Anyone read - \"Theory of Doubling Up\" by Chen and Ankenman?
 

In the latest volume of "The Intelligent Gambler" there is an article by William Chen and Jerrod Ankenman called "The Theory of Doubling Up" where they present a mathematical model that "can be utilized to estimate tournament equity on a skill-adjusted basis".

It's a slightly mathematical article (not bad really) and very thought provoking. I am not an expert in tournamnent by any means so I am wondering if any of you experts have read it and if so what you thought.

*I just got this article today so maybe you haven't read it yet. I can't find a link to it on the net, if someone can please post it.

Re: Anyone read - \"Theory of Doubling Up\" by Chen and Ankenman?
 

Is this the first you double up and then you triple up theory presented in Tilt? I'm proud to say I employed the alternative:
1.triple up
2.double up
theory in a 120 person B&M tourney in Shreveport recently and won $1600.

I knew if I looked hard enough I would find an excuse to post this shamless brag!

[img]/images/graemlins/laugh.gif[/img]

Re: Anyone read - \"Theory of Doubling Up\" by Chen and Ankenman?
 

1st everyone getting HOH2 before me. Then going to Borders only to see 75 books by Tom McEvoy, and only one copy of TOP, and HOH1. Now an article I would love to read without a link.

Today is forever going to be known as the day the poker died.

Im going to go join RGP now......

Re: Anyone read - \"Theory of Doubling Up\" by Chen and Ankenman?
 

I would link to the article if I could find it but it's just on a pamphlet. It's very good. I'm going to try and summarize it but I'm very busy today and I don't want to plagarize too much of the article.

Re: Anyone read - \"Theory of Doubling Up\" by Chen and Ankenman?
 

[ QUOTE ]
Is this the first you double up and then you triple up theory presented in Tilt? I'm proud to say I employed the alternative:
1.triple up
2.double up
theory in a 120 person B&M tourney in Shreveport recently and won $1600.

I knew if I looked hard enough I would find an excuse to post this shamless brag!


[/ QUOTE ]

I don't know what you are referring too but I am quite sure this is much different. This is very well explained mathematical model.

Re: Anyone read - \"Theory of Doubling Up\" by Chen and Ankenman?
 

I didnt even know they put out that pamphlet anymore, how do I get on the mailing list?

Re: Anyone read - \"Theory of Doubling Up\" by Chen and Ankenman?
 

I have no idea. I think I ordered HOH1 from Conjelco last summer and now I'm on the mailing list. I'm not even sure if that is why or not though. I'm really going to try and write a summary sometime today if I can find the time.

In the article he provides an example of a play that in a vacumn is chip +EV for a neutral player. Then he redoes the calculation for a player with an advantage over the field and that exact play becomes -EV for the good player (in terms of overall tournament equity).

Article Summary w an Example
 

The formulas derived in the article presented here without the derivation (I’m too lazy):

C: chance of our hero doubling his stack before going bust
P: number of players in the tournament
N: number of doubles (of your original buy-in) required to attain all of the chips in play
S: stack size (as a ratio of the initial stack… so if the initial stack is 10k and you currently have 20k then S = 2)

E: equity
Eo: a priori overall equity
d = ? (it’s introduced right at the end without being defined and I don’t have time to go back through and try and figure out what it is supposed to be)

(also, for those non-math guys when I write log2 it means (log base 2) and you can refer logarithms on the internet somewhere if you want)
(note: most calculators have a log button. This is log base 10. if you want to calculate log2 just apply this relationship:
log2(y) = log(y)/log(2))

From the summary:
“At the beginning of of the tournament each player has a chance of doubling up C, which is related to his a priori overall equity Eo in the tournament by the following equation:
<font class="small">Code:</font><hr /><pre>
Eo = C ^ (log2 (P*d)) [1]
</pre><hr />

“Then from any stack size S, C needs to double his stack a certain number of times in order to have all the chips,”
<font class="small">Code:</font><hr /><pre>
N = log2 (P/S) [2]
</pre><hr />

“And finally, for any stack size,
<font class="small">Code:</font><hr /><pre>
E = C^N [3]
</pre><hr />
“These equations can be utilized to estimate tournament equity on a skill adjusted basis.”

Ok, that is the summary from the article. Upon further examination I am quite sure that [1] is wrong and that the d term is a typo (it is not explained at all, and throughout the article I can see how [1] applies, but there is no need for the d term) so I present you with the formula I am quite sure is what he meant
Eo = C^(log2(P)) [1a]

An example:
So… let’s take a 100 man tournament with an above average player who’s equity is estimated at 2 buy-ins.

Substituting into formala [1a]:
2/100 = C^log2(100)
solving for C yeilds a 55.5% chance of doubling his stack before going bust.

Now that we know C we can combine formula’s [2] and [3] and figure out our players chances of winning the tournament from any stack size (let’s say he is at 5 times his initial buy in)
E = C^(log2(P/S))
E = 0.555^log2(100/5)
E = 7.8%

So, we see that our player, who has an overall equity in the tournament of 2 buy-ins and has since increased his stack to 5 times it’s initial starting size now has a 7.8% chance of winning the tournament (initially it was 2%).

The main question that I have is:
- How does this theory apply when half the players have been eliminated and we have a stack of S?

I\'ve answered my own question
 

[ QUOTE ]
The main question that I have is:
- How does this theory apply when half the players have been eliminated and we have a stack of S?

[/ QUOTE ]

Answer: It doesn't change anything.

Let's expand on our original 100 man tournament example. Our player has 5 times his initial stack. Let's say there are only 50 players left in the tournament at this time. That means the average stack is now twice the orignal stack and our hero's stack is only 2.5 times the current average stack.

Therefor:
<font class="small">Code:</font><hr /><pre>
P = 50
S = 5/2 (since the average stack is 2 times the original stack our players S ratio should be adjusted to 2.5 instead of 5)
Thus: the ratio of P/S remains the same as it was before (20).


[/ QUOTE ]

It appears that this "Theory of Doubling Up" can be applied universally at any point in the tournament to calculated a players probability of winning the tournament given an assumed initial equity. It's an interesting result, but probably only really important in winner takes all type of tournaments.

Re: Anyone read - \"Theory of Doubling Up\" by Chen and Ankenman?
 

[ QUOTE ]
For the purpose of this discussion we will ignore the value of time and assume your goal is simply to increase your equity in the tournament

[/ QUOTE ]

This is key. If you get knocked out of the tournament you get something precious, you get some of your life back.

For most people I believe the model needs modification.