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#1
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2 become 1 - final table question!
Here's a question I can't find the answer to on homepokertourney.com ...
We're expecting a few more than 10 people to turn up to the pub game I help run on Sunday, so we're going to have to have a 2-table tournament. I figured that the easiest way to do it was to have (if, say, we get 16 people) 2 tables of 8 players, with the top 4 on each making the final table. But if one table gets down to 4 players at 300/600 and the other table finishes at 400/800, then what level should the final table start up at? Or is there an even simpler way of running a 2-table game? Rob |
#2
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Re: 2 become 1 - final table question!
Typically whenever there is more than a 1 seat difference between tables you move someone from the fuller table to the emptier table. Once the hand on the higher table finishes, have the dealer wash the cards and everyone picks one.. low card moves to the other table. Easy.
All tables are on the same blind level, this should be based on time, not on hands dealt/number of players. Check out www.thetournamentdirector.net for a great peice of software (donation-ware) to help you run your tourney. It works great if you have an old laptop you can use. |
#3
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Re: 2 become 1 - final table question!
Yeah, I agree with this method in principle, but I was trying to simplify it so that I didn't have to monitor two tables and order people to change seats when they might be settled with a drink and a comfy spot.
This is a game full of casual players and their friends and girlfriends, so although I want it to be professional and fairly-run, I don't want it to be TOO professional and hard to explain ... |
#4
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Re: 2 become 1 - final table question!
It is more complicated to have different blind levels on different tables. Also, it is unfair to the players as those on the first table to reach 4 players would not get the same opportunity to grow their stacks as a big stack on the table with 5 or 6 players left would get. Other reasons such as blinds going around faster if you were to get to say an 8 and 5 split or something like that just corrupt the fairness of the game if you do not keep the tables as close to even as possible.
It is a lot easier to merely state that if at any time one table has two fewer people than the other one, the player getting the low card is moving to the other table. One of the tables is going to have to move at somepoint anyway once you combine for the final table. |
#5
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Re: 2 become 1 - final table question!
for our 2 table tourys..
if table 1 starts with say 9 players, and table 2 starts with 8... if table 2 loses a player, that means they have 2 less players than table 1. We will send a player from table 1 over to table 2. We determine who is moved by position. If the person who busted was 3 spots left of the dealer.. then we move the player 3 to the left of the dealer on table 1. This was actually covered on homepokertourney.com ... thats where I got the idea from. |
#6
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Re: 2 become 1 - final table question!
[ QUOTE ]
Also, it is unfair to the players as those on the first table to reach 4 players would not get the same opportunity to grow their stacks as a big stack on the table with 5 or 6 players left would get. [/ QUOTE ] Actually, the logic of your argument on this point should be just the opposite. Starting a table with 8 players at, say $1,000 each, and ending with 4 will always give them an average stack of $2,000. Moving players during play will pump more money from the slower table into the faster table. This is not necessarily wrong but will effect strategy on the rate of play. I like to keep the tables balanced and keep the clock going with the blinds to stay on schedule. Combine the tables at 4 and 4 and if one gets there faster than they can take a break. Generally if a table finishes early, it is because a maniac is driving the action and not the blinds. It is more difficult for a table to finish late against the blind schedule driving the action. |
#7
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Re: 2 become 1 - final table question!
[ QUOTE ]
[ QUOTE ] Also, it is unfair to the players as those on the first table to reach 4 players would not get the same opportunity to grow their stacks as a big stack on the table with 5 or 6 players left would get. [/ QUOTE ] Actually, the logic of your argument on this point should be just the opposite. Starting a table with 8 players at, say $1,000 each, and ending with 4 will always give them an average stack of $2,000. Moving players during play will pump more money from the slower table into the faster table. This is not necessarily wrong but will effect strategy on the rate of play. I like to keep the tables balanced and keep the clock going with the blinds to stay on schedule. Combine the tables at 4 and 4 and if one gets there faster than they can take a break. Generally if a table finishes early, it is because a maniac is driving the action and not the blinds. It is more difficult for a table to finish late against the blind schedule driving the action. [/ QUOTE ] My argument about the slow table being put at a disadvantage is that if you force the table to stop play earlier than the other one, you are depriving any big stack there the opportunity to apply pressure and grow his stack any more heading into the final table. meanwhile, on the table that still has 5 or 6 tables, a big stack could easily apply pressure and pick up bigger and bigger blinds as peopel tighten up on the final table "bubble", thus allowing this player a definitive advantage. The only fair way in my opinion to do a play to four at each table with no table switching if neccesary would be to take the 4 remaining players of each tbale and starting them fresh on a new table. That way, no one was put at an advantage or a disadvantage because their particular table played faster. |
#8
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Re: 2 become 1 - final table question!
Balancing two tables of eight is not that difficult when you think it through. You start with two tables of eight. You only need to move someone if you get to a situation where you have 6 and 8. Then just wait for the button to hit the first empty seat at the 6 tables and pull the button from the 8 table over to the 6 table so they remain in the same position. Now you have 7 and 7. The next move is at 7 and 5, then 6 and 4. At this point if your table is large enough you can combine for a final table of 10.
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#9
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Re: 2 become 1 - final table question!
[ QUOTE ]
Balancing two tables of eight is not that difficult when you think it through. You start with two tables of eight. You only need to move someone if you get to a situation where you have 6 and 8. Then just wait for the button to hit the first empty seat at the 6 tables and pull the button from the 8 table over to the 6 table so they remain in the same position. Now you have 7 and 7. The next move is at 7 and 5, then 6 and 4. At this point if your table is large enough you can combine for a final table of 10. [/ QUOTE ] Good point! When you put it like this, it's not as much hassle as I imagined. Although, why wait for the button to hit the empty seat? [ QUOTE ] Regarding moving players. I have seen a number of suggestions. Homepokertourney recommends that you move someone in relation to the dealer. So if the small blind gets knocked out on table two, then the current small blind on table one is the person to move to table two. [/ QUOTE ] Thanks mookid. That's how I'm gonna do it. Rob |
#10
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Re: 2 become 1 - final table question!
[ QUOTE ]
Although, why wait for the button to hit the empty seat? [/ QUOTE ] If it's your first time balancing tables it's the easiest. Also you miss any "hey I just paid the blinds" situations. |
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