Author Topic: Hand For Hand elimination at different tables  (Read 3283 times)

D.C.

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Hand For Hand elimination at different tables
« on: January 13, 2011, 09:40:34 AM »
Hello fellow TDs.

First of all I hope you all had a great 2010 and wish an even better 2011.

I'd like the help of this community to clarify the thought around a rule and maybe share experiences how you deal with simultaneous eliminations durinhg hand-for-hand play at different tables.

My tournament team and I were meeting last night in preparation for the 2011 season of the BSOP and we were debating about WSOP hand-for-hand rule (Section VII, rule 105 - http://www.wsop.com/pdfs/2010/2010-WSOP-Rules.pdf).

Their text says:
Quote
(...) During the Hand-for-Hand process,
more than one player may be eliminated during the same hand. If two players are eliminated during the same hand at different tables, both
players will “tie” for that place finish. If two players are eliminated during the same hand at the same table, the player who began that hand
with the highest chip count will receive the higher place finish.

Can anyone explain why is this procedure done and not follow the chip stack definition for the hight place finisher?

Thanks for the help.

DC
Devanir "D.C." Campos
Brazilian Series of Poker Tournament Director

Stuart Murray

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Re: Hand For Hand elimination at different tables
« Reply #1 on: January 13, 2011, 11:00:34 AM »
Hi DC,

I was not aware of this rule, I remember an interview (attached) where Jack talks about the elimination process during the bubble at the WSOP which was based on chips, perhaps this is a new rule they are implementing.  For me though I will be keeping it the same as I have always done, based on chips accross tables and if they have the same chips, base it on hands, only then if these are all the same would the players tie or chop for that position(s).

http://www.youtube.com/watch?v=Tsusx2cgVUU

Regards
Stuart
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National Tournament Director

chet

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Re: Hand For Hand elimination at different tables
« Reply #2 on: January 13, 2011, 06:17:44 PM »
My thought would be that if there are numerous tables remaining, it may be difficult to accurately determine the exact chip count for each player at the start of the hand.  If both players are at the same table, determining chip count is not an issue.

Hope this helps!

WSOPMcGee

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Re: Hand For Hand elimination at different tables
« Reply #3 on: February 04, 2011, 11:24:44 PM »
Hi D.C.

To answer one question, no this is not a new WSOP rule, it's been a rule for a long time.

To add to Chet's comment further, the reason is as stated, it's not easy to be able to determine chip counts at all tables when the money is reached. In early tournament history, there was not even a hand-for-hand scenario. Play was just continued until a player was eliminated, with each table playing a different amount of hands. Thus, because of stalling, hand-for-hand was implemented.

More to your question though, the reason for chip counts being used if at one table vs multiple tables has some to do with hierarchy and some to do with organization and most to do with competitive opportunity.

First hierarchy / One Table Order - In order for Player A and Player B to go out in the same hand, there has to be another factor in the equation, Player C. In this equation Player C chips > Player B chips > Player A chips. That is for both Players A and B to be knocked out, Player C must have them both covered and win.

The logic is as follows in this all-in situation:
  • If Player A wins the hand, he can not be knocked out and only Player B can be knocked out by Player C. Thus, if Player C beats Player B, you have one finisher in Player B.
  • If Player B wins the hand, he can not be knocked out and only Player A can be knocked out by Player B. Thus, by Player B beating Player A, you have one finisher in Player A.
  • If Player A wins the hand, and Player B's hand beats Player C's hand, no one is knocked out.
In all situations only Player A is at complete risk. Player A can be knocked out by Player B and by Player C. Therefore, because Player A is "covered" by all, they will finish lower should multiple players be knocked out. Even if Player A beat Player B's hand, the argument is mute for splitting equally because if not for Player C, Player A would still be playing and so would Player B.

Otherwise you'd have a situation where in a heads-up situation you only have one of two possible outcomes, A) Player A beats Player B and remains in the tournament or B) Player B beats Player A and a player is eliminated and placed in finishing order.

Organizational / Multiple Table Order - For multiple tables things become more complex. First you have to the task of knowing all players chips counts at all tables. The larger the tournament the more difficult the task. The task becomes even tougher the more players away from the money (or money change) that you are.

Ex#1 - Imagine you have 4 tables in action and pay 27. You have 28 players left. You are hand for hand. You have 1 player get knocked out at table #1. Easy enough.
Ex#2 - Imagine you have 4 tables in action and pay 27. You have 28 players left. You are hand for hand. You have 2 players get knocked out at table #1. Using the above rule, lowest chips is eliminated first. This is also easy to follow.
Ex#3 - Imagine you have 4 tables in action and pay 27. You have 28 players left. You are hand for hand. You have 2 players get knocked out at table #1 and 1 player get knocked out on table #2. Player A has lower chips than Player B at table #1. Player C is at table #2. Because Players A and B could not compete against nor eliminte Player C due to being at separate tables, so you have to treat them as equals. They split evenly.
Ex#4 - Imagine you have 4 tables in action and pay 27. You have 30 players left. You are NOT hand for hand. You have 4 players get knocked out at table #1. Using the one table rule, you place the player with the most chips in 27th place.
Ex#5 - Imagine you have 4 tables in action and pay 27. You have 30 players left. You are NOT hand for hand. You have 4 players get knocked out. One at table #1, two  at table #2 and one at table #3. Again, Players A, B, C and D are all on separate tables and have no chance to compete against and eliminate each other. Thus, you have to treat them as equals. Moreover, when you're not hand-for-hand, chances are you are not paying close attention to chip stacks and neither are your dealers, as compared to other competitors.

Now expand those scenarios to paying 100+ players and even like in the case of the WSOP, paying 500+ players.

Lastly as noted in the organizational example, players must have the competitive opportunity to eliminate other players. Meaning you can't penalize a player for finishing at the exact same time as another player and place them lower in ranking based on chip count when that player never had an opportunity to compete against the player with more chips because of being at separate tables.
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Brian Vickers

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Re: Hand For Hand elimination at different tables
« Reply #4 on: March 09, 2011, 10:24:58 AM »
Hi D.C.

To answer one question, no this is not a new WSOP rule, it's been a rule for a long time.

To add to Chet's comment further, the reason is as stated, it's not easy to be able to determine chip counts at all tables when the money is reached. In early tournament history, there was not even a hand-for-hand scenario. Play was just continued until a player was eliminated, with each table playing a different amount of hands. Thus, because of stalling, hand-for-hand was implemented.

More to your question though, the reason for chip counts being used if at one table vs multiple tables has some to do with hierarchy and some to do with organization and most to do with competitive opportunity.

First hierarchy / One Table Order - In order for Player A and Player B to go out in the same hand, there has to be another factor in the equation, Player C. In this equation Player C chips > Player B chips > Player A chips. That is for both Players A and B to be knocked out, Player C must have them both covered and win.

The logic is as follows in this all-in situation:
  • If Player A wins the hand, he can not be knocked out and only Player B can be knocked out by Player C. Thus, if Player C beats Player B, you have one finisher in Player B.
  • If Player B wins the hand, he can not be knocked out and only Player A can be knocked out by Player B. Thus, by Player B beating Player A, you have one finisher in Player A.
  • If Player A wins the hand, and Player B's hand beats Player C's hand, no one is knocked out.
In all situations only Player A is at complete risk. Player A can be knocked out by Player B and by Player C. Therefore, because Player A is "covered" by all, they will finish lower should multiple players be knocked out. Even if Player A beat Player B's hand, the argument is mute for splitting equally because if not for Player C, Player A would still be playing and so would Player B.

Otherwise you'd have a situation where in a heads-up situation you only have one of two possible outcomes, A) Player A beats Player B and remains in the tournament or B) Player B beats Player A and a player is eliminated and placed in finishing order.

Organizational / Multiple Table Order - For multiple tables things become more complex. First you have to the task of knowing all players chips counts at all tables. The larger the tournament the more difficult the task. The task becomes even tougher the more players away from the money (or money change) that you are.

Ex#1 - Imagine you have 4 tables in action and pay 27. You have 28 players left. You are hand for hand. You have 1 player get knocked out at table #1. Easy enough.
Ex#2 - Imagine you have 4 tables in action and pay 27. You have 28 players left. You are hand for hand. You have 2 players get knocked out at table #1. Using the above rule, lowest chips is eliminated first. This is also easy to follow.
Ex#3 - Imagine you have 4 tables in action and pay 27. You have 28 players left. You are hand for hand. You have 2 players get knocked out at table #1 and 1 player get knocked out on table #2. Player A has lower chips than Player B at table #1. Player C is at table #2. Because Players A and B could not compete against nor eliminte Player C due to being at separate tables, so you have to treat them as equals. They split evenly.
Ex#4 - Imagine you have 4 tables in action and pay 27. You have 30 players left. You are NOT hand for hand. You have 4 players get knocked out at table #1. Using the one table rule, you place the player with the most chips in 27th place.
Ex#5 - Imagine you have 4 tables in action and pay 27. You have 30 players left. You are NOT hand for hand. You have 4 players get knocked out. One at table #1, two  at table #2 and one at table #3. Again, Players A, B, C and D are all on separate tables and have no chance to compete against and eliminate each other. Thus, you have to treat them as equals. Moreover, when you're not hand-for-hand, chances are you are not paying close attention to chip stacks and neither are your dealers, as compared to other competitors.

