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  #11  
Old 10-28-2005, 05:07 PM
pkrPlayerBr pkrPlayerBr is offline
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Default Re: The Multi-table Grind

or a moron playing 23s before the flop where there was 3 bets. It happened to me the other night and after showing a nice 3bb/100 win rate for 800 hands, in 200 i gave all of my profits back.
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  #12  
Old 10-28-2005, 05:09 PM
shant shant is offline
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Default Re: The Multi-table Grind

SWEET ONE TIME MY WINRATE WAS 14.8BB/100 FOR 800 HANDS
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  #13  
Old 10-28-2005, 05:11 PM
SeaEagle SeaEagle is offline
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Default Re: The Multi-table Grind

nh
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  #14  
Old 10-28-2005, 05:21 PM
W. Deranged W. Deranged is offline
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Default Re: The Multi-table Grind

This is why I like multitabling. It helps to remove the huge "I just lost $800 in 45 minutes and didn't win a single hand" kind of punch-you-in-the-face poker moments.

I have been 5-tabling 5/10 full recently and find that I rarely have momentary swings greater than $200 and this is very helpful for the poker psyche.
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  #15  
Old 10-28-2005, 11:36 PM
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Default Re: The Multi-table Grind

[ QUOTE ]
f(x)= C(1/x)
f(x)= the amount of cash you win/lose ($)
x = the amount of tabels
C = konstant, depends on how fast your brains works [img]/images/graemlins/smile.gif[/img] etc.

^ f($)
|.............
|\................
|..\...............
|....\.............
|.....\............
|......\...........
--------------> x (tabels)

I hope this solved your problems [img]/images/graemlins/wink.gif[/img]

[/ QUOTE ]

I'm not sure that winrate is inversely related to number of tables played. For one thing, the constant C is determined by only one thing, not lots of factors....That is; what is your winrate at one table?

This is because f(1)= C.

Moreover, it must be the case that at some point you will begin losing money by trying to play too many tables....say 50. But the model 1/x approaches 0 as x -> infinity. So this says that if I play 10,000 tables I am essentially breaking even, which is clearly not true.

Also, winrate = 1/x => that by playing 2 tables you half your winrate, and by playing 4 you half that, etc. Which does not happen.

I wasn't sure if this was meant as more of an analogue or a specific model. In the case of the former it's apt enough, for the latter it doesn't work.

I believe a more accurate model is (C+k) - k*x, where C is your winrate at one table, and k is determined by how fast your brain works , etc. (x>= 1 is again # of tables)

Even this is too simplistic though as the loss is definitely not linear, in fact there is virtually no drop off from 1 to 2 tables. A better model (I'm too lazy to make this perfect) would be something that looked like:

2 * exp( -.025 * (x-1)^2 ) - (x-1) / 20

Where exp(*)=e^* and again x>=1. Plot this in matlab or maple and I think you'll see it's a more accurate picture of what happens.
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