the general concept. I never claimed my first rev would be exactly right. I'm hoping to get some feedback to help me refine it.

[AtticusFinch]

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You double up on the first hand. So now, according to your formula, your chances of winning have doubled to 2%.

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Per his formula, your chances of winning would be a bit less than 2%, he "cheated" when he calculated the double up case here:

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Now, Let's say you double up early, and have ~2q chips.

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you'd actually have a bit less than 2q since your doubling increased Q, the net affect is your probability of winning increases by just a tad under a double

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Exactly. I fudged it a little for simplicity. Despite this, I still think it's a little high.

[curtains]

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I wanted logarithmic decay for below-average stacks as well as logarithmic growth for above-average . . . 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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It seems to me that you created a formula to correspond with how you think that the probability of winning should be. There seems to be no theoretical basis for your formula.

As to your formula, you say that doubling up exactly doubles your chances of winning. So let's take a theoretical tournament with 100 players. It's the first hand. At this point, everybody has a 1% chance of winning.

You double up on the first hand. So now, according to your formula, your chances of winning have doubled to 2%. One player is out of the tournament. There are 98 other players left. These 98 players have a 98% chance total of winning the tournament (since you have a 2% chance of winning), which is exactly what they had before you busted that guy. In other words, the fact that you busted out one player does not in any way help the other 98 players. You accrued all the benefits of busting that guy. That does not seem right to me.

Actually, I believe if you used your formula at this point to calculate the chances of all 99 players (yourself included), you will see that the total adds up to something greater than 100%, I believe.

Now let's assume that you busted out 2 players. Your chances of winning the tournament are about 2.69%. There are 97 players left. They have a 97.31% chance of winning the tournament, which is a little bit better than what they had before you busted those two guys. So busting one guy didn't help them, but busting two guys did. Doesn't seem logical to me.

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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)

[Exitonly]

i think it might increase your EV by 100%, or close, you're twice as likely to win, and considerably more likely to cash..