Now expand those scenarios to paying 100+ players and even like in the case of the WSOP, paying 500+ players.

Lastly as noted in the organizational example, players must have the competitive opportunity to eliminate other players. Meaning you can't penalize a player for finishing at the exact same time as another player and place them lower in ranking based on chip count when that player never had an opportunity to compete against the player with more chips because of being at separate tables.

Frontrunner for "Post of the Year"?   :D

Nick C

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Re: Hand For Hand elimination at different tables
« Reply #5 on: March 10, 2011, 01:17:13 AM »
Hey D C,
 Aren't you glad you asked? You better believe Thomas on this one. Wow, I second Brian's vote for post of the year!

WSOPMcGee

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Re: Hand For Hand elimination at different tables
« Reply #6 on: July 13, 2011, 09:45:26 AM »
Bumping for reference to the subject discussed during the summit.

Plus I want to stay in the running for "Post of the Year" :)
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Jim LeVoir

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Re: Hand For Hand elimination at different tables
« Reply #7 on: December 20, 2017, 01:57:38 PM »
I’ve been a TD for almost 4 years now, and I have to say that I find myself at odds with this discussion and the subsequent rulings as described.  Having attended the 2017 Summit, I am familiar with our process, and yes, I can live with it…., but for the sake all things being said I will submit the following:
-   The size of one’s chip stack is paramount.  It is a fundamental and core component of the game and is included in every last calculation and decision made by a player.  I feel that if we are to say that it matters sometimes, then we should say that it matters all of the time and without exception – a point that I will also make in regards to when a player’s stack is zero.
-   Additionally, saying that a player’s stack is only measurable against another player’s at the same table because they have had the chance to play against each other is an arbitrary condition.  Furthermore, it is not the correct measurement to be using in that particular moment – whether or not players bust at the same table or different tables.  Here’s why:

When any player busts out, there are likely to be other players who had started that hand with less chips, but yet are still in the tournament.  This leads to a logical conclusion which is that - IT IS WHEN A PLAYER BUSTS - that is far more significant than what their stack may have been prior to busting. 

-   Consider other professional individual tournaments and how they are paid out.  Let’s use professional golf as an example.  When two or more players are tied for any position other than first, the values of their places are added together and then divided by the number of players tied.
o   Example:
-   How is this example relatable to poker finishes? 
o   Back to the golf example –
  •    say the best score in the clubhouse is -9 held by Player A
       Player B and Player C get to the 18th tee and Player B’s score is actually -10, and Player C’s score is -8.
       Player B shoots +2 (double bogey) and finishes with -8.
       Player C shoots par and is still -8

It would be absolutely unheard of for anyone to say that Player B deserves a full second place because they had a better score going into the 18th or final hole.  And yet, this is exactly what we are doing in poker.
Player B had more to lose (a better score, or what could be thought of as a larger stack), which means that they actually made a bigger mistake than Player C, and yet we are using that measurement to argue for paying them more?  We might as well argue for using preflop probability calculations to determine who took the worse beat and then pay them according to that...

And related to my comment that chip stacks mean everything, the simple fact that we are all overlooking is that regardless of how many people bust on a given hand, on how many tables, or on the same table – they all have the same sized stack at the end of the hand, which is zero, nothing, zip.  They all made the same decision which was to put their entire tournament life on the line and they all achieved the same net result. 

Additionally, golfers face varying conditions such as changing wind or rain which can dramatically change their ability to score.  When we say that players have not had the chance to directly compete against each other, what we are really saying is that they have not had the chance to face the same opponents, or,  to face the same conditions. 

And so, if on a final table of 10, and 4 players bust on a single hand, should they chop 10th, 9th, 8th, and 7th places?  100% that is how that should work. 

I just cannot understand why we are saying that anything matters prior to the end of the hand.  You do not measure scores on the 18th tee box, you measure scores in the clubhouse, and we should be no different.

~my two cents