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-   -   curious about the math of deal-making (http://archives2.twoplustwo.com/showthread.php?t=46585)

doormat 09-24-2003 11:31 AM

curious about the math of deal-making
 
I rarely play in tournaments and neither condemn nor endorse deal-making, but purely from a math standpoint I am curious as to the fairest way that players should split the prize money if they choose to make a deal. The simplest case would be 2 players remaining, one has 100k and the other 200k. Assuming equal skill, what would be the fair deal on splitting the prize money? How would it differ in NL from limit? What if there were 3 or more players involved? Forgive me if this is a trivial matter covered in tournament books, which I have not read.

doormat

Bozeman 09-24-2003 01:56 PM

Re: curious about the math of deal-making
 
The two player one is trivial, but more players is still debated.

For any # of players of equal ability, your chance of first is equal to your fraction of the total chips. So in your example, one player has 1/3 chance of winning. His "fair" share of the prize money is thus 2nd+1/3*(1st-2nd).

For more players, I have a scheme that is correct for fairly reasonable assumptions (Computing tournament finish place probability) . Others have made various approximations to get reasonably close (to these) results.

irchans 09-24-2003 10:30 PM

Re: curious about the math of deal-making
 
Hi Bozeman,

The method you propose in your 08/26/03 post is rather well known I think (among those that discuss the mathematics of deal-making). I believe that Malmuth proposed the same method and something similar was used in Ziemba's book for horse racing ("Efficiency of Racetrack Betting Markets" (1994) see Harville's article page 213.)

I wish someone would take the time to survey the posts on this subject in the old 2+2 and rec.gambling.poker archives. I know it has been discussed several times.


Bozeman 09-24-2003 11:27 PM

Re: curious about the math of deal-making
 
Thanks for the info, I thought so, but no-one pointed me to it and I failed to find any. For the record, irrelevant that it is, I first posted this 07/23/02 (original reply), and had been using the idea for several months. No clue about when/how to find RGP thread?

Thanks,
Craig

Wake up CALL 09-25-2003 12:45 AM

Re: curious about the math of deal-making
 
[ QUOTE ]
Thanks for the info, I thought so, but no-one pointed me to it and I failed to find any. For the record, irrelevant that it is, I first posted this 07/23/02 (original reply), and had been using the idea for several months. No clue about when/how to find RGP thread?

Thanks,
Craig

[/ QUOTE ]

This is a very interesting thread and I finally have something useful to add. Below are some links to RGP posts:

Three Way Tournament Settlements this links to a post by Tom Weidman. After reading Tom's post you should click on the read entire thread link and read all the responses, pretty interesting.

Another is the Landrum/Burns Duscussion and just one more I will recommend is the Idiots Guide to Tournament Deals . As before click on the Read Entire Thread link in the upper right hand corner of each link.

I hope you find these interesting and useful.



M.B.E. 09-25-2003 02:54 AM

Re: curious about the math of deal-making
 
Here's two other discussions:

Malmuth's settling up in tournaments

Deal-making and the concept of inverse equity

M.B.E. 09-25-2003 03:07 AM

Re: curious about the math of deal-making
 
[ QUOTE ]
For any # of players of equal ability, your chance of first is equal to your fraction of the total chips.

[/ QUOTE ]

What is the proof for this statement, in the abstract? Sklansky discusses two approaches to a proof at TPFAP pp. 104-06, for the two-player situation, but I don't find them compelling.

It is the conventional wisdom, but I'm not persuaded that it's true, especially for more than two players. For example, it seems to me that if there are three players left, all good tournament players and equally skilled, and I have 10% of the chips while my opponents have 45% each, then my chance of winning is probably less than 10%.

But in this age of online poker, we don't have to rely on theoretical arguments. It should be possible to do a rigorous study, using statistical methods, of a large enough sample of sit-n-go tournaments at some online site, to assess whether the conventional wisdom is borne out.

Bozeman 10-12-2003 07:04 PM

Re: curious about the math of deal-making
 
Thank you very much, everyone!

This is exactly the sort of thing I have long hoped to see.

For the record, I am also the Craig Howald in the old 2+2 posts.

I have done a summary of the available work in a new thread.

Craig

rockoon 10-14-2003 01:35 AM

Re: curious about the math of deal-making
 
[ QUOTE ]
[ QUOTE ]
For any # of players of equal ability, your chance of first is equal to your fraction of the total chips.

[/ QUOTE ]

What is the proof for this statement, in the abstract? Sklansky discusses two approaches to a proof at TPFAP pp. 104-06, for the two-player situation, but I don't find them compelling.

It is the conventional wisdom, but I'm not persuaded that it's true, especially for more than two players. For example, it seems to me that if there are three players left, all good tournament players and equally skilled, and I have 10% of the chips while my opponents have 45% each, then my chance of winning is probably less than 10%.

But in this age of online poker, we don't have to rely on theoretical arguments. It should be possible to do a rigorous study, using statistical methods, of a large enough sample of sit-n-go tournaments at some online site, to assess whether the conventional wisdom is borne out.

[/ QUOTE ]

For two players of equal ability, it is absolutely correct. The problem with the 3 player scenario is that there is money going to 2nd else it would also be absolutely correct. Not saying its incorrect. My simulations seem to indicate it is by its hardly a proof.

doormat 10-14-2003 01:56 AM

Re: curious about the math of deal-making
 
Thanks, Craig - very informative!

doormat


